Stir a spoonful of salt into a glass of water and the grains vanish — and every sip tastes equally salty, top, middle, and bottom. Scoop a glass of water from a puddle instead, and specks drift through it; by morning a layer of grime sits on the bottom. Both are mixtures of water and something else — so why do chemists treat them so differently?
You've already seen mixtures. One kind of mixture matters so much in chemistry that it gets its own name.
The idea
Filtered salt water is the same throughout: every drop carries the same amount of dissolved salt, so every sip tastes the same.
A mixture that is the same throughout is called a 'solution'.
A solution is the same throughout because the dissolved particles are spread evenly among the liquid's particles.
Muddy water is not the same throughout: its specks hang unevenly in the water and settle to the bottom over time.
In a solution the dissolved particles are spread evenly among the liquid's particles; in muddy water the specks hang unevenly and settle.
So muddy water is a mixture, but it is not a solution.
To classify a mixture, ask one question: is it the same throughout?
Worked examples
Worked example 1. A pitcher of sweet tea was stirred until all the sugar disappeared. A sip from the top and a sip from the bottom taste equally sweet. Is the sweet tea a solution?
Step 1
The tea tastes the same at the top and at the bottom, so the mixture is the same throughout.
Step 2
A mixture that is the same throughout is a solution — its dissolved particles are spread evenly among the liquid's particles.
Step 3
The sweet tea is a solution.
Worked example 2. A bottle of Italian salad dressing sits with a layer of oil resting on top of a layer of vinegar. Is the dressing a solution?
Step 1
The dressing is not the same throughout — the top of the bottle is oil and the bottom is vinegar.
Step 2
A mixture that is not the same throughout is not a solution.
Step 3
The dressing is a mixture, but it is not a solution.
You can now classify a mixture as a solution or not a solution from whether it is the same throughout, because in a solution the dissolved particles are spread evenly among the liquid's particles.
Check your understanding
Which mixture is a solution?
AClear apple juice that tastes the same in every sip.correct
BOrange juice with pulp drifting through it.
This option is wrong — you counted drifting pulp as dissolved — visible bits hanging in a liquid mean the mixture is not the same throughout.
CHot chocolate with undissolved powder resting on the bottom.
This option is wrong — you ignored the settled powder — particles resting on the bottom are not spread evenly among the liquid's particles.
DPond water whose specks settle into a layer overnight.
This option is wrong — you overlooked the settling — a solution stays the same throughout on its own, and settling shows it never was.
Ask one question: is the mixture the same throughout? Clear apple juice tastes the same in every sip, so its dissolved particles are spread evenly among the liquid's particles. Drifting pulp, settled powder, and settling specks all show a mixture that is not the same throughout — not a solution.
Check your understanding
Jar A holds a sports drink — clear, and it tastes the same in every sip. Jar B holds snow-globe water, with white flakes that swirl and then settle. Which classification is correct?
AJar A holds a solution; Jar B does not.correct
BJar B holds a solution; Jar A does not.
This option is wrong — you swapped the two jars — settling flakes mean Jar B is not the same throughout, while Jar A's even taste means it is.
CBoth jars hold solutions.
This option is wrong — you counted every liquid mixture as a solution — Jar B's flakes settle, so it is not the same throughout.
DNeither jar holds a solution.
This option is wrong — you required a pure substance — a solution IS a mixture, as long as it is the same throughout, and Jar A qualifies.
A solution is a mixture that is the same throughout. Jar A tastes the same everywhere — its dissolved particles are spread evenly among the liquid's particles, so it is a solution. Jar B's flakes swirl and settle — it is not the same throughout, so it is not a solution.
Check your understanding
Which observation shows that a mixture is NOT a solution?
ASpecks hang in the liquid and settle to the bottom over time.correct
BThe mixture tastes the same in every sip.
This option is wrong — you flipped the test — tasting the same everywhere is exactly what a mixture that is the same throughout does.
CThe mixture has a color.
This option is wrong — you used color as the test — a mixture can be colored and still be the same throughout in every drop.
DThe mixture contains two different substances.
This option is wrong — you confused mixture with not-a-solution — every solution contains two or more substances; the question is whether they are spread evenly.
The one question is: is the mixture the same throughout? Settling specks show particles that were never spread evenly among the liquid's particles. Color and having two substances say nothing against it — a solution is a mixture, and it can be colored.
Lesson 2 of 45 · SOL-002
Solute and solvent
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Did You Know?
You've already seen that a solution is a mixture that is the same throughout. The two parts of a solution have their own names.
The idea
In salt water, the water does the dissolving, and the salt gets dissolved.
The substance that does the dissolving is called the 'solvent'.
The solvent is the substance present in the larger amount.
The substance that gets dissolved is called the 'solute'.
So in salt water, water is the solvent and salt is the solute.
The solute gets dissolved; the solvent does the dissolving and is present in the larger amount; together they make the solution.
Worked examples
Worked example 1. A packet of powdered drink mix is stirred into a pitcher of water. Name the solute and the solvent.
Step 1
The water does the dissolving and is present in the larger amount, so water is the solvent.
Step 2
The drink-mix powder gets dissolved, so the powder is the solute.
Step 3
Solvent: water. Solute: drink-mix powder.
Worked example 2. Soda water is made by dissolving carbon dioxide gas in water. Name the solute and the solvent.
Step 1
The water does the dissolving and is present in the larger amount, so water is the solvent.
Step 2
The carbon dioxide gets dissolved, so the gas is the solute.
Step 3
Solvent: water. Solute: carbon dioxide.
You can now identify the solute and the solvent in a given solution, where the solvent is the substance present in the larger amount that does the dissolving and the solute is the substance that gets dissolved.
Check your understanding
Epsom salt is dissolved in a tub of warm water. Which statement names the two parts correctly?
AEpsom salt is the solute, and water is the solvent.correct
BEpsom salt is the solvent, and water is the solute.
This option is wrong — you swapped the two names — the substance that GETS dissolved is the solute, and Epsom salt is the one being dissolved.
CThe finished salty water is the solute, and water is the solvent.
This option is wrong — you named the whole solution as one of its own parts — the solute is the Epsom salt that went in, not the mixture that came out.
DEpsom salt is the solute, and the finished salty water is the solvent.
This option is wrong — you named the whole solution as the solvent — the solvent is the water that did the dissolving, not the mixture it produced.
The substance that gets dissolved is the solute: Epsom salt. The substance that does the dissolving, present in the larger amount, is the solvent: water. The salty water is neither part — it is the solution the two parts make.
Check your understanding
Baking soda is stirred into a glass of water until it disappears. Which substance is the solvent?
AThe water.correct
BThe baking soda.
This option is wrong — you named the substance that gets dissolved — that one is the solute.
CThe finished baking-soda-and-water mixture.
This option is wrong — you named the whole solution — the solvent is one part of it, the water that does the dissolving.
DBoth substances equally.
This option is wrong — you gave both parts the same role — the substance that gets dissolved is the solute, so only the water is the solvent.
Ask which substance does the dissolving. The water does the dissolving and is present in the larger amount, so water is the solvent. The baking soda gets dissolved — it is the solute.
Check your understanding
A chemist stirs 5 mL of glycerin into 200 mL of water, and the glycerin dissolves completely. Which statement names the two parts correctly?
AWater is the solvent, and glycerin is the solute.correct
BGlycerin is the solvent, and water is the solute.
This option is wrong — you swapped the roles — the water is present in the larger amount and does the dissolving, so water is the solvent.
CGlycerin is the solvent, because it is the thicker liquid.
This option is wrong — you used thickness to assign the role — the solvent is the substance present in the larger amount that does the dissolving.
DWater is the solute, because the glycerin was poured into it.
This option is wrong — you used pouring order to assign the role — the solute is the substance that gets dissolved, however the two were combined.
Compare the amounts: 200 mL of water against 5 mL of glycerin. The water is present in the larger amount and does the dissolving, so water is the solvent. The glycerin gets dissolved, so glycerin is the solute.
Lesson 3 of 45 · SOL-003
Aqueous solutions and (aq)
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Did You Know?
You've already named the solute and the solvent in a solution. In a huge share of the solutions chemists use, the solvent is water — so that case gets its own name and its own symbol.
The idea
A solution whose solvent is water is called an 'aqueous solution'.
Chemists mark an aqueous solution by writing (aq) after the chemical formula of the dissolved substance.
NaCl(aq) means sodium chloride dissolved in water.
The formula before the (aq) names the solute; the (aq) itself says the solute is dissolved in water.
The formula names the solute; (aq) says it is dissolved in water.
Worked examples
Worked example 1. What does KBr(aq) mean?
Step 1
Answer: potassium bromide dissolved in water.
Worked example 2. Magnesium chloride, MgCl₂, is dissolved in water. Write the formula with the symbol that shows this.
Step 1
Answer: MgCl₂(aq).
You can now state that a solution whose solvent is water is called an aqueous solution and is shown after a chemical formula with the symbol (aq).
Check your understanding
What does CaCl₂(aq) mean?
ACalcium chloride dissolved in water.correct
BCalcium chloride melted into a pure liquid.
This option is wrong — you read (aq) as melted — the melted pure substance is written (l); (aq) means dissolved in water.
CCalcium chloride dissolved in any convenient liquid.
This option is wrong — you widened (aq) to every solvent — (aq) means the solvent is water specifically.
DWater dissolved in calcium chloride.
This option is wrong — you flipped the two roles — the formula before the (aq) names the solute, and the water is the solvent.
The formula before the (aq) names the solute: calcium chloride. The (aq) says the solute is dissolved in water. CaCl₂(aq) is an aqueous solution of calcium chloride.
Check your understanding
Silver nitrate, AgNO₃, is dissolved in water. Write the formula with the symbol that shows it is dissolved in water.
Accepted answer: AgNO₃(aq)
The solute is silver nitrate, so its formula comes first: AgNO₃. Dissolved in water is marked with (aq). AgNO₃(aq).
Check your understanding
Which mixture is an aqueous solution?
ATable sugar dissolved in water.correct
BIodine dissolved in alcohol (tincture of iodine).
This option is wrong — you saw a clear liquid and assumed water — the solvent here is alcohol, so the solution is not aqueous.
CFood flavoring dissolved in vegetable oil.
This option is wrong — you counted any liquid solvent — aqueous means the solvent is water, and here it is oil.
DMelted candle wax poured into a mold.
This option is wrong — you called a pure melted substance a solution — nothing is dissolved and no water is present.
Aqueous means the solvent is water. Sugar dissolved in water has water as its solvent — an aqueous solution. Alcohol and oil solvents make solutions, but not aqueous ones; melted wax is not a solution at all.
Lesson 4 of 45 · SOL-004
Why water dissolves so many substances
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Have You Ever Wondered?
Wonder this:
Of all the liquids in your kitchen, water is the one that dissolves the most things — salt, sugar, drink mix, plant food. Cooking oil dissolves almost none of them. What is special about the water molecule?
You've already classified the water molecule as polar: its oxygen end is partially negative, and its hydrogen ends are partially positive. Those charged ends are the answer.
The idea
An ionic solid like table salt is built from positive ions and negative ions.
Water is such a good solvent for ionic substances because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions.
In salt water, every Na⁺ ion feels a pull from water's oxygen ends, and every Cl⁻ ion feels a pull from water's hydrogen ends.
Water's partially negative oxygen end attracts positive ions; its partially positive hydrogen ends attract negative ions.
Water's partially charged ends also attract other polar particles, so polar substances dissolve well in water too.
Either way, the pattern is the same: the water molecule has a charged end to offer whatever charge the solute carries.
Worked examples
Worked example 1. Road crews spread calcium chloride, CaCl₂ — an ionic compound made of Ca²⁺ and Cl⁻ ions — on icy roads, and it dissolves readily in the meltwater. Explain why water dissolves calcium chloride.
Step 1
Water dissolves ionic compounds because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions.
Step 2
The oxygen ends attract the Ca²⁺ ions, and the hydrogen ends attract the Cl⁻ ions.
Step 3
Every ion in the compound feels a pull from one end of the water molecule, so calcium chloride dissolves readily in water.
Worked example 2. Rubbing alcohol is made of polar molecules, and it mixes completely with water. Explain why.
Step 1
Water's partially charged ends attract other polar particles.
Step 2
Each polar alcohol molecule feels a pull from the water molecules' partially charged ends.
Step 3
The attraction between water's charged ends and the polar alcohol molecules lets the two liquids mix completely.
You can now explain why water is such a good solvent for ionic and polar substances, because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions, and those same ends attract other polar particles.
Check your understanding
Potassium iodide dissolves in water. Which end of the water molecule attracts the K⁺ ions?
AThe oxygen end, because it is partially negative.correct
BThe hydrogen ends, because they are partially positive.
This option is wrong — you sent the positive ion to the positive ends — opposite charges attract, so the partially negative oxygen end pulls on K⁺.
CBoth ends equally, because the molecule is neutral overall.
This option is wrong — you dropped the partial charges — the molecule is neutral overall, but each END carries a partial charge, and only the negative end attracts a positive ion.
DThe hydrogen ends, because hydrogen atoms are smaller.
This option is wrong — you used size to assign the attraction — the pull comes from the partial charges, not from atom size.
K⁺ is a positive ion. Water's partially negative oxygen end attracts positive ions. So the oxygen end of each water molecule pulls on the K⁺ ions.
Check your understanding
Magnesium bromide, an ionic compound made of Mg²⁺ and Br⁻ ions, dissolves readily in water. Why is water so good at dissolving it?
ABecause water's partially negative oxygen end attracts the positive Mg²⁺ ions and its partially positive hydrogen ends attract the negative Br⁻ ions.correct
BBecause water molecules are small enough to slip into the spaces between the ions and physically push the ions apart.
This option is wrong — you made dissolving a matter of size — the water molecule's power comes from its partially charged ends attracting the ions.
CBecause water reacts chemically with the compound and turns the Mg²⁺ and Br⁻ ions into entirely new substances.
This option is wrong — you called dissolving a chemical reaction — the ions are unchanged; water's charged ends only attract them.
DBecause water's partially negative oxygen end attracts the Br⁻ ions and its partially positive hydrogen ends attract the Mg²⁺ ions.
This option is wrong — you matched each ion to the like-charged end — opposite charges attract, so the negative oxygen end pulls positive ions.
Water's oxygen end is partially negative; its hydrogen ends are partially positive. Water dissolves ionic compounds because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions. Oxygen ends pull the Mg²⁺; hydrogen ends pull the Br⁻.
Check your understanding
Glycerin is made of polar molecules, and it dissolves completely in water. Why does water dissolve glycerin so well?
AWater's partially charged ends attract the polar glycerin molecules.correct
BThe glycerin molecules break apart into ions, which water then attracts.
This option is wrong — you turned a polar molecular substance into ions — the glycerin molecules are attracted whole, by water's partially charged ends.
CWater warms the glycerin until its molecules spread out on their own.
This option is wrong — you swapped attraction for heating — the dissolving is driven by the pull between water's charged ends and the polar molecules.
DGlycerin is a thick liquid, and thick liquids blend into water.
This option is wrong — you used thickness — the reason is water's partially charged ends attracting the polar particles.
Glycerin's molecules are polar — they have partially charged ends of their own. Water's partially charged ends attract other polar particles. That attraction pulls the glycerin molecules in among the water molecules.
Lesson 5 of 45 · SOL-005
How water takes an ionic crystal apart
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Did You Know?
You've already seen that a solid ionic compound is a lattice — a repeating stack of positive and negative ions — and that water's charged ends attract those ions. Here is what happens, step by step, when the crystal meets the water.
The idea
When an ionic crystal drops into water, water molecules crowd against the ions at the crystal's surface.
The water molecules attach to those surface ions, because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions.
Water molecules attach to the surface ions, pull them away one by one, and surround each freed ion.
The attached water molecules pull the surface ions away from the lattice one by one.
Each freed ion ends up surrounded by a shell of water molecules.
Chemists call this surrounding of freed ions by water molecules 'hydration'.
Around a positive ion, the water molecules point their oxygen ends inward; around a negative ion, they point their hydrogen ends inward.
In salt water, the crystal's Na⁺ and Cl⁻ ions end up separated, hydrated, and spread through the water.
Worked examples
Worked example 1. A crystal of potassium bromide, KBr — a lattice of K⁺ and Br⁻ ions — is dropped into water. Describe, step by step, how the water dissolves it.
Step 1
Water molecules attach to the K⁺ and Br⁻ ions at the crystal's surface, because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions.
Step 2
The attached water molecules pull the surface ions away from the lattice one by one.
Step 3
Each freed K⁺ and Br⁻ ion is surrounded by a shell of water molecules — the ions are hydrated.
Step 4
The crystal shrinks from the surface inward as separated, hydrated K⁺ and Br⁻ ions spread through the water.
Worked example 2. When lithium chloride dissolves, water molecules surround each freed Li⁺ ion. Which end of each water molecule points toward the Li⁺, and why?
Step 1
Li⁺ is a positive ion.
Step 2
The water molecules aim their oxygen ends at it, because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions.
Step 3
The oxygen ends point inward, toward the Li⁺ ion.
You can now explain how water dissolves an ionic compound: water molecules attach to the ions at the crystal surface, pull them away one by one, and surround each freed ion — chemists call this hydration — because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions.
Check your understanding
A crystal of sodium iodide, a lattice of Na⁺ and I⁻ ions, is dropped into water. What happens FIRST?
AWater molecules attach to the ions at the crystal's surface.correct
BThe whole crystal splits instantly into separate ions, and the water arrives afterward.
This option is wrong — you let the crystal fall apart on its own — the ions leave only because attached water molecules pull them away.
CThe ions at the crystal's center leave before the ions at the surface.
This option is wrong — you started dissolving from the inside — water can only reach the surface, so the crystal shrinks from the outside in.
DThe water molecules split into hydrogen and oxygen pieces that enter the crystal.
This option is wrong — you broke the solvent apart — water molecules stay whole while their charged ends attract the surface ions.
Dissolving starts at the crystal's surface, where the water can touch. Water molecules attach there, because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions. Only then do the attached water molecules pull the surface ions away, one by one.
Check your understanding
Potassium fluoride dissolves in water, and water molecules surround each freed F⁻ ion. Which way do the water molecules point?
AHydrogen ends inward, toward the F⁻ ion.correct
BOxygen ends inward, toward the F⁻ ion.
This option is wrong — you aimed the negative end at a negative ion — water's partially positive hydrogen ends are the ones that attract negative ions.
CWhichever way each molecule happens to land — the direction is random.
This option is wrong — you removed the attraction — the shell has a set orientation because each water end attracts the opposite charge.
DHydrogen ends inward around half the shell and oxygen ends inward around the other half.
This option is wrong — you split the shell — every water molecule around a negative ion aims its partially positive hydrogen ends inward.
F⁻ is a negative ion. Water's partially positive hydrogen ends attract negative ions. So every water molecule in the shell points its hydrogen ends inward, toward the F⁻.
Check your understanding
Magnesium chloride dissolves in water, and each freed ion ends up surrounded by a shell of water molecules. What do chemists call this surrounding of freed ions by water molecules?
AHydration.correct
BMelting.
This option is wrong — you named the change of a pure solid into a liquid by heat — no water shell is involved in melting.
CCrystallization.
This option is wrong — you named the reverse process — crystallization is ions joining a solid, not being surrounded by water.
DA chemical reaction that forms a new compound.
This option is wrong — you called hydration a chemical change — the ions and water molecules are unchanged; they are only attracted and rearranged.
Each freed ion is surrounded by a shell of water molecules. Chemists call that surrounding 'hydration'. Nothing new forms — the ions and the water molecules are unchanged, only rearranged.
Lesson 6 of 45 · SOL-006
Sketch dissolving at the particle level
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Did You Know?
You've already followed hydration step by step. Chemists compress that whole story into one particle sketch — and drawing it yourself is the fastest way to own it.
The idea
Draw each ion as a circle labeled with its symbol and charge, like Na⁺ or Cl⁻.
Draw the ions separated — no two ions touching, each one out on its own in the water.
Draw each water molecule as one large circle labeled O with two small circles labeled H attached to one side.
Around each positive ion, draw three or four water molecules with their O sides facing the ion.
Around each negative ion, draw three or four water molecules with their H sides facing the ion.
The orientation is the attraction made visible, because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions.
A finished sketch of dissolved salt shows three separated Na⁺ ions and three separated Cl⁻ ions, each surrounded by correctly aimed water molecules.
Dissolved salt: separated, labeled ions, each hydrated — O sides toward Na⁺, H sides toward Cl⁻.
Worked examples
Worked example 1. Sketch a particle diagram of dissolved potassium chloride: two K⁺ ions and two Cl⁻ ions, hydrated.
Step 1
Draw two circles labeled K⁺ and two circles labeled Cl⁻, all four separated from one another.
Step 2
Around each K⁺, draw three water molecules with their O sides facing the ion.
Step 3
Around each Cl⁻, draw three water molecules with their H sides facing the ion.
Step 4
The sketch shows four separated, hydrated ions — oxygen ends toward every K⁺, hydrogen ends toward every Cl⁻.
Worked example 2. The figure shows a classmate's sketch of dissolved lithium bromide. What needs fixing?
Step 1
Check each shell against the attraction, because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions.
Step 2
The Br⁻ shells face their H sides inward — correct for a negative ion.
Step 3
But the water molecules around each Li⁺ also face their H sides inward, and Li⁺ is a positive ion.
Step 4
Flip the water molecules around each Li⁺ so their O sides face the ion.
You can now sketch a particle diagram of an ionic compound dissolved in water, showing separated ions each surrounded by water molecules with oxygen ends facing positive ions and hydrogen ends facing negative ions.
Check your understanding
Sodium fluoride, NaF, is dissolved in water. Which sketch shows the dissolved mixture correctly?
Acorrect
B
This option is wrong — you aimed the hydrogen ends at the positive ions — water's partially negative oxygen end attracts positive ions, so the O sides face Na⁺.
C
This option is wrong — you left the lattice intact — dissolving separates the ions, so no drawn ion should touch another ion.
D
This option is wrong — you dissolved the compound as ion pairs — each ion is hydrated separately, inside its own shell of water molecules.
A correct sketch shows every ion separated and labeled, each inside its own shell of water molecules — O sides facing Na⁺ and H sides facing F⁻, because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions. Reversed shells, an intact lattice, and touching ion pairs each break one of those rules.
Your turn
On paper, sketch a particle diagram of dissolved potassium iodide: two K⁺ ions and two I⁻ ions, each hydrated. Label every ion, and draw each water molecule as one large O circle with two small H circles. Then select Continue to compare your sketch with the model answer.
Model answer. Four well-separated circles — two labeled K⁺ and two labeled I⁻. Each K⁺ sits inside a ring of three or four water molecules with their O sides facing the ion; each I⁻ sits inside a ring of three or four water molecules with their H sides facing the ion. No ion touches another ion.
Every ion is a labeled circle with its symbol and charge — K⁺ or I⁻.
No two ions touch — all four are out on their own in the water.
Each water molecule is one large O circle with two small H circles on one side.
Around each K⁺, the water molecules face their O sides inward.
Around each I⁻, the water molecules face their H sides inward.
Your turn
On paper, sketch a particle diagram of dissolved magnesium chloride, MgCl₂: one Mg²⁺ ion and two Cl⁻ ions, each hydrated. Label every ion with its symbol and full charge. Then select Continue to compare your sketch with the model answer.
Model answer. Three well-separated circles — one labeled Mg²⁺ and two labeled Cl⁻. The Mg²⁺ sits inside a ring of four water molecules with O sides facing it; each Cl⁻ sits inside a ring of three water molecules with H sides facing it. No ion touches another ion.
The ion count matches the formula — one Mg²⁺ and two Cl⁻.
Every ion is labeled with its symbol and full charge, including the 2+ on magnesium.
No two ions touch.
The water molecules around Mg²⁺ face their O sides inward.
The water molecules around each Cl⁻ face their H sides inward.
Lesson 7 of 45 · SOL-007
Molecular substances dissolve as whole molecules
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Did You Know?
You've already watched water pull an ionic crystal apart into separate ions. Sugar dissolves in water too — but sugar is a molecular compound, and its story is different.
The idea
When a molecular substance dissolves in water, its molecules stay whole.
The whole molecules simply spread out among the water molecules.
No ions form: each dissolved molecule keeps all of its atoms and carries no charge.
Dissolved sugar spreads through water as whole sugar molecules — every molecule still a complete, neutral cluster of its atoms.
So an ionic compound dissolves as separated ions, but a molecular compound dissolves as whole molecules.
An ionic compound dissolves as separated ions; a molecular compound dissolves as whole, neutral molecules.
Worked examples
Worked example 1. Ethanol is a molecular compound that dissolves completely in water. Describe what the ethanol particles do as it dissolves.
Step 1
Ethanol is molecular, so its molecules stay whole when it dissolves.
Step 2
The whole ethanol molecules spread out among the water molecules.
Step 3
The solution contains whole, neutral ethanol molecules spread evenly among the water molecules.
Worked example 2. Glucose, a molecular compound, dissolves in water. Does the solution contain glucose ions?
Step 1
Glucose is molecular, and a molecular substance's molecules stay whole when it dissolves.
Step 2
No ions form — each dissolved glucose molecule keeps all of its atoms and carries no charge.
Step 3
No — the solution contains whole, neutral glucose molecules, not ions.
You can now state that when a molecular substance dissolves in water, its molecules stay whole and spread out among the water molecules instead of breaking into ions.
Check your understanding
Acetone, a molecular compound, dissolves completely in water. What form does the dissolved acetone take?
AWhole acetone molecules spread among the water molecules.correct
BSeparate positive and negative acetone ions spread through the water.
This option is wrong — you gave a molecular compound the ionic story — a molecular substance's molecules stay whole and carry no charge.
CSingle atoms split off from the acetone molecules.
This option is wrong — you broke the molecules into atoms — dissolving spreads molecules out; it does not take them apart.
DClumps of acetone molecules gathered at the bottom of the container.
This option is wrong — you described undissolved acetone — dissolved particles are spread evenly among the liquid's particles.
Acetone is molecular, so its molecules stay whole when it dissolves. The whole molecules spread out among the water molecules. No ions, no loose atoms, no clumps — just whole, neutral molecules spread evenly.
Check your understanding
Urea, a molecular compound, is dissolved in water. Which particle diagram shows the mixture correctly?
Acorrect
B
This option is wrong — you split the urea into charged pieces — a molecular compound's molecules stay whole and carry no charge.
C
This option is wrong — you left the urea gathered at the bottom — dissolved molecules spread evenly among the water molecules.
D
This option is wrong — you floated the urea as a separate layer — dissolved molecules mix in among the water molecules instead of sitting apart.
A dissolved molecular compound appears as whole, identical molecules spread evenly among the water molecules. Split-up charged pieces, a pile on the bottom, and a separate layer each contradict what dissolving a molecular substance does.
Check your understanding
Antifreeze, a molecular compound, and calcium chloride, an ionic compound, are each dissolved in separate beakers of water. Which statement compares the two solutions correctly?
AThe antifreeze spreads as whole molecules; the calcium chloride spreads as separated ions.correct
BThe antifreeze spreads as separated ions; the calcium chloride spreads as whole molecules.
This option is wrong — you swapped the two stories — ionic compounds are the ones that separate into ions, and molecular compounds stay whole.
CBoth spread as separated ions.
This option is wrong — you gave every solute the ionic story — a molecular compound's molecules stay whole and neutral in solution.
DBoth spread as whole molecules.
This option is wrong — you gave every solute the molecular story — an ionic compound has no molecules; its lattice separates into ions.
Match the story to the compound type. Molecular antifreeze dissolves as whole, neutral molecules spread among the water molecules. Ionic calcium chloride dissolves as separated, hydrated Ca²⁺ and Cl⁻ ions.
Lesson 8 of 45 · SOL-008
Electrolytes and nonelectrolytes
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Did You Know?
Wonder this:
Two beakers on a bench hold clear, colorless liquids that look identical. Dip a simple electrical tester into the first, and its bulb glows. Dip it into the second — nothing. Something invisible separates the two.
Both beakers hold solutions. Whether a solution carries electricity turns out to sort dissolved substances into two named groups.
The idea
A conductivity tester dips two metal strips, wired to a battery and a bulb, into a liquid — if the liquid conducts electricity, the bulb lights.
Dip the tester into salt water, and the bulb lights.
A substance whose water solution conducts electricity is called an 'electrolyte' — table salt is an electrolyte.
The same tester, two solutions: salt water conducts and lights the bulb; sugar water does not.
Dip the tester into sugar water, and the bulb stays dark.
A substance whose water solution does not conduct electricity is called a 'nonelectrolyte' — sugar is a nonelectrolyte.
Worked examples
Worked example 1. A baking-soda solution lights the conductivity bulb. Is baking soda an electrolyte or a nonelectrolyte?
Step 1
Answer: an electrolyte — its water solution conducts electricity.
Worked example 2. A glycerin solution leaves the conductivity bulb dark. Is glycerin an electrolyte or a nonelectrolyte?
Step 1
Answer: a nonelectrolyte — its water solution does not conduct electricity.
You can now state that an electrolyte is a substance whose water solution conducts electricity and a nonelectrolyte is a substance whose water solution does not.
Check your understanding
In a lab test, a washing-soda solution lights the conductivity bulb and a corn-syrup solution leaves it dark. Which classification is correct?
AWashing soda is an electrolyte; corn syrup is a nonelectrolyte.correct
BWashing soda is a nonelectrolyte; corn syrup is an electrolyte.
This option is wrong — you attached each name to the wrong result — the lit bulb marks the conductor, and a substance whose solution conducts is the electrolyte.
CBoth are electrolytes.
This option is wrong — you counted every dissolved substance as an electrolyte — the corn-syrup solution left the bulb dark, so corn syrup is a nonelectrolyte.
DBoth are nonelectrolytes.
This option is wrong — you counted neither as an electrolyte — the washing-soda solution lit the bulb, so washing soda is an electrolyte.
Read each bulb. The washing-soda solution conducts — washing soda is an electrolyte. The corn-syrup solution does not conduct — corn syrup is a nonelectrolyte.
Check your understanding
What is an electrolyte?
AA substance whose water solution conducts electricity.correct
BA substance whose water solution does not conduct electricity.
This option is wrong — you stated the opposite group — that is the nonelectrolyte's definition.
CAny substance that dissolves in water.
This option is wrong — you made dissolving the test — plenty of substances dissolve and still leave the bulb dark; conducting is the test.
DA metal that carries current as a solid wire.
This option is wrong — you moved the definition to solid metals — electrolyte describes a dissolved substance whose SOLUTION conducts.
The definition hangs on one test: does the water solution conduct electricity? If it conducts, the dissolved substance is an electrolyte. If it does not, the dissolved substance is a nonelectrolyte.
Check your understanding
Epsom salt dissolves completely in one beaker of water and lights the tester's bulb; methanol dissolves completely in another beaker and leaves the bulb dark. Which classification is correct?
AEpsom salt is an electrolyte; methanol is a nonelectrolyte.correct
BEpsom salt is a nonelectrolyte; methanol is an electrolyte.
This option is wrong — you swapped the names — the substance whose solution lights the bulb is the electrolyte.
CBoth are electrolytes, because both dissolved completely.
This option is wrong — you used dissolving as the test — dissolving is required for the test, but only conducting makes a substance an electrolyte.
DNeither can be classified from a single bulb test.
This option is wrong — you dismissed the result — one clear conduct-or-not observation is exactly what the two definitions are built on.
Both substances dissolved, so both solutions give a fair test. The Epsom-salt solution conducts — Epsom salt is an electrolyte. The methanol solution does not conduct — methanol is a nonelectrolyte.
Lesson 9 of 45 · SOL-009
Why some solutions conduct
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Did You Know?
You've already seen the two test results — salt water lights the bulb, sugar water leaves it dark — and you've already seen what each dissolved compound looks like at the particle level. Put those together and the bulb explains itself.
The idea
A solution conducts only when charged particles are free to move through it.
Dissolved ions are exactly that: charged, and free to move — so dissolved ions carry the current.
Salt water conducts: its dissolved Na⁺ and Cl⁻ ions are charged particles moving freely through the water, and they carry the current between the tester's strips.
Neutral whole molecules carry no charge, so they cannot carry a current.
Free-moving charged ions carry the current; neutral whole molecules cannot.
Sugar water stays dark: its dissolved sugar molecules are neutral and whole, so nothing charged moves through the solution.
So an electrolyte's solution conducts because it is full of free-moving ions, and a nonelectrolyte's solution does not because it holds only neutral molecules.
Worked examples
Worked example 1. A potassium bromide solution lights the conductivity bulb. Explain why.
Step 1
A solution conducts only when charged particles are free to move through it.
Step 2
Dissolved potassium bromide is separated K⁺ and Br⁻ ions — charged, and free to move through the water.
Step 3
The free-moving K⁺ and Br⁻ ions carry the current, so the bulb lights.
Worked example 2. An ethanol solution leaves the conductivity bulb dark. Explain why.
Step 1
A solution conducts only when charged particles are free to move through it.
Step 2
Dissolved ethanol spreads as neutral whole molecules, which carry no charge.
Step 3
Nothing charged moves through the solution, so no current flows and the bulb stays dark.
You can now explain why some solutions conduct electricity and others do not, because a solution conducts only when charged particles are free to move through it: dissolved ions carry the current, and neutral whole molecules cannot.
Check your understanding
A calcium chloride solution lights the conductivity bulb. Why does the solution conduct electricity?
ABecause a solution conducts only when charged particles are free to move through it, and the dissolved Ca²⁺ and Cl⁻ ions are exactly that.correct
BBecause the water molecules themselves carry the current once anything at all has been dissolved in it.
This option is wrong — you gave the water the job — water molecules are neutral; the dissolved ions are the charged particles that carry the current.
CBecause the dissolved calcium chloride stays as whole neutral molecules that pass the current along from one to the next.
This option is wrong — you kept the compound whole and neutral — neutral particles cannot carry current, and dissolved ionic compounds separate into ions.
DBecause the electric current itself breaks the dissolved compound into ions as it passes through the solution.
This option is wrong — you reversed the order — the ions are already there from dissolving; the current flows because they are free to move.
A solution conducts only when charged particles are free to move through it. Dissolved calcium chloride is separated Ca²⁺ and Cl⁻ ions — charged and free-moving. Those ions carry the current, so the bulb lights.
Check your understanding
An acetone solution leaves the conductivity bulb dark. Why does the solution NOT conduct electricity?
ABecause its dissolved particles are neutral whole molecules, and a solution conducts only when charged particles are free to move through it.correct
BBecause the dissolved acetone ions are too large and heavy to squeeze between the water molecules and reach the metal strips.
This option is wrong — you gave acetone ions — a molecular compound dissolves as neutral whole molecules, so there are no ions of any size.
CBecause the dissolved acetone molecules soak up the electric current before it can cross the solution.
This option is wrong — you made the molecules absorb electricity — they simply carry no charge, so there is nothing to carry the current.
DBecause too little acetone was dissolved in the water to complete the tester's circuit.
This option is wrong — you blamed the amount — no quantity of neutral molecules conducts; the particle KIND is what decides.
A solution conducts only when charged particles are free to move through it. Dissolved acetone is neutral whole molecules — nothing charged. With nothing charged free to move, no current flows and the bulb stays dark.
Check your understanding
What must be true of the particles in ANY solution that conducts electricity?
ASome of the dissolved particles are charged and free to move.correct
BSome of the dissolved particles are charged, but held firmly in place.
This option is wrong — you dropped the movement half — charge alone is not enough; the charged particles must be free to move through the solution.
CAll of the dissolved particles are neutral and moving quickly.
This option is wrong — you dropped the charge half — motion alone is not enough; neutral particles cannot carry a current however fast they move.
DSome of the dissolved particles come from a metal.
This option is wrong — you required a metal origin — any free-moving charged particles carry current, whatever substance supplied them.
The mechanism has two requirements joined together. A solution conducts only when charged particles are free to move through it. Charged AND free-moving — drop either half and the bulb stays dark.
Lesson 10 of 45 · SOL-010
Classify electrolytes from compound type
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You've already seen why solutions conduct: free-moving ions. That turns compound type into a prediction tool — you can classify a dissolved substance as an electrolyte or a nonelectrolyte before ever dipping a tester.
The idea
Ionic compounds separate into free-moving ions when they dissolve, so dissolved ionic compounds are electrolytes.
Most molecular compounds stay as neutral whole molecules when they dissolve, so most dissolved molecular compounds are nonelectrolytes.
To classify a dissolved compound, first decide its type: ionic or molecular.
Predicting conductivity from compound type
Compound type
Dissolved particles
Classification
ionic
separated, free-moving ions
electrolyte
molecular (most)
neutral whole molecules
nonelectrolyte
Ionic compounds dissolve as free-moving ions — electrolytes; most molecular compounds dissolve as neutral whole molecules — nonelectrolytes.
KBr is a metal (potassium) combined with a nonmetal (bromine) — an ionic compound — so dissolved KBr is an electrolyte.
Worked examples
Worked example 1. Calcium chloride, CaCl₂, dissolves in water. Classify it as an electrolyte or a nonelectrolyte.
Step 1
CaCl₂ is a metal (calcium) combined with a nonmetal (chlorine), so it is an ionic compound.
Step 2
Ionic compounds separate into free-moving ions when they dissolve.
Step 3
CaCl₂ is an electrolyte.
Worked example 2. Glucose, a molecular compound, dissolves in water. Classify it as an electrolyte or a nonelectrolyte.
Step 1
Glucose is a molecular compound.
Step 2
Most molecular compounds stay as neutral whole molecules when they dissolve.
Step 3
Glucose is a nonelectrolyte.
You can now classify a dissolved compound as an electrolyte or a nonelectrolyte from its compound type: ionic compounds separate into free-moving ions in water, while most molecular compounds stay as neutral whole molecules.
Check your understanding
Each of these four compounds is dissolved in water. Which one is an electrolyte?
ASodium sulfate, Na₂SO₄ — an ionic compound.correct
BEthanol — a molecular compound.
This option is wrong — you classified a molecular compound as an electrolyte — its dissolved molecules stay neutral and whole, so its solution does not conduct.
CFructose — a molecular compound.
This option is wrong — you picked a molecular compound — dissolved fructose is neutral whole molecules, a nonelectrolyte.
DAntifreeze — a molecular compound.
This option is wrong — you picked a molecular compound — dissolved antifreeze stays as neutral whole molecules, a nonelectrolyte.
Classify by compound type first. Na₂SO₄ is ionic — it separates into free-moving ions when it dissolves, so it is an electrolyte. The three molecular compounds dissolve as neutral whole molecules — nonelectrolytes.
Check your understanding
Magnesium bromide, MgBr₂ — an ionic compound — and methanol — a molecular compound — are each dissolved in water. Which classification pair is correct?
AMagnesium bromide is an electrolyte; methanol is a nonelectrolyte.correct
BMagnesium bromide is a nonelectrolyte; methanol is an electrolyte.
This option is wrong — you swapped the types' outcomes — the ionic compound is the one that dissolves into free-moving ions and conducts.
CBoth are electrolytes.
This option is wrong — you classified the molecular compound as an electrolyte too — most molecular compounds dissolve as neutral whole molecules.
DBoth are nonelectrolytes.
This option is wrong — you classified the ionic compound as a nonelectrolyte — dissolved ionic compounds separate into free-moving ions and conduct.
MgBr₂ is ionic: it dissolves as free-moving Mg²⁺ and Br⁻ ions — an electrolyte. Methanol is molecular: it dissolves as neutral whole molecules — a nonelectrolyte. Compound type in, classification out.
Check your understanding
Each of these four compounds is dissolved in water. Which one is a NONELECTROLYTE?
APropylene glycol — a molecular compound.correct
BPotassium iodide, KI — an ionic compound.
This option is wrong — you classified an ionic compound as a nonelectrolyte — dissolved KI is free-moving K⁺ and I⁻ ions, an electrolyte.
CSodium phosphate, Na₃PO₄ — an ionic compound.
This option is wrong — you picked an ionic compound — it separates into free-moving ions in water, an electrolyte.
DCalcium bromide, CaBr₂ — an ionic compound.
This option is wrong — you picked an ionic compound — dissolved CaBr₂ is free-moving ions, an electrolyte.
Three of the four are ionic — metal ions with nonmetal or polyatomic partners — and dissolve as free-moving ions. Propylene glycol is molecular: it dissolves as neutral whole molecules. Most dissolved molecular compounds are nonelectrolytes.
Lesson 11 of 45 · SOL-011
Like dissolves like
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Wonder this:
Water is not the only solvent. Candle wax laughs at water — but drop it into mineral oil and it dissolves away. Solvents play favorites, and the pattern of favorites is one you can learn in a sentence.
You've already classified substances as polar, nonpolar, or ionic. That classification is all the pattern needs.
The idea
Polar and ionic substances generally dissolve well in polar solvents.
Nonpolar substances generally dissolve well in nonpolar solvents.
Chemists sum this up as 'like dissolves like': a solute generally dissolves well when its polarity matches the solvent's.
Water is a polar solvent, so ionic table salt dissolves well in it.
Like dissolves like
Solute
Polar solvent
Nonpolar solvent
polar or ionic
dissolves well
dissolves poorly
nonpolar
dissolves poorly
dissolves well
Matching polarity generally dissolves well; mismatched polarity generally dissolves poorly.
Mineral oil is a nonpolar solvent, so nonpolar candle wax dissolves well in it.
Mismatched pairs generally dissolve poorly: nonpolar candle wax barely dissolves in polar water.
Worked examples
Worked example 1. Grease is nonpolar. In which kind of solvent does grease generally dissolve well?
Step 1
Like dissolves like: nonpolar substances generally dissolve well in nonpolar solvents.
Step 2
Grease generally dissolves well in nonpolar solvents.
Worked example 2. Corn syrup is polar. In which kind of solvent does corn syrup generally dissolve well?
Step 1
Like dissolves like: polar substances generally dissolve well in polar solvents.
Step 2
Corn syrup generally dissolves well in polar solvents.
You can now state the pattern 'like dissolves like': polar and ionic substances generally dissolve well in polar solvents, and nonpolar substances generally dissolve well in nonpolar solvents.
Check your understanding
Motor oil is nonpolar. According to 'like dissolves like', in which kind of solvent does it generally dissolve well?
AIn nonpolar solvents.correct
BIn polar solvents.
This option is wrong — you matched the solute to the opposite kind — like dissolves like means nonpolar goes with nonpolar.
CIn both kinds of solvent equally.
This option is wrong — you dropped the pattern — matching polarity is what predicts good dissolving.
DIn neither kind — oils do not dissolve in anything.
This option is wrong — you made oil undissolvable — a nonpolar solvent generally dissolves a nonpolar substance well.
Name the solute's polarity: motor oil is nonpolar. Nonpolar substances generally dissolve well in nonpolar solvents. So motor oil generally dissolves well in nonpolar solvents.
Check your understanding
Which pairing follows the pattern 'like dissolves like'?
AA nonpolar solute in a nonpolar solvent.correct
BA nonpolar solute in a polar solvent.
This option is wrong — you paired mismatched polarities — the pattern matches nonpolar with nonpolar.
CAn ionic solute in a nonpolar solvent.
This option is wrong — you sent the ionic solute to the wrong side — ionic substances generally dissolve well in POLAR solvents.
DA thick solute in a thick solvent.
This option is wrong — you matched thickness — the pattern matches POLARITY, not how thick the liquids are.
'Like' in the pattern means like POLARITY. Polar and ionic solutes pair with polar solvents; nonpolar solutes pair with nonpolar solvents. Nonpolar-in-nonpolar is the matched pair here.
Check your understanding
Road tar is nonpolar, and Epsom salt is ionic. Which solvent-kind assignment fits 'like dissolves like'?
ATar in a nonpolar solvent; Epsom salt in a polar solvent.correct
BTar in a polar solvent; Epsom salt in a nonpolar solvent.
This option is wrong — you swapped both assignments — nonpolar goes with nonpolar, and ionic goes with polar.
CBoth in a polar solvent.
This option is wrong — you sent the nonpolar tar to a polar solvent — mismatched polarity generally dissolves poorly.
DBoth in a nonpolar solvent.
This option is wrong — you sent the ionic salt to a nonpolar solvent — ionic substances generally dissolve well in polar solvents.
Assign each solute by its own polarity. Nonpolar tar matches a nonpolar solvent. Ionic Epsom salt matches a polar solvent.
Lesson 12 of 45 · SOL-012
Why like dissolves like
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Did You Know?
You've already seen the pattern — like dissolves like — and you've already seen that particles attract their neighbors. The attractions are the reason the pattern works.
The idea
Before anything dissolves, the solute's particles are attracting each other, and so are the solvent's.
Dissolving asks each solute particle to leave its own kind and move in among the solvent's particles.
Particles dissolve when the attractions they can form with the solvent are similar in kind to the attractions they leave behind.
Oil in water fails that test: water's polar particles attract each other strongly, and a nonpolar oil particle can offer water only weak attractions of a different kind.
So the water particles stay with each other, the oil particles stay with each other, and the oil does not dissolve.
Water's particles attract each other strongly; an oil particle offers water only weak attractions of a different kind — so each group stays with its own.
Worked examples
Worked example 1. Sugar is polar, and it dissolves well in polar water. Use attractions to explain why.
Step 1
Particles dissolve when the attractions they can form with the solvent are similar in kind to the attractions they leave behind.
Step 2
A polar sugar molecule leaves polar-to-polar attractions behind and forms the same kind of attractions with the polar water molecules.
Step 3
The attractions match in kind, so sugar dissolves well in water.
Worked example 2. Butter's fat is nonpolar, and it dissolves in nonpolar cooking oil. Use attractions to explain why.
Step 1
Particles dissolve when the attractions they can form with the solvent are similar in kind to the attractions they leave behind.
Step 2
A nonpolar fat particle leaves weak nonpolar attractions behind and forms the same weak kind with the nonpolar oil particles.
Step 3
The attractions match in kind on both sides, so the fat dissolves in the oil.
You can now explain the reason behind 'like dissolves like', because particles dissolve when the attractions they can form with the solvent are similar in kind to the attractions they leave behind.
Check your understanding
A wax crayon mark is nonpolar, and it does not dissolve in polar water. Why not?
ABecause the attractions its particles can form with water are different in kind from the attractions the water particles form with each other.correct
BBecause polar and nonpolar particles actively push each other away, the way two like magnetic poles push apart when brought together.
This option is wrong — you invented a repulsion — nothing pushes; the possible attractions are simply mismatched in kind, so each group stays with its own.
CBecause the crayon's particles are far too heavy to float among the water particles, so they sink instead of spreading.
This option is wrong — you used particle weight — heavy particles dissolve fine when the attractions match; the mismatch is what stops the crayon.
DBecause nobody stirred the mixture long enough or hard enough for the crayon's particles to spread out.
This option is wrong — you blamed stirring — stirring cannot fix attractions that are mismatched in kind.
Water's polar particles attract each other strongly. A nonpolar crayon particle can offer water only weak attractions of a different kind. Particles dissolve when the attractions they can form with the solvent are similar in kind to the attractions they leave behind — here they are not, so the crayon stays put.
Check your understanding
Rubbing alcohol is polar, and it mixes completely with polar water. Why?
ABecause the polar alcohol molecules form the same kind of attractions with water that they leave behind in the alcohol.correct
BBecause the alcohol molecules break apart into charged ions, which the water molecules then carry away.
This option is wrong — you turned a molecular liquid into ions — the alcohol molecules stay whole; their polar attractions with water are what let them mix.
CBecause the alcohol particles and the water particles happen to be almost exactly the same size.
This option is wrong — you matched size — the match that matters is the KIND of attraction the particles form, not their size.
DBecause any two liquids poured into the same container will always mix completely on their own.
This option is wrong — you made mixing automatic — oil and water are two liquids that refuse; matching attractions are what let liquids mix.
Both liquids are polar, so both form polar-to-polar attractions. Each alcohol molecule leaves attractions of one kind and forms the same kind with water. Matched kinds on both sides — the liquids mix completely.
Check your understanding
Epsom salt is ionic. It dissolves in polar water but not in nonpolar lamp oil. Why does it fail to dissolve in the lamp oil?
ABecause the attractions lamp oil could offer the ions are nothing like the ones they leave behind in the lattice.correct
BBecause the lamp oil is too thick and slow-flowing for the ions to push their way through it.
This option is wrong — you used thickness — even a thin nonpolar solvent fails, because it cannot offer attractions like the ones holding the lattice.
CBecause ions are only ever able to move through water, and never through any other liquid.
This option is wrong — you made water magical — the real test is whether the solvent's possible attractions match the ones the ions leave behind.
DBecause the charged ions actively repel the nonpolar oil particles and get pushed back into the solid crystal.
This option is wrong — you invented a repulsion — the oil's attractions are simply too weak and different in kind to replace the lattice's.
The ions sit in a lattice held by strong attractions between opposite charges. Particles dissolve when the attractions they can form with the solvent are similar in kind to the attractions they leave behind. Nonpolar lamp oil offers nothing like the lattice's attractions, so the ions stay in the crystal.
Lesson 13 of 45 · SOL-013
Predict whether it will dissolve
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Did You Know?
You've already seen the pattern 'like dissolves like' and the reason behind it. Now the pattern goes to work: compare two polarities, and the comparison predicts whether a solute dissolves.
The idea
To predict whether a solute will dissolve well in a solvent, compare the polarity of the two substances.
If the solute and the solvent are alike in polarity, predict that the solute dissolves well.
If one substance is polar and the other is nonpolar, predict that the solute dissolves poorly.
Predicting with like dissolves like
Solute
Solvent
Prediction
polar
polar
dissolves well
nonpolar
nonpolar
dissolves well
polar
nonpolar
dissolves poorly
nonpolar
polar
dissolves poorly
ionic
polar
dissolves well
ionic
nonpolar
dissolves poorly
Alike polarities predict good dissolving; unlike polarities predict poor dissolving.
An ionic solute counts as the strongest polar case: predict that it dissolves well in a polar solvent and poorly in a nonpolar one.
The prediction works because particles dissolve when the attractions they can form with the solvent are similar in kind to the attractions they leave behind.
Iodine is a nonpolar solid, hexane is a nonpolar liquid, and water is a polar liquid.
The polarities match in hexane and clash in water, so iodine dissolves better in hexane than in water.
Worked examples
Worked example 1. Table sugar is a polar solid and water is a polar liquid. Predict whether sugar dissolves well in water.
Step 1
Sugar is polar and water is polar — the polarities are alike.
Step 2
Predict that sugar dissolves well in water.
Worked example 2. Motor grease is a nonpolar solid. Predict whether it dissolves better in water, a polar liquid, or in mineral spirits, a nonpolar liquid.
Step 1
Grease and mineral spirits are both nonpolar — the polarities are alike.
Step 2
Grease and water are unlike — nonpolar against polar.
Step 3
Predict that grease dissolves better in mineral spirits than in water.
You can now predict whether a given solute will dissolve well in a given solvent by comparing the polarity of the two substances.
Check your understanding
Naphthalene, the ingredient in mothballs, is a nonpolar solid. Water is a polar liquid and hexane is a nonpolar liquid. What is the correct prediction for naphthalene?
AIt dissolves better in hexane than in water.correct
BIt dissolves better in water than in hexane.
This option is wrong — you treated water as able to dissolve anything — water is polar, and its polarity clashes with a nonpolar solute.
CIt dissolves equally well in both liquids.
This option is wrong — you ignored the polarity comparison — the match with hexane and the clash with water make the two liquids behave very differently.
DIt dissolves in neither liquid, because it is a solid.
This option is wrong — you used the state of matter instead of polarity — solids dissolve well wherever the polarities are alike.
Compare the polarities: naphthalene is nonpolar, hexane is nonpolar, water is polar. Alike polarities predict good dissolving; unlike polarities predict poor dissolving. Naphthalene matches hexane and clashes with water, so it dissolves better in hexane.
Check your understanding
Candle wax is a nonpolar solid. Four liquids are available: turpentine (nonpolar), water (polar), salt water (polar), and vinegar (polar). Which liquid dissolves the wax best?
ATurpentinecorrect
BWater
This option is wrong — you skipped the polarity comparison — polar water clashes with a nonpolar solute.
CSalt water
This option is wrong — you expected the dissolved salt to help — salt water is still polar, and it still clashes with a nonpolar solute.
DVinegar
This option is wrong — you matched wax to a sharp-smelling cleaner instead of to a polarity — vinegar is polar and clashes with nonpolar wax.
Compare the polarities: wax is nonpolar, and only turpentine is nonpolar too. Alike polarities predict good dissolving. Turpentine matches the wax, so it dissolves the wax best.
Check your understanding
Vitamin C is a polar molecule. Cooking oil is a nonpolar liquid and water is a polar liquid. What is the correct prediction for vitamin C?
AIt dissolves better in water than in cooking oil.correct
BIt dissolves better in cooking oil than in water.
This option is wrong — you matched the vitamin to the food instead of to a polarity — polar vitamin C clashes with nonpolar oil.
CIt dissolves equally well in both liquids.
This option is wrong — you ignored the polarity comparison — the match with water and the clash with oil make the two liquids behave very differently.
DIt dissolves in neither liquid, because vitamins do not dissolve.
This option is wrong — you gave vitamins a rule of their own — the polarity comparison covers them like any other solute.
Compare the polarities: vitamin C is polar, water is polar, cooking oil is nonpolar. Alike polarities predict good dissolving; unlike polarities predict poor dissolving. Vitamin C matches water and clashes with oil, so it dissolves better in water.
Summary video — What solutions are, how dissolving works, and electrolytes
Stir spoonful after spoonful of salt into a glass of water. For a while every spoonful vanishes — then one refuses, and no amount of stirring makes it disappear. Water's capacity for salt has a hard limit, and chemists measure it.
You've already seen which solutes dissolve well in which solvents. The next question is how much can dissolve.
The idea
The maximum amount of a solute that can dissolve in a set amount of solvent at a given temperature is called the solute's 'solubility'.
Chemists usually report solubility as the grams of solute that dissolve in 100 g of water.
The temperature must be named, because the maximum changes when the temperature changes.
The solubility of KCl at 20 °C is about 34 g per 100 g of water.
That means 100 g of water at 20 °C can hold at most 34 g of dissolved KCl — a 35th gram would stay solid.
A solubility names three things: the maximum solute, the amount of solvent, and the temperature.
Worked examples
Worked example 1. The solubility of NaNO₃ at 20 °C is 88 g per 100 g of water. What is the most NaNO₃ that 100 g of water at 20 °C can hold in solution?
Step 1
Answer: 88 g.
Worked example 2. A handbook entry reads 'solubility of NH₄Cl: 37 g per 100 g of water.' Which required condition is missing from the entry?
Step 1
Answer: the temperature.
You can now state that solubility is the maximum amount of a solute that can dissolve in a set amount of solvent at a given temperature, usually reported as grams of solute per 100 g of water.
Check your understanding
The solubility of baking soda at 20 °C is about 10 g per 100 g of water. What does this statement say?
AAt 20 °C, 100 g of water can dissolve at most 10 g of baking soda.correct
BAny amount of water at 20 °C can dissolve at most 10 g of baking soda.
This option is wrong — you dropped the set amount of solvent — the 10 g maximum belongs to 100 g of water, and more water can hold more.
C100 g of water can dissolve at most 10 g of baking soda at any temperature.
This option is wrong — you dropped the named temperature — the maximum changes when the temperature changes.
DAt 20 °C, 10 g of water can dissolve at most 100 g of baking soda.
This option is wrong — you swapped the solute and the solvent — the 10 g is the dissolved baking soda and the 100 g is the water.
Solubility is the maximum amount of solute that can dissolve in a set amount of solvent at a given temperature. Here the solute maximum is 10 g, the solvent amount is 100 g of water, and the temperature is 20 °C. So 100 g of water at 20 °C can hold at most 10 g of dissolved baking soda.
Check your understanding
What does a solute's solubility measure?
AThe maximum amount of it that can dissolve in a set amount of solvent at a given temperature.correct
BThe amount of it that happens to be dissolved in a particular solution.
This option is wrong — you measured what IS dissolved instead of what CAN dissolve — solubility is the ceiling, not the current contents.
CHow quickly it disappears when stirred into a solvent.
This option is wrong — you measured speed instead of amount — solubility caps how much dissolves, however fast or slow the dissolving runs.
DThe temperature at which it starts to dissolve in a solvent.
This option is wrong — you turned the stated condition into the quantity — temperature is named alongside a solubility, but the solubility itself is an amount in grams.
Solubility is the maximum amount of a solute that can dissolve in a set amount of solvent at a given temperature. It is an amount — usually grams per 100 g of water — not a speed and not a temperature. A solution may hold less than the maximum; the solubility is the ceiling.
Check your understanding
The solubility of table sugar at 20 °C is about 200 g per 100 g of water. A cook stirs 150 g of sugar into 100 g of water at 20 °C. Can all of it dissolve?
AYes — 150 g is below the 200 g maximum.correct
BNo — only 100 g can dissolve, to match the mass of the water.
This option is wrong — you used the water's mass as the ceiling — the ceiling is the solubility, 200 g, not the 100 g of water.
CYes — any amount of sugar dissolves if it is stirred long enough.
This option is wrong — you removed the ceiling — stirring time never lifts the maximum, and past 200 g the extra sugar stays solid.
DIt cannot be decided without knowing how fast the cook stirs.
This option is wrong — you brought speed into an amount question — the solubility caps how much dissolves regardless of the stirring.
Compare the added amount with the solubility: 150 g against a 200 g maximum. 150 g is under the ceiling, so all of it can dissolve at 20 °C. The ceiling is set by the solubility, not by the water's mass or the stirring.
Lesson 15 of 45 · SOL-015
Temperature and solid solubility
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Did You Know?
You've already seen that a solubility is only complete with its temperature named. Here is what the temperature actually does to solids.
The idea
For most solid solutes in water, solubility increases as temperature increases.
More sugar dissolves in hot tea than in iced tea.
The numbers can move a long way: 100 g of water dissolves 32 g of KNO₃ at 20 °C but 110 g at 60 °C.
A few solid solutes are exceptions — their solubility decreases as temperature increases.
What warming does to solid solubility
Solid solute
20 °C
60 °C
Direction
KNO₃ (typical)
32 g
110 g
increases
lithium sulfate (exception)
35 g
32 g
decreases
Solubility in grams per 100 g of water. Most solids climb with temperature; a few exceptions fall.
Lithium sulfate is one of the exceptions: less of it dissolves in hot water than in cold.
So unless a solid is named as an exception, predict that warming the water raises its solubility.
Worked examples
Worked example 1. Potassium bromide is a typical solid solute. Predict how its solubility at 60 °C compares with its solubility at 20 °C.
Step 1
KBr is not named as an exception, so the pattern for most solids applies: solubility increases as temperature increases.
Step 2
Its solubility at 60 °C is higher — more KBr can dissolve in the warmer water.
Worked example 2. Cerium sulfate is one of the exception solids. Predict how its solubility at 60 °C compares with its solubility at 20 °C.
Step 1
For an exception solid, the direction reverses: solubility decreases as temperature increases.
Step 2
Its solubility at 60 °C is lower — less cerium sulfate can dissolve in the warmer water.
You can now state that the solubility of most solid solutes in water increases as temperature increases, and that a few solid solutes are exceptions whose solubility decreases instead.
Check your understanding
Alum is a typical solid solute. In which water can more alum dissolve: water at 15 °C or water at 75 °C?
AThe water at 75 °Ccorrect
BThe water at 15 °C
This option is wrong — you ran the pattern backwards — that direction belongs to the few exception solids, and alum is a typical solid.
CThe same amount in both
This option is wrong — you treated solubility as fixed — for most solids the maximum rises with temperature.
DIt cannot be predicted from the temperatures alone
This option is wrong — you asked for data the pattern already supplies — a typical solid dissolves more in warmer water.
For most solid solutes in water, solubility increases as temperature increases. Alum is a typical solid, not an exception. So the 75 °C water can dissolve more alum than the 15 °C water.
Check your understanding
Which statement gives the temperature pattern for solids dissolving in water?
AMost solid solutes dissolve more at higher temperature, and a few exceptions dissolve less.correct
BEvery solid solute without exception dissolves more at higher temperature.
This option is wrong — you erased the exceptions — a few solids, like lithium sulfate, dissolve less as the water warms.
CMost solid solutes dissolve less at higher temperature, and a few exceptions dissolve more.
This option is wrong — you flipped the whole pattern — the increase is the majority direction and the decrease is the exception.
DThe temperature of the water does not change how much of a solid can dissolve.
This option is wrong — you froze the maximum in place — KNO₃ alone climbs from 32 g to 110 g between 20 °C and 60 °C.
For most solid solutes in water, solubility increases as temperature increases. A few solids are exceptions whose solubility decreases instead. Most, not all — that is the whole pattern.
Check your understanding
A student needs to dissolve a large scoop of copper(II) sulfate, a typical solid solute, in 100 g of water. Which water gives the scoop the best chance of fully dissolving?
AWater at 80 °Ccorrect
BWater at 5 °C
This option is wrong — you ran the pattern backwards — cold water holds less of a typical solid, not more.
CWater at 20 °C
This option is wrong — you treated room temperature as the best a solvent can do — warming past it keeps raising a typical solid's solubility.
DWater at any temperature — the maximum is the same in each
This option is wrong — you treated the maximum as fixed — for most solids it rises with temperature.
For most solid solutes in water, solubility increases as temperature increases. Copper(II) sulfate is a typical solid, so the hottest water holds the most. Water at 80 °C gives the scoop the best chance of fully dissolving.
Lesson 16 of 45 · SOL-016
Temperature and gas solubility
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Did You Know?
Solids are one story. Dissolved gases run the other way.
The idea
Gases dissolve in water too: the fizz in soda water is dissolved carbon dioxide, and pond water carries the dissolved oxygen its fish breathe.
The solubility of a gas in water decreases as temperature increases.
Cold soda water holds more dissolved CO₂ than warm soda water.
A dissolved gas as water warms
Temperature
Dissolved CO₂ (g per 100 g of water)
0 °C
0.34
20 °C
0.17
40 °C
0.10
Carbon dioxide's solubility in water falls as the temperature rises.
The direction is the opposite of the solid pattern: warming lets most solids dissolve more, and warming lets every gas dissolve less.
The gas pattern needs no exception list — the common gases all follow it.
Worked examples
Worked example 1. A trout stream runs cold in spring and warm in late summer. Predict how the maximum dissolved oxygen the water can hold compares between the two seasons.
Step 1
Oxygen is a gas, and the solubility of a gas in water decreases as temperature increases.
Step 2
The cold spring water can hold more dissolved oxygen than the warm late-summer water.
Worked example 2. Bottled water is stored either in a hot car at 40 °C or in a cool cellar at 12 °C. Predict which storage lets the water keep more dissolved air.
Step 1
Air is a mixture of gases, and gas solubility decreases as temperature increases.
Step 2
The cool cellar water can keep more dissolved air than the hot-car water.
You can now state that the solubility of a gas in water decreases as temperature increases.
Check your understanding
An aquarium heater raises the tank water from 20 °C to 30 °C. What happens to the maximum amount of oxygen the water can hold in solution?
AIt decreases.correct
BIt increases.
This option is wrong — you applied the solid pattern to a gas — gas solubility runs the other way, falling as the water warms.
CIt stays the same.
This option is wrong — you treated gas solubility as temperature-proof — warming lowers the maximum a gas can stay dissolved.
DIt falls to zero.
This option is wrong — you emptied the water completely — warming lowers the maximum, it does not remove every dissolved gas particle.
Oxygen is a gas. The solubility of a gas in water decreases as temperature increases. So the warmer tank water can hold less dissolved oxygen — the maximum decreases.
Check your understanding
Which statement gives the temperature pattern for gases dissolved in water?
AA gas's solubility decreases as the water gets warmer.correct
BA gas's solubility increases as the water gets warmer.
This option is wrong — you gave gases the solid pattern's direction — for gases, warming always lowers the maximum.
CA gas's solubility does not change as the water gets warmer.
This option is wrong — you treated gas solubility as temperature-proof — CO₂ alone falls from 0.34 g to 0.10 g between 0 °C and 40 °C.
DA gas's solubility decreases for a few gases and increases for most.
This option is wrong — you imported the solid pattern's exception structure — the gas pattern has no exception list; every common gas dissolves less when warm.
The solubility of a gas in water decreases as temperature increases. Unlike the solid pattern, this direction has no everyday exceptions. Warmer water always holds less dissolved gas.
Check your understanding
A pot of cold tap water is slowly warmed to 45 °C without boiling. Compare the dissolved air the water can hold at 45 °C with what it could hold cold.
AIt can hold less dissolved air at 45 °C.correct
BIt can hold more dissolved air at 45 °C.
This option is wrong — you applied the solid pattern to the gases in air — gas solubility falls as the water warms.
CIt can hold the same amount at both temperatures.
This option is wrong — you treated gas solubility as temperature-proof — warming lowers the maximum for every gas in the air.
DIt can hold no dissolved air until it cools again.
This option is wrong — you emptied the warm water completely — 45 °C water still holds dissolved air, just less of it.
Air is a mixture of gases. The solubility of a gas in water decreases as temperature increases. So the 45 °C water can hold less dissolved air than the cold water could.
Lesson 17 of 45 · SOL-017
Why warming drives gases out
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You've already seen that a gas's solubility falls as water warms. Here is why the warmth pushes the gas out.
The idea
Warm soda water goes flat because its dissolved CO₂ leaves the liquid.
A dissolved gas particle keeps moving among the water molecules.
At the liquid's surface, a moving gas particle can escape back out of the liquid.
At higher temperature the dissolved gas particles move faster and escape the liquid more easily, so fewer stay dissolved.
Gases become less soluble as temperature increases because at higher temperature dissolved gas particles move faster and escape the liquid more easily.
Faster escape means fewer gas particles stay dissolved, so the maximum the warm water can hold is smaller.
Worked examples
Worked example 1. In summer, the warm surface water of a lake holds less dissolved oxygen than it held in winter. Explain why.
Step 1
Oxygen is a dissolved gas, and gases become less soluble as temperature increases because at higher temperature dissolved gas particles move faster and escape the liquid more easily.
Step 2
In the warm surface water, the dissolved oxygen particles escape more easily than they did in winter.
Step 3
The warm water holds less oxygen because its faster-moving oxygen particles escape the liquid more easily.
Worked example 2. As a kettle of water warms — long before it boils — tiny bubbles of dissolved air collect on the kettle's wall. Explain why the air comes out of solution.
Step 1
The bubbles are dissolved gas leaving the liquid.
Step 2
At higher temperature dissolved gas particles move faster and escape the liquid more easily, so the warming water can hold less air than it holds when cold.
Step 3
The warming water cannot keep all its dissolved air, and the escaping air gathers into bubbles.
You can now explain why gases become less soluble as temperature increases, because at higher temperature dissolved gas particles move faster and escape the liquid more easily.
Check your understanding
Which change in the gas particles causes a gas's solubility to drop as water warms?
AThey move faster and escape the liquid more easily.correct
BThey swell in size until the water can no longer hold them.
This option is wrong — you resized the particles — warming changes their speed, not their size.
CThey break apart into smaller particles that leak away.
This option is wrong — you broke the particles apart — the gas particles stay whole; they simply move faster and escape more easily.
DThey become heavier and settle out of the liquid.
This option is wrong — you made the particles heavier and sent them the wrong way — escaping gas leaves through the surface because it moves faster, not because it sinks.
Gases become less soluble as temperature increases because at higher temperature dissolved gas particles move faster and escape the liquid more easily. The particles themselves do not change size or break apart — only their speed changes. Faster particles escape more easily, so fewer stay dissolved.
Check your understanding
A cold mountain stream holds more dissolved oxygen than the warm lowland river it feeds. Why does the warm river keep less oxygen?
ABecause at higher temperature dissolved gas particles move faster and escape the liquid more easily.correct
BBecause warm water has less room between its molecules for oxygen to fit.
This option is wrong — you explained the drop with space — the water still has room; the faster-moving oxygen particles simply escape more easily.
CBecause the oxygen reacts with the warm water and forms a new substance.
This option is wrong — you turned dissolving into a chemical change — the oxygen particles stay oxygen; they escape, they do not react.
DBecause cold water locks the oxygen particles in place so they cannot leave.
This option is wrong — you froze the dissolved particles — gas particles keep moving in cold water too; they just move more slowly and escape less easily.
Dissolved oxygen particles keep moving among the water molecules at every temperature. Gases become less soluble as temperature increases because at higher temperature dissolved gas particles move faster and escape the liquid more easily. The warm river's oxygen escapes faster than the cold stream's, so the river keeps less.
Check your understanding
Watering cans left in a hot greenhouse hold less dissolved air than identical cans in a cool shed. What explains the difference?
AIn the warmer water, the dissolved gas particles move faster and escape the liquid more easily.correct
BIn the warmer water, the air dissolves deeper and hides below the surface.
This option is wrong — you sent the gas downward — dissolved gas leaves warm water through the surface; it does not retreat into the depths.
CThe greenhouse air pushes extra air into the cans, crowding the old air out.
This option is wrong — you invented an air-pressure story — the difference comes from the water's temperature, which sets how easily the dissolved particles escape.
DThe warm water's molecules grip the gas particles more tightly.
This option is wrong — you ran the grip backwards — if warm water held gas more tightly, it would hold MORE dissolved air, not less.
Both cans hold dissolved air, and its particles keep moving. Gases become less soluble as temperature increases because at higher temperature dissolved gas particles move faster and escape the liquid more easily. The greenhouse cans are warmer, so their dissolved air escapes more easily and less remains.
Lesson 18 of 45 · SOL-018
Saturated and unsaturated
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Did You Know?
You've already seen that solubility caps how much solute a set amount of water can hold at a given temperature. Solutions get names from where they sit against that cap.
The idea
When a solution holds as much dissolved solute as its solubility allows at that temperature, chemists call it 'saturated'.
When a solution can still dissolve more solute at that temperature, chemists call it 'unsaturated'.
Add a little more solute and watch: if it keeps dissolving, the solution was unsaturated.
If the added solute sinks and stays solid however long you stir, the solution was already saturated.
The solubility of NH₄Cl at 20 °C is 37 g per 100 g of water.
So a solution holding 37 g of NH₄Cl in 100 g of water at 20 °C is saturated, and one holding 25 g is unsaturated.
A solution at the cap is saturated; a solution below the cap is unsaturated.
Worked examples
Worked example 1. The solubility of KBr at 20 °C is 65 g per 100 g of water. Classify a solution holding 65 g of KBr in 100 g of water at 20 °C.
Step 1
Compare the dissolved amount with the cap: 65 g held against a 65 g solubility.
Step 2
The solution holds as much dissolved solute as its solubility allows.
Step 3
The solution is saturated.
Worked example 2. The solubility of table sugar at 20 °C is about 200 g per 100 g of water. Classify a solution holding 120 g of sugar in 100 g of water at 20 °C.
Step 1
Compare the dissolved amount with the cap: 120 g held against a 200 g solubility.
Step 2
The solution can still dissolve more sugar at 20 °C.
Step 3
The solution is unsaturated.
You can now classify a solution as saturated when it holds as much dissolved solute as its solubility allows at that temperature, or unsaturated when it can still dissolve more.
Check your understanding
The solubility of KNO₃ at 20 °C is 32 g per 100 g of water. Beaker 1 holds 25 g of KNO₃ dissolved in 100 g of water at 20 °C; beaker 2 holds 32 g dissolved in the same conditions. Classify the two solutions.
ABeaker 1 is unsaturated and beaker 2 is saturated.correct
BBeaker 1 is saturated and beaker 2 is unsaturated.
This option is wrong — you attached each label to the wrong beaker — saturated means holding the full cap, and only beaker 2 holds all 32 g.
CBoth beakers are saturated.
This option is wrong — you treated any dissolved solute as saturation — below the 32 g cap, beaker 1 can still dissolve more.
DBoth beakers are unsaturated.
This option is wrong — you required leftover solid before calling a solution saturated — holding the full 32 g is exactly what saturated means.
Compare each dissolved amount with the 32 g cap. Beaker 1 holds 25 g — below the cap, so it can still dissolve more: unsaturated. Beaker 2 holds 32 g — the full cap: saturated.
Check your understanding
A pinch of KCl is added to a KCl solution at constant temperature, and the pinch dissolves completely. What was the solution before the pinch was added?
AUnsaturated — it could still dissolve more KCl at that temperature.correct
BSaturated — it held as much KCl as its solubility allowed at that temperature.
This option is wrong — you called it saturated even though it took more solute — a saturated solution would have left the pinch sitting as solid.
CUnsaturated — it held no solid KCl on the bottom.
This option is wrong — you classified by the missing solid instead of by the dissolving — a saturated solution can also show a clean bottom; the dissolving pinch is what proves room remained.
DSaturated — it already had some KCl dissolved in it.
This option is wrong — you treated any dissolved solute as saturation — holding some KCl is not the same as holding the full cap.
Add a little more solute and watch: if it keeps dissolving, the solution was unsaturated. The pinch dissolved completely, so room remained below the cap. The solution was unsaturated — it could still dissolve more KCl.
Check your understanding
What makes a solution saturated?
AIt holds as much dissolved solute as its solubility allows at that temperature.correct
BIt holds some dissolved solute, however small the amount.
This option is wrong — you counted any dissolved solute as saturation — the label belongs only to a solution at its cap.
CIt has solid solute resting on the bottom of its container.
This option is wrong — you made the leftover solid the definition — resting solid is evidence the cap was passed, but a solution at its cap with no solid is saturated too.
DIt holds a greater mass of dissolved solute than of solvent.
This option is wrong — you compared solute with solvent — saturation compares the dissolved amount with the solubility, whatever the solvent's mass.
Saturated names one condition: the solution holds as much dissolved solute as its solubility allows at that temperature. It is a comparison with the cap, not with the solvent and not with zero. Solid on the bottom is a clue, never the definition.
Lesson 19 of 45 · SOL-019
Supersaturated solutions
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Did You Know?
Wonder this:
Click the metal disc inside a reusable hand warmer and the clear liquid crystallizes solid in seconds, pouring out heat. Until that click, the liquid was holding more dissolved solute than it had any right to.
You've already seen saturated and unsaturated solutions. There is a third state — a strange, temporary one.
The idea
A solution can temporarily hold MORE dissolved solute than its solubility allows — such a solution is 'supersaturated'.
A supersaturated solution is made in two steps.
First, saturate the solution at a high temperature, where the solubility of a typical solid is large.
Two steps make a supersaturated solution: saturate at high temperature, then cool slowly without disturbing it.
Then cool it slowly, without disturbing it.
Cooling lowers the cap, but the extra solute stays dissolved anyway — for a while.
The hand warmer's liquid is a supersaturated solution of sodium acetate.
Worked examples
Worked example 1. A solution holds more dissolved solute than its solubility allows at its temperature. What is this solution called?
Step 1
Answer: supersaturated.
Worked example 2. Name the two steps that make a supersaturated solution.
Step 1
Answer: saturate the solution at a high temperature, then cool it slowly without disturbing it.
You can now state that a supersaturated solution temporarily holds more dissolved solute than its solubility allows, and that it is made by saturating a solution at high temperature and then cooling it slowly without disturbing it.
Check your understanding
Which solution is supersaturated?
AOne holding more dissolved solute than its solubility allows at its temperature.correct
BOne holding exactly as much dissolved solute as its solubility allows.
This option is wrong — you described a saturated solution — supersaturated means the cap is exceeded, not met.
COne that can still dissolve more solute at its temperature.
This option is wrong — you described an unsaturated solution — a supersaturated solution is past its cap, not under it.
DOne with solid solute resting on the bottom.
This option is wrong — you used leftover solid as the mark — a supersaturated solution is clear, holding its extra solute dissolved.
Supersaturated means holding MORE dissolved solute than the solubility allows. It is a temporary state, and the solution looks clear. At the cap is saturated; under the cap is unsaturated; past the cap is supersaturated.
Check your understanding
Which procedure produces a supersaturated solution?
ASaturate the solution at a high temperature, then cool it slowly without disturbing it.correct
BStir extra solute into a saturated solution at constant temperature.
This option is wrong — you tried to push past the cap directly — at a fixed temperature the extra solute just stays solid, however hard the stirring.
CCool a hot saturated solution quickly while stirring it hard.
This option is wrong — you disturbed the cooling — stirring or shaking during the cool-down makes the extra solute crystallize out instead of staying dissolved.
DDissolve a small amount of solute in a large amount of cold water.
This option is wrong — you made an unsaturated solution — a small amount in cold water sits far below the cap, not above it.
Two steps: saturate at a high temperature, where the cap is large. Then cool slowly, without disturbing it. The cap falls as it cools, and the extra solute stays dissolved anyway — supersaturated.
Check your understanding
Honey is mostly sugar dissolved in a little water — it holds far more sugar than its solubility allows, and a jar of it slowly grows crystals in the cupboard. Classify clear honey before any crystals form.
ASupersaturatedcorrect
BSaturated
This option is wrong — you placed honey exactly at the cap — it holds far MORE sugar than the solubility allows, which is what makes it supersaturated.
CUnsaturated
This option is wrong — you placed honey under the cap — an unsaturated solution could dissolve still more sugar and would never grow crystals on its own.
DNone of the above
This option is wrong — you left the state unnamed — a clear solution holding more than its cap is exactly what supersaturated means.
Compare what honey holds with what its solubility allows: far more. A solution temporarily holding more dissolved solute than its solubility allows is supersaturated. The slow crystal growth in the cupboard is the temporary state ending.
Lesson 20 of 45 · SOL-020
Classify the saturation state from behavior
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Did You Know?
Wonder this:
Three flasks, three clear liquids, same solute, same temperature — one unsaturated, one saturated, one supersaturated. By eye they are identical. One small crystal tells them apart.
You've already seen all three saturation states. This is the test that identifies them.
The idea
To identify a solution's state, drop in one small crystal of the solute and watch what happens.
If the crystal dissolves, the solution was unsaturated — it still had room below its cap.
One small crystal of the solute reveals the state: it dissolves, rests unchanged, or triggers rapid crystal growth.
If the crystal rests on the bottom unchanged, the solution was saturated — it holds exactly its cap.
If the crystal sets off rapid growth of new crystals, the solution was supersaturated — the extra solute crystallizes out onto it.
One crystal, three possible outcomes, and each outcome names the state.
Worked examples
Worked example 1. A small crystal of KCl dropped into a KCl solution slowly shrinks and disappears. Identify the solution's state.
Step 1
The crystal dissolved, so the solution still had room below its cap.
Step 2
The solution was unsaturated.
Worked example 2. A small crystal of sodium thiosulfate dropped into a clear sodium thiosulfate solution sets off a spreading mass of new crystals. Identify the solution's state.
Step 1
The crystal triggered rapid growth of new crystals, so extra solute beyond the cap was waiting to crystallize out.
Step 2
The solution was supersaturated.
You can now classify a solution as unsaturated, saturated, or supersaturated from how it responds when one small crystal of the solute is added: the crystal dissolves, the crystal rests unchanged, or extra solute rapidly crystallizes out.
Check your understanding
One small KNO₃ crystal is dropped into a KNO₃ solution. After ten minutes it sits on the bottom, unchanged. What was the solution?
ASaturatedcorrect
BUnsaturated
This option is wrong — you expected the resting crystal outcome from a solution with room to spare — in an unsaturated solution the crystal dissolves.
CSupersaturated
This option is wrong — you matched the calm outcome to the wrong state — a supersaturated solution answers the crystal with rapid new crystal growth.
DNone of the above
This option is wrong — you decided the test gave no answer — resting unchanged IS an answer, and it is the saturated outcome.
Run the one-crystal test's outcomes: dissolves, rests unchanged, or triggers growth. This crystal rested on the bottom unchanged. A solution that neither takes in the crystal nor crystallizes onto it holds exactly its cap — saturated.
Check your understanding
A small crystal of copper(II) sulfate dropped into a copper(II) sulfate solution shrinks steadily and vanishes. What was the solution?
AUnsaturatedcorrect
BSaturated
This option is wrong — you let a solution at its cap dissolve more — a saturated solution leaves the crystal resting unchanged.
CSupersaturated
This option is wrong — you matched dissolving to the overfilled state — a supersaturated solution grows crystals onto the dropped one; it never dissolves it.
DNone of the above
This option is wrong — you decided the test gave no answer — a dissolving crystal is the unsaturated outcome.
The crystal dissolved. Room below the cap is the only way a solution takes in more solute. The solution was unsaturated.
Check your understanding
A sodium acetate solution is known to be supersaturated. Predict what one small sodium acetate crystal does when dropped in.
AIt sets off rapid growth of new crystals.correct
BIt dissolves completely.
This option is wrong — you gave the overfilled solution room it does not have — dissolving is the unsaturated outcome.
CIt rests on the bottom unchanged.
This option is wrong — you predicted the saturated outcome — a supersaturated solution cannot sit calmly with a crystal; its extra solute crystallizes out onto it.
DIt floats on the surface without changing.
This option is wrong — you judged by floating — the crystal's density is beside the point; the state shows itself in dissolving, resting, or triggering growth.
A supersaturated solution temporarily holds more solute than its cap allows. The dropped crystal gives the extra solute a place to crystallize out. So the crystal sets off rapid growth of new crystals.
Lesson 21 of 45 · SOL-021
Read a solubility curve
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Did You Know?
Wonder this:
A substance's solubility at every temperature from 0 °C to 100 °C would fill a table a hundred rows long. One curved line carries all of it.
You've already seen that solubility changes with temperature. A graph shows the whole relationship at once.
The idea
A 'solubility curve' is the graph of a substance's solubility against temperature.
Temperature runs along the bottom axis, in °C.
Solubility curves of KNO₃ and KCl. The dashed guide reads the KNO₃ curve at 40 °C: 64 g per 100 g of water.
Solubility runs up the side axis, in grams of solute per 100 g of water.
Every point on the curve is a solubility: the maximum grams that dissolve in 100 g of water at that temperature.
To read the solubility at a temperature, find the temperature on the bottom axis, go straight up to the curve, then straight across to the side axis.
On the KNO₃ curve, starting at 40 °C and going up and across lands on 64 — the solubility of KNO₃ at 40 °C is 64 g per 100 g of water.
Worked examples
Worked example 1. Using the KCl curve on the graph, find the solubility of KCl at 30 °C.
Step 1
Find 30 °C on the bottom axis.
Step 2
Go straight up to the KCl curve.
Step 3
Go straight across to the side axis: it reads 37.
Step 4
The solubility of KCl at 30 °C is 37 g per 100 g of water.
Worked example 2. Using the same KCl curve, find the solubility of KCl at 70 °C.
Step 1
Find 70 °C on the bottom axis.
Step 2
Go straight up to the KCl curve, then straight across: the side axis reads 48.
Step 3
The solubility of KCl at 70 °C is 48 g per 100 g of water.
You can now identify the solubility of a substance at a given temperature by reading its solubility curve, the graph of its solubility against temperature.
Check your understanding
The graph shows the solubility curve of NH₄Cl. Read the solubility of NH₄Cl at 60 °C, in g per 100 g of water.
Answer: 55g per 100 g of water(tolerance ±2)
Find 60 °C on the bottom axis. Go straight up to the NH₄Cl curve. Go straight across to the side axis: it reads 55. The solubility of NH₄Cl at 60 °C is 55 g per 100 g of water.
Check your understanding
The graph shows the solubility curve of KClO₃. Read the solubility of KClO₃ at 40 °C, in g per 100 g of water.
Answer: 14g per 100 g of water(tolerance ±2)
Find 40 °C on the bottom axis. Go straight up to the KClO₃ curve. Go straight across to the side axis: it reads 14. The solubility of KClO₃ at 40 °C is 14 g per 100 g of water.
Check your understanding
The graph shows the solubility curve of NaNO₃. Read the solubility of NaNO₃ at 20 °C, in g per 100 g of water.
Answer: 88g per 100 g of water(tolerance ±2)
Find 20 °C on the bottom axis. Go straight up to the NaNO₃ curve. Go straight across to the side axis: it reads 88. The solubility of NaNO₃ at 20 °C is 88 g per 100 g of water.
Lesson 22 of 45 · SOL-022
Saturation state from a point on the graph
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Did You Know?
You've already read solubilities off a curve, and you've met the three saturation states. Put together, the curve sorts any plotted solution into its state at a glance.
The idea
Any solution can be plotted as a point on the solubility graph: its temperature across, its dissolved grams per 100 g of water up.
The curve itself is the cap, so a point ON the curve is a saturated solution.
The KCl curve is the cap: below it unsaturated, on it saturated, above it supersaturated.
A point BELOW the curve holds less than the cap — an unsaturated solution.
A point ABOVE the curve holds more than the cap — a supersaturated solution.
A point below the KCl curve, for instance, is an unsaturated KCl solution.
So the state reads straight off the point's position: below, on, or above the curve.
Worked examples
Worked example 1. On the graph, an NH₄Cl solution is plotted at 30 °C holding 41 g per 100 g of water. Classify its state.
Step 1
Find the point: 30 °C across, 41 g up.
Step 2
The NH₄Cl curve passes through 41 g at 30 °C, so the point sits ON the curve.
Step 3
The solution is saturated.
Worked example 2. On the graph, a NaNO₃ solution is plotted at 40 °C holding 120 g per 100 g of water. Classify its state.
Step 1
Find the point: 40 °C across, 120 g up.
Step 2
The NaNO₃ curve reads 104 g at 40 °C, so the point sits ABOVE the curve.
Step 3
The solution is supersaturated.
You can now classify a plotted solution as unsaturated, saturated, or supersaturated from where its point sits relative to the solubility curve: below the curve, on the curve, or above the curve.
Check your understanding
The graph shows the KNO₃ solubility curve and one plotted KNO₃ solution, point P at 60 °C holding 70 g per 100 g of water. Classify the solution at P.
AUnsaturatedcorrect
BSaturated
This option is wrong — you placed the point on the curve — at 60 °C the curve sits at 110 g, well above P's 70 g.
CSupersaturated
This option is wrong — you read above and below backwards — points ABOVE the curve hold more than the cap; P sits below it.
DNone of the above
This option is wrong — you left the point unclassified — every plotted solution is below, on, or above the curve, and P is below.
Find P: 60 °C across, 70 g up. The KNO₃ curve reads 110 g at 60 °C, so P sits below the curve. A point below the curve holds less than the cap — the solution is unsaturated.
Check your understanding
The graph shows the KClO₃ solubility curve and one plotted KClO₃ solution, point P at 50 °C holding 19 g per 100 g of water. Classify the solution at P.
ASaturatedcorrect
BUnsaturated
This option is wrong — you placed the point below the curve — at 50 °C the curve passes exactly through 19 g, and P sits on it.
CSupersaturated
This option is wrong — you placed the point above the curve — P sits on the curve itself, holding exactly the cap.
DNone of the above
This option is wrong — you left the point unclassified — a point on the curve is the saturated case.
Find P: 50 °C across, 19 g up. The KClO₃ curve reads 19 g at 50 °C, so P sits ON the curve. The curve is the cap — a solution on it is saturated.
Check your understanding
The graph shows the KBr solubility curve and one plotted KBr solution, point P at 20 °C holding 80 g per 100 g of water. Classify the solution at P.
ASupersaturatedcorrect
BUnsaturated
This option is wrong — you read above and below backwards — P holds 80 g against a 65 g cap, which is more than the curve allows.
CSaturated
This option is wrong — you placed the point on the curve — the KBr curve reads 65 g at 20 °C, and P sits 15 g above it.
DNone of the above
This option is wrong — you left the point unclassified — a point above the curve is the supersaturated case.
Find P: 20 °C across, 80 g up. The KBr curve reads 65 g at 20 °C, so P sits above the curve. A point above the curve holds more than the cap — the solution is supersaturated.
Lesson 23 of 45 · SOL-023
Compare substances on one graph
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Did You Know?
You've already read one curve at a time. Graphs usually carry several substances' curves together, and that is where they earn their keep.
The idea
One graph often shows several substances' solubility curves on the same grid.
To compare solubilities, pick the temperature first, then read every curve at that same temperature.
Reading both curves at 50 °C: KNO₃ sits at 85 g and NaCl at 37 g, so KNO₃ is the more soluble at 50 °C.
At the chosen temperature, the higher curve belongs to the more soluble substance.
At 50 °C the KNO₃ curve (85 g) runs far above the NaCl curve (37 g), so KNO₃ is the more soluble of the two at 50 °C.
The order can change with temperature, because curves can cross.
So a ranking is only ever a ranking AT a temperature — always name it.
Worked examples
Worked example 1. Using the graph, rank KBr and KClO₃ by solubility at 30 °C.
Step 1
Read both curves at 30 °C: KBr reads 70 g and KClO₃ reads 10 g per 100 g of water.
Step 2
The higher curve at that temperature belongs to the more soluble substance.
Step 3
At 30 °C, KBr is more soluble than KClO₃.
Worked example 2. Using the graph, rank NaNO₃, NH₄Cl, and KClO₃ from most to least soluble at 20 °C.
Step 1
Read all three curves at 20 °C: NaNO₃ 88 g, NH₄Cl 37 g, KClO₃ 7 g per 100 g of water.
Step 2
Order the readings from highest to lowest.
Step 3
At 20 °C: NaNO₃ is most soluble, then NH₄Cl, then KClO₃.
You can now rank two or more substances by their solubility at a given temperature by comparing their solubility curves.
Check your understanding
The graph shows the solubility curves of NH₄Cl, KCl, and KClO₃. Which substance is the most soluble at 40 °C?
ANH₄Clcorrect
BKCl
This option is wrong — you picked the middle curve — at 40 °C KCl reads 40 g, below NH₄Cl's 46 g.
CKClO₃
This option is wrong — you picked the lowest curve — at 40 °C KClO₃ reads only 14 g, the least soluble of the three.
DThe graph cannot rank them at a single temperature.
This option is wrong — you treated the curves as trend-only — reading each curve at 40 °C gives three exact values to rank.
Read every curve at 40 °C: NH₄Cl 46 g, KCl 40 g, KClO₃ 14 g per 100 g of water. The higher curve at that temperature belongs to the more soluble substance. NH₄Cl is the most soluble at 40 °C.
Check your understanding
The graph shows the solubility curves of NaNO₃, KBr, and KCl. Which substance is the least soluble at 60 °C?
AKClcorrect
BNaNO₃
This option is wrong — you gave the top curve the bottom rank — at 60 °C NaNO₃ reads 124 g, the most soluble of the three.
CKBr
This option is wrong — you picked the middle curve — at 60 °C KBr reads 86 g, well above KCl's 46 g.
DThey are equally soluble at 60 °C.
This option is wrong — you skipped the readings — the three curves sit at 124, 86, and 46 g at 60 °C, nowhere near equal.
Read every curve at 60 °C: NaNO₃ 124 g, KBr 86 g, KCl 46 g per 100 g of water. The lowest curve at that temperature belongs to the least soluble substance. KCl is the least soluble at 60 °C.
Check your understanding
The graph shows the solubility curves of NaNO₃, KBr, and KNO₃. Which order ranks them from most to least soluble at 10 °C?
ANaNO₃, then KBr, then KNO₃correct
BKNO₃, then KBr, then NaNO₃
This option is wrong — you ranked at the hot end of the graph — at 80 °C KNO₃ leads, but the question names 10 °C, where it trails.
CKBr, then NaNO₃, then KNO₃
This option is wrong — you swapped the top two — at 10 °C NaNO₃ reads 80 g and KBr reads 60 g.
DNaNO₃, then KNO₃, then KBr
This option is wrong — you swapped the bottom two — at 10 °C KBr reads 60 g and KNO₃ only 21 g.
Read every curve at 10 °C: NaNO₃ 80 g, KBr 60 g, KNO₃ 21 g per 100 g of water. Order the readings from highest to lowest. At 10 °C: NaNO₃, then KBr, then KNO₃.
Lesson 24 of 45 · SOL-024
How much crystallizes or dissolves on a temperature change
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Did You Know?
You've already classified a cooling solution's state from the curve. The curve can also say how many grams the change moves.
The idea
A saturated solution holds exactly the cap at its temperature.
When the solution cools, the cap falls with the temperature.
The solute above the new, lower cap cannot stay dissolved — it crystallizes out of the solution.
mass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature
Both solubilities are read off the curve, and both are per 100 g of water — so the answer is too.
Cooling a saturated KNO₃ solution from 60 °C to 30 °C: the cap falls from 110 g to 46 g, and the 64 g difference crystallizes out.
Warming a saturated solution runs the same subtraction the other way: the difference between the caps is the extra mass that can then dissolve.
A saturated KNO₃ solution in 100 g of water cools from 60 °C to 30 °C; the curve reads 110 g and 46 g.
110 − 46 = 64, so 64 g of KNO₃ crystallizes out of solution.
Worked examples
Worked example 1. A saturated NaNO₃ solution in 100 g of water cools from 50 °C to 20 °C. Using the NaNO₃ curve on the graph, find the mass of NaNO₃ that crystallizes out.
Step 1
Write down the values in the question
A saturated solution holds exactly the cap at its temperature.
solubility at 50 °C = 114 g per 100 g of water (from the curve)
solubility at 20 °C = 88 g per 100 g of water (from the curve)
Step 2
Write down the equation
mass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature
Step 3
Substitute in the values, and calculate
mass that crystallizes = 114 − 88
mass that crystallizes = 26 g of NaNO₃
Worked example 2. A saturated KClO₃ solution in 100 g of water is warmed from 20 °C to 60 °C. Using the KClO₃ curve on the graph, find the extra mass of KClO₃ that can now dissolve.
Step 1
Write down the values in the question
A saturated solution holds exactly the cap at its temperature, and warming raises the cap.
solubility at 60 °C = 24 g per 100 g of water (from the curve)
solubility at 20 °C = 7 g per 100 g of water (from the curve)
Step 2
Write down the equation
extra mass that can dissolve = solubility at the warmer temperature − solubility at the cooler temperature
Step 3
Substitute in the values, and calculate
extra mass that can dissolve = 24 − 7
extra mass that can dissolve = 17 g of KClO₃
You can now calculate the mass of solute that crystallizes out when a saturated solution cools, or the extra mass that can dissolve when it warms, by subtracting two readings from the solubility curve.
Check your understanding
A saturated NH₄Cl solution in 100 g of water cools from 60 °C to 20 °C. Using the NH₄Cl curve on the graph, find the mass of NH₄Cl that crystallizes out, in grams.
Answer: 18g(tolerance ±2)
A saturated solution holds exactly the cap at its temperature. Read the two caps from the curve: solubility at 60 °C = 55 g per 100 g of water solubility at 20 °C = 37 g per 100 g of water Write down the equation: mass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature Substitute in the values, and calculate: mass that crystallizes = 55 − 37 mass that crystallizes = 18 g of NH₄Cl
Check your understanding
A saturated KBr solution in 100 g of water cools from 80 °C to 30 °C. Using the KBr curve on the graph, find the mass of KBr that crystallizes out, in grams.
Answer: 25g(tolerance ±2)
A saturated solution holds exactly the cap at its temperature. Read the two caps from the curve: solubility at 80 °C = 95 g per 100 g of water solubility at 30 °C = 70 g per 100 g of water Write down the equation: mass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature Substitute in the values, and calculate: mass that crystallizes = 95 − 70 mass that crystallizes = 25 g of KBr
Check your understanding
A saturated KCl solution in 100 g of water is warmed from 10 °C to 70 °C. Using the KCl curve on the graph, find the extra mass of KCl that can now dissolve, in grams.
Answer: 17g(tolerance ±2)
A saturated solution holds exactly the cap at its temperature, and warming raises the cap. Read the two caps from the curve: solubility at 70 °C = 48 g per 100 g of water solubility at 10 °C = 31 g per 100 g of water Write down the equation: extra mass that can dissolve = solubility at the warmer temperature − solubility at the cooler temperature Substitute in the values, and calculate: extra mass that can dissolve = 48 − 31 extra mass that can dissolve = 17 g of KCl
Lesson 25 of 45 · SOL-025
Stirring and dissolving speed
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Have You Ever Wondered?
Wonder this:
Drop a spoonful of sugar into a glass of water and wait. Ten minutes later, most of it still sits at the bottom. Stir for a few seconds and it is gone. The water is the same, the sugar is the same — what did the spoon change?
You've already seen how water takes a solute apart particle by particle. That picture explains what stirring does.
The idea
In still water, sugar dissolves only where water touches it — at the sugar's outside surface.
As sugar dissolves, the water right next to that surface becomes crowded with dissolved sugar.
Crowded water takes in new sugar only slowly, so dissolving slows down.
Without stirring, crowded water lingers at the solute's surface. Stirring brings fresh water to the surface, where dissolving happens.
Stirring sweeps the crowded water away and brings fresh water to the sugar's surface.
So stirring or shaking makes a solute dissolve faster, because dissolving happens where moving solvent particles contact the solute surface, and stirring keeps bringing fresh solvent to that surface.
Stirring, shaking, and swirling all do the same job — chemists call all of them 'agitation'.
Worked examples
Worked example 1. A cook adds one spoonful of salt to each of two identical pots of water at the same temperature, stirs the first pot, and leaves the second still. In which pot does the salt dissolve faster?
Step 1
Dissolving happens where moving solvent particles contact the solute surface.
Step 2
Stirring keeps bringing fresh water to the salt's surface in the first pot.
Step 3
In the still pot, crowded water lingers at the salt's surface and dissolving slows.
Step 4
The salt in the stirred pot dissolves faster.
Worked example 2. A hiker pours drink powder into a water bottle, caps it, and shakes it hard. Why does shaking make the powder dissolve faster?
Step 1
Shaking is agitation, the same job as stirring.
Step 2
Shaking keeps bringing fresh water into contact with the powder's surface.
Step 3
The powder dissolves faster in the shaken bottle, because dissolving happens where moving solvent particles contact the solute surface, and shaking keeps bringing fresh solvent to that surface.
You can now predict that stirring or shaking makes a solute dissolve faster, because dissolving happens where moving solvent particles contact the solute surface, and stirring keeps bringing fresh solvent to that surface.
Check your understanding
Two identical mugs hold water at the same temperature. A student adds one spoonful of instant coffee to each, stirs the first mug, and leaves the second untouched. In which mug does the coffee dissolve faster?
AThe stirred mugcorrect
BThe untouched mug
This option is wrong — you flipped the effect — stirring keeps bringing fresh water to the coffee's surface, so the stirred mug dissolves its coffee faster.
CBoth mugs, at exactly the same speed
This option is wrong — you treated stirring as changing nothing — in the untouched mug, water crowded with dissolved coffee lingers at the surface and slows dissolving.
DNeither mug, because coffee cannot dissolve without heating
This option is wrong — you made heating a requirement — the coffee dissolves in both mugs; stirring just makes it happen sooner.
Dissolving happens where moving solvent particles contact the solute surface. Stirring keeps bringing fresh water to the coffee's surface. So the stirred mug dissolves its coffee faster.
Check your understanding
A bather pours epsom salt into a full bathtub and swirls the water with one hand. Why does swirling make the epsom salt dissolve faster?
ASwirling keeps bringing fresh water to the salt's surface, where dissolving happens.correct
BSwirling warms the water, and warm water works on the salt more strongly.
This option is wrong — you credited swirling with heating — a hand swirling bathwater does not raise its temperature; it moves fresh water to the salt's surface.
CSwirling breaks each salt crystal into much smaller pieces.
This option is wrong — you turned swirling into crushing — gentle swirling leaves the crystals' size unchanged; it changes which water touches them.
DSwirling squeezes the salt crystals until they fall apart.
This option is wrong — you invented a pressure effect — dissolving happens where moving solvent particles contact the solute surface, and swirling refreshes that contact.
Dissolving happens where moving solvent particles contact the solute surface. Without swirling, water crowded with dissolved salt lingers at the crystals' surfaces. Swirling sweeps that crowded water away and brings fresh water in, so the salt dissolves faster.
Check your understanding
Plant fertilizer granules are dissolving in a watering can of still water. The water right next to the granules has become crowded with dissolved fertilizer. What does this crowded layer do to the dissolving?
AIt slows the dissolving, because crowded water takes in new solute only slowly.correct
BIt speeds the dissolving, because dissolved fertilizer pulls more fertilizer off the granules.
This option is wrong — you ran the effect backwards — water already crowded with dissolved solute takes in new solute slowly, which is exactly why stirring helps.
CIt stops the dissolving completely and permanently.
This option is wrong — you turned a slowdown into a full stop — dissolving continues slowly as the dissolved fertilizer drifts away on its own.
DIt changes nothing, because dissolving depends only on the granules themselves.
This option is wrong — you left the solvent out of the story — dissolving happens where moving solvent particles contact the solute surface, so what that nearby water holds matters.
As solute dissolves, the water next to the surface becomes crowded with dissolved solute. Crowded water takes in new solute only slowly, so dissolving slows down. Stirring sweeps the crowded water away and brings fresh water to the surface — that is why agitation speeds dissolving.
Lesson 26 of 45 · SOL-026
Temperature and dissolving speed
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Did You Know?
You've already seen that stirring speeds up dissolving by bringing fresh solvent to the solute's surface. Temperature speeds it up a different way — by changing the solvent particles themselves.
The idea
Sugar dissolves faster in hot water than in cold water.
The reason starts with motion: particles move faster at higher temperature.
Faster water particles strike the sugar's surface more often.
Particles move faster at higher temperature, so solvent particles strike the solute surface more often and harder.
Each strike is also harder, so it knocks sugar particles loose from the surface more easily.
So a solute dissolves faster at higher temperature, because particles move faster at higher temperature, so solvent particles strike the solute surface more often and harder.
Worked examples
Worked example 1. A student dissolves 10 g of potassium chloride in water at 80 °C and another 10 g in water at 20 °C, with no stirring. In which beaker does the KCl dissolve faster?
Step 1
Particles move faster at higher temperature.
Step 2
In the 80 °C beaker, the water particles strike the KCl surface more often and harder.
Step 3
The KCl in the 80 °C beaker dissolves faster.
Worked example 2. Dishwasher detergent powder dissolves quickly in a hot wash cycle but slowly in a cold rinse. Why?
Step 1
The hot water's particles move faster than the cold water's particles.
Step 2
They strike the powder's surface more often and harder, knocking detergent particles loose sooner.
Step 3
The powder dissolves faster in the hot cycle, because particles move faster at higher temperature, so solvent particles strike the solute surface more often and harder.
You can now predict that a solute dissolves faster at higher temperature, because particles move faster at higher temperature, so solvent particles strike the solute surface more often and harder.
Check your understanding
A cook stirs one packet of gelatin dessert powder into hot water and an identical packet into cold water, stirring both the same way. In which water does the powder dissolve faster?
AThe hot watercorrect
BThe cold water
This option is wrong — you flipped the direction — hot water's particles move faster, striking the powder's surface more often and harder, so the powder dissolves sooner.
CBoth waters, at exactly the same speed, since the stirring is identical
This option is wrong — you gave stirring the whole job — with equal stirring, the temperature difference still matters, because particles move faster at higher temperature.
DNeither — gelatin powder only dissolves in boiling water
This option is wrong — you invented a boiling requirement — the powder dissolves in both, just faster where the water particles strike its surface harder and more often.
Particles move faster at higher temperature. The hot water's particles strike the powder's surface more often and harder. So the powder dissolves faster in the hot water.
Check your understanding
Why does baking soda dissolve faster in warm water than in cold water?
AWarm water's particles move faster, so they strike the baking soda's surface more often and harder.correct
BWarm water is thinner, so the baking soda sinks through it and dissolves on the way down.
This option is wrong — you told a sinking story — dissolving speed comes from how often and how hard water particles strike the solute surface, and faster particles strike more.
CWarm water contains more water particles in the same space.
This option is wrong — you added particles instead of speeding them up — warming changes how fast the particles move, not how many there are.
DWarmth softens the baking soda until it falls apart on its own.
This option is wrong — you left the water out of the story — the solvent's faster-moving particles do the work, knocking solute particles loose from the surface.
A solute dissolves faster at higher temperature, because particles move faster at higher temperature, so solvent particles strike the solute surface more often and harder. The baking soda's surface gets struck more often, and each strike is harder. Solute particles are knocked loose sooner, so the baking soda dissolves faster.
Check your understanding
A scoop of borax dissolves noticeably faster in a 60 °C laundry soak than in a 15 °C one. What are the water particles in the 60 °C soak doing differently?
AMoving faster, so they strike the borax surface more often and hardercorrect
BSticking to each other more tightly, which squeezes the borax apart
This option is wrong — you described a squeeze that is not there — warming makes water particles move faster, and the faster strikes on the borax surface are what speed dissolving.
CMoving more slowly, which gives each one more time at the borax surface
This option is wrong — you reversed the motion change — particles move faster at higher temperature, not slower.
DSwelling to a larger size, so each one covers more of the borax surface
This option is wrong — you resized the particles — temperature changes the particles' speed, never their size.
Particles move faster at higher temperature. Faster water particles strike the borax surface more often, and each strike is harder. So the borax dissolves faster in the 60 °C soak.
Lesson 27 of 45 · SOL-027
Surface area and dissolving speed
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Did You Know?
Stirring refreshes the water at a solute's surface, and heat makes the water strike that surface harder. The third speed factor changes the surface itself.
The idea
A solute dissolves only at its surface, where water touches it — the inside of a lump has to wait its turn.
Crushing a lump into smaller pieces uncovers surfaces that were hidden inside.
More exposed surface means more places where water contacts solute at the same time.
Crushing exposes surfaces that were hidden inside — more places for water to contact the solute at once.
So a crushed or powdered solute dissolves faster than one large piece, because dissolving happens where moving solvent particles contact the solute surface, and crushing exposes more of that surface.
That is why powdered sugar vanishes into water while a whole sugar cube dissolves slowly, from the outside in.
Worked examples
Worked example 1. A cook has coarse rock salt and an equal mass of fine table salt, and dissolves each in its own glass of water at the same temperature, unstirred. Which dissolves faster?
Step 1
The fine salt's many small grains expose far more total surface than the coarse lumps.
Step 2
More exposed surface means more places where water contacts salt at the same time.
Step 3
The fine table salt dissolves faster.
Worked example 2. A student drops one whole hard candy into a glass of water and an identical candy, crushed to fragments, into another. Why do the fragments dissolve faster?
Step 1
Crushing uncovered candy surfaces that were hidden inside the whole candy.
Step 2
Water now contacts candy at many more places at once.
Step 3
The fragments dissolve faster, because dissolving happens where moving solvent particles contact the solute surface, and crushing exposes more of that surface.
You can now predict that a crushed or powdered solute dissolves faster than one large piece, because dissolving happens where moving solvent particles contact the solute surface, and crushing exposes more of that surface.
Check your understanding
A student has two equal-mass samples of copper(II) sulfate: large blue crystals and the same substance ground to a powder. Each is added, unstirred, to its own beaker of water at the same temperature. Which sample dissolves faster?
AThe ground powdercorrect
BThe large crystals
This option is wrong — you flipped the effect — grinding exposes surfaces that were hidden inside, giving water more places to contact the solute at once.
CBoth samples, at exactly the same speed, because the masses are equal
This option is wrong — you counted only the mass — equal masses can expose very different amounts of surface, and dissolving happens only at the surface.
DNeither — copper(II) sulfate dissolves only in stirred water
This option is wrong — you made stirring a requirement — both samples dissolve in still water; the powder just finishes sooner because it exposes more surface.
Dissolving happens where moving solvent particles contact the solute surface. Grinding exposes surfaces that were hidden inside the crystals. More exposed surface means more places dissolving happens at once, so the powder dissolves faster.
Check your understanding
Why does a crushed medicine tablet dissolve in water faster than the same tablet left whole?
ACrushing exposes more of the tablet's surface, giving water more places to contact it at once.correct
BCrushing warms the tablet as it breaks apart, and warmer solutes always dissolve faster.
This option is wrong — you credited crushing with heating — crushing changes the exposed surface, not the temperature.
CCrushing grinds the tablet's particles down until they are smaller than water's particles.
This option is wrong — you shrank the particles themselves — crushing breaks the tablet into smaller PIECES, but each piece still contains enormous numbers of unchanged particles.
DA crushed tablet weighs less than a whole one, and lighter samples dissolve faster.
This option is wrong — you lost mass that never left — crushing rearranges the same mass into more pieces; the gain is exposed surface, not lightness.
A solute dissolves only at its surface, where water touches it. Crushing uncovers surfaces that were hidden inside the whole tablet. More exposed surface means more places where dissolving happens at the same time.
Check your understanding
Bath shops sell the same bath salt as chunky lumps and as fine grains. Equal masses go into identical basins of water at the same temperature, unstirred. Which statement about the fine grains is correct?
AThey dissolve faster, because together they expose more surface to the water.correct
BThey dissolve more slowly, because small grains pack together and shut water out.
This option is wrong — you imagined the grains sealing together — loose grains in water stay bathed in it, and their many faces give water more contact points, not fewer.
CThey dissolve at the same speed as the lumps, because the substance is identical.
This option is wrong — you judged by identity alone — same substance, but dissolving happens where moving solvent particles contact the solute surface, and the grains expose far more surface.
DThey dissolve faster, because fine grains are warmer than lumps.
This option is wrong — you gave the right speed with a temperature story — the grains and lumps start at the same temperature; the difference is exposed surface.
Dissolving happens where moving solvent particles contact the solute surface. Fine grains expose far more total surface than chunky lumps of the same mass. So the fine grains dissolve faster.
Lesson 28 of 45 · SOL-028
Faster dissolving is not more dissolving
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Did You Know?
You've already seen three ways to speed up dissolving — stirring, heating, crushing — and you've seen that solubility caps how much can dissolve at a given temperature. Those are two different things, and mixing them up is the classic mistake.
The idea
Two different questions hide inside 'dissolves better': how FAST the solute dissolves, and how MUCH can dissolve.
Stirring changes only the speed — the solute reaches its dissolved state sooner, but the amount that can dissolve stays the same.
Crushing also changes only the speed.
Two different questions
Change
How fast it dissolves
How much can dissolve
Stirring
faster
unchanged
Crushing
faster
unchanged
Raising the temperature
faster
rises for most solid solutes
Stirring and crushing change only the speed. Temperature changes the speed and, for most solid solutes, the amount.
The amount that CAN dissolve at a given temperature is the solubility, and stirring and crushing leave the solubility untouched.
Temperature is different: raising it makes solutes dissolve faster, and for most solid solutes it also raises how much can dissolve.
So classify any change by asking two questions: did the speed change, and did the solubility change?
Worked examples
Worked example 1. A student stirs a drink-mix solution vigorously. Does the stirring change how fast the mix dissolves, how much can dissolve, or both?
Step 1
Ask the speed question: stirring brings fresh water to the mix's surface, so the speed rises.
Step 2
Ask the solubility question: stirring leaves the solubility at that temperature untouched.
Step 3
Stirring changes only how fast the mix dissolves.
Worked example 2. A student warms water from 20 °C to 60 °C before adding potassium nitrate, a typical solid solute. Does the warming change how fast it dissolves, how much can dissolve, or both?
Step 1
Ask the speed question: hotter water particles strike the solute surface more often and harder, so the speed rises.
Step 2
Ask the solubility question: for most solid solutes, including potassium nitrate, solubility rises with temperature.
Step 3
Warming changes both the speed and the amount that can dissolve.
You can now classify a change as affecting how fast a solute dissolves, how much solute can dissolve, or both: stirring and crushing change only the speed, while temperature changes both the speed and, for most solutes, the amount.
Check your understanding
A pharmacist grinds coarse sea-salt crystals to a fine powder before dissolving them. What does the grinding change about the dissolving?
AOnly how fast the solute dissolvescorrect
BOnly how much solute can dissolve
This option is wrong — you gave grinding power over solubility — crushing exposes more surface, so dissolving is faster, but the amount that can dissolve at that temperature is unchanged.
CBoth the speed and the amount
This option is wrong — you let the speed gain drag the amount with it — grinding speeds dissolving, but only temperature changes how much can dissolve.
DNeither the speed nor the amount
This option is wrong — you dismissed grinding entirely — it genuinely speeds dissolving by exposing more solute surface; it just leaves the solubility untouched.
Ask the speed question: grinding exposes more surface, so the salt dissolves faster. Ask the solubility question: grinding leaves the solubility at that temperature untouched. So grinding changes only how fast the solute dissolves.
Check your understanding
A photographer warms the water before dissolving fixer crystals, a typical solid solute. What does the warming change about the dissolving?
ABoth the speed and the amountcorrect
BOnly how fast the solute dissolves
This option is wrong — you stopped at the speed — warming also raises how much of a typical solid solute can dissolve.
COnly how much solute can dissolve
This option is wrong — you skipped the speed — warming also makes the crystals dissolve faster, because faster water particles strike their surfaces more often and harder.
DNeither the speed nor the amount
This option is wrong — you dismissed temperature entirely — it is the one change on the list that moves both the speed and, for most solid solutes, the amount.
Ask the speed question: hotter water dissolves the crystals faster. Ask the solubility question: for most solid solutes, solubility rises with temperature. So warming changes both the speed and the amount that can dissolve.
Check your understanding
A runner shakes a capped bottle of water and glucose powder instead of letting it sit. What does the shaking change about the dissolving?
AOnly how fast the solute dissolvescorrect
BOnly how much solute can dissolve
This option is wrong — you gave shaking power over solubility — agitation refreshes the water at the powder's surface, changing speed only.
CBoth the speed and the amount
This option is wrong — you let the speed gain drag the amount with it — shaking speeds dissolving, but the solubility at that temperature stays fixed.
DNeither the speed nor the amount
This option is wrong — you dismissed shaking entirely — shaking is agitation, and it genuinely speeds dissolving while leaving the solubility untouched.
Shaking is agitation — the same job as stirring. It speeds the dissolving by bringing fresh water to the powder's surface. The solubility at that temperature is untouched, so only the speed changes.
Summary video — Solubility, solubility curves, and dissolving speed
Stir a spoonful of table salt into water and it vanishes. Stir in a spoonful of powdered chalk — another ionic compound — and you get a gritty white cloud that never clears. Looking at the two powders side by side, you could not have told which would dissolve. How do chemists know in advance?
Chemists answered this the hard way: they tested compound after compound in water and wrote down what happened.
The idea
The results compress into a short list called the 'solubility rules'.
The first rule names the ions that never let you down: if a compound contains one of them, the compound is soluble in water.
Group 1 metal ions — Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺ — are on the list.
The ammonium ion, NH₄⁺, is on the list.
Always soluble — no common exceptions
Ion family
Ions
Group 1 metal ions
Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺
Ammonium
NH₄⁺
Nitrate
NO₃⁻
Acetate
C₂H₃O₂⁻
If a compound contains one of these ions, the compound is soluble in water.
The nitrate ion, NO₃⁻, and the acetate ion, C₂H₃O₂⁻, are on the list.
These four families have no common exceptions — every common compound containing one of these ions is soluble.
That is why every common nitrate compound is soluble, whatever the other ion is: KNO₃, Ca(NO₃)₂, and Pb(NO₃)₂ all dissolve freely.
Worked examples
Worked example 1. Sodium acetate, NaC₂H₃O₂, contains the group 1 metal ion Na⁺. Is it soluble in water?
Step 1
Answer: Yes — compounds of group 1 metal ions are soluble, with no common exceptions.
Worked example 2. Ammonium chloride, NH₄Cl, contains the ammonium ion. Is it soluble in water?
Step 1
Answer: Yes — compounds of the ammonium ion are soluble, with no common exceptions.
You can now state that ionic compounds containing group 1 metal ions, ammonium (NH₄⁺), nitrate (NO₃⁻), or acetate (C₂H₃O₂⁻) are soluble in water, with no common exceptions.
Check your understanding
Which ion guarantees that any common compound containing it is soluble in water?
ANO₃⁻correct
BSO₄²⁻
This option is wrong — you added sulfate to the no-exceptions list — the guaranteed ions are the group 1 metals, ammonium, nitrate, and acetate.
CCO₃²⁻
This option is wrong — you added carbonate to the no-exceptions list — the guaranteed ions are the group 1 metals, ammonium, nitrate, and acetate.
DCl⁻
This option is wrong — you added chloride to the no-exceptions list — chloride compounds are common, but chloride is not one of the four guaranteed families.
The always-soluble list holds exactly four families: group 1 metal ions, ammonium, nitrate, and acetate. Nitrate, NO₃⁻, is on the list. Every common nitrate compound is soluble, whatever the other ion is.
Check your understanding
Rubidium sulfate, Rb₂SO₄, contains the group 1 metal ion Rb⁺. What do the solubility rules say about it?
AIt is soluble in water.correct
BIt is insoluble in water.
This option is wrong — you overrode the group 1 guarantee — a compound containing a group 1 metal ion is soluble, with no common exceptions.
CThe rules say nothing about compounds like this one.
This option is wrong — you missed the ion that settles it — Rb⁺ is a group 1 metal ion, and the first rule covers every compound containing one.
DIt is soluble only in hot water.
This option is wrong — you attached a temperature condition the rule does not have — the group 1 guarantee holds for water at ordinary temperatures, no heating required.
Rb⁺ is a group 1 metal ion. Compounds of group 1 metal ions are soluble, with no common exceptions. So Rb₂SO₄ is soluble in water.
Check your understanding
Which list contains only ions from the always-soluble families?
AK⁺, NH₄⁺, NO₃⁻correct
BNa⁺, Ca²⁺, NO₃⁻
This option is wrong — you treated calcium as a group 1 metal — Ca²⁺ is a group 2 ion, and group 2 is not one of the always-soluble families.
CNH₄⁺, Cl⁻, C₂H₃O₂⁻
This option is wrong — you slipped chloride into the list — the guaranteed negative ions are nitrate and acetate only.
DLi⁺, Mg²⁺, C₂H₃O₂⁻
This option is wrong — you treated magnesium as a group 1 metal — Mg²⁺ is a group 2 ion, and group 2 is not one of the always-soluble families.
The always-soluble families are the group 1 metal ions, ammonium, nitrate, and acetate. K⁺ is a group 1 metal ion, NH₄⁺ is ammonium, and NO₃⁻ is nitrate. So K⁺, NH₄⁺, NO₃⁻ is the list with no outsiders.
Lesson 30 of 45 · SOL-030
Usually-soluble ions and their exceptions
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The always-soluble list settles any compound containing a group 1 metal ion, ammonium, nitrate, or acetate. The next two rules cover big anion families that are usually — not always — soluble.
The idea
Most chloride, bromide, and iodide compounds are soluble in water.
The exceptions are the chlorides, bromides, and iodides of silver and lead: AgCl, AgBr, AgI, PbCl₂, PbBr₂, and PbI₂ are insoluble.
Usually soluble — with named exceptions
Ion family
Usually
Exceptions (insoluble)
Chloride, bromide, iodide (Cl⁻, Br⁻, I⁻)
soluble
those of silver and lead
Sulfate (SO₄²⁻)
soluble
those of barium, lead, and calcium
Silver breaks the halide rule; lead breaks both.
Most sulfate compounds are soluble in water.
The exceptions are the sulfates of barium, lead, and calcium: BaSO₄, PbSO₄, and CaSO₄ are insoluble.
Two metals to watch, then: silver breaks the halide rule, and lead breaks both.
Worked examples
Worked example 1. Which two metals form insoluble chlorides, bromides, and iodides?
Step 1
Answer: Silver and lead.
Worked example 2. Calcium chloride, CaCl₂, is spread on icy roads. Is it soluble in water?
Step 1
Answer: Yes — most chlorides are soluble, and calcium is not one of the chloride exceptions. Calcium breaks only the sulfate rule.
You can now state that most chloride, bromide, and iodide compounds are soluble except those of silver and lead, and that most sulfate compounds are soluble except those of barium, lead, and calcium.
Check your understanding
Which bromide compound is insoluble in water?
AAgBrcorrect
BNaBr
This option is wrong — you marked a soluble bromide insoluble — sodium is not a bromide exception, and most bromides are soluble.
CCaBr₂
This option is wrong — you carried calcium's sulfate exception over to bromide — calcium breaks only the sulfate rule, so CaBr₂ is soluble.
DMgBr₂
This option is wrong — you marked a soluble bromide insoluble — magnesium is not a bromide exception, and most bromides are soluble.
Most chloride, bromide, and iodide compounds are soluble. The exceptions are those of silver and lead. AgBr is a silver bromide — an exception — so it is insoluble.
Check your understanding
Which sulfate compound is soluble in water?
AMgSO₄correct
BBaSO₄
This option is wrong — you marked a named exception soluble — barium is one of the three sulfate exceptions, so BaSO₄ is insoluble.
CPbSO₄
This option is wrong — you marked a named exception soluble — lead is one of the three sulfate exceptions, so PbSO₄ is insoluble.
DCaSO₄
This option is wrong — you marked a named exception soluble — calcium is one of the three sulfate exceptions, so CaSO₄ is insoluble.
Most sulfate compounds are soluble. The exceptions are the sulfates of barium, lead, and calcium. Magnesium is not on the exception list, so MgSO₄ — epsom salt — is soluble.
Check your understanding
According to this course's solubility rules, which set names the metals whose sulfates are insoluble?
ABarium, lead, calciumcorrect
BSilver, lead, barium
This option is wrong — you carried silver over from the halide exceptions — silver breaks the halide rule, not the sulfate rule.
CBarium, lead, magnesium
This option is wrong — you swapped calcium out for magnesium — magnesium sulfate is the soluble epsom salt; calcium is the third sulfate exception.
DSodium, barium, calcium
This option is wrong — you put a group 1 metal on an exception list — sodium compounds are soluble with no common exceptions; lead is the missing sulfate exception.
Most sulfates are soluble. The exceptions are the sulfates of barium, lead, and calcium. Silver belongs to the halide exceptions instead, and lead appears on both lists.
Lesson 31 of 45 · SOL-031
Usually-insoluble ions and their exceptions
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Halides and sulfates are usually soluble with short exception lists. The last four families flip the pattern: for them, insoluble is the norm.
The idea
Most carbonate, phosphate, hydroxide, and sulfide compounds are insoluble in water.
The exceptions flip the other way: when the positive ion is a group 1 metal or ammonium, the compound is soluble.
Usually insoluble — with rescue exceptions
Ion family
Usually
Exceptions (soluble)
Carbonate (CO₃²⁻)
insoluble
group 1 metal or ammonium compounds
Phosphate (PO₄³⁻)
insoluble
group 1 metal or ammonium compounds
Hydroxide (OH⁻)
insoluble
group 1 metal or ammonium compounds
Sulfide (S²⁻)
insoluble
group 1 metal or ammonium compounds
For these four families, insoluble is the norm — a group 1 metal or ammonium partner makes the compound soluble.
Na₂CO₃ is a soluble carbonate, because sodium is a group 1 metal.
Chalk — calcium carbonate, CaCO₃ — sits in this family with no rescue ion, which is why it stays a gritty cloud in water.
Worked examples
Worked example 1. Ammonium sulfide, (NH₄)₂S, contains the ammonium ion. Is it soluble in water?
Step 1
Answer: Yes — sulfides are usually insoluble, but ammonium is one of the two rescue ions.
Worked example 2. Magnesium hydroxide, Mg(OH)₂, is the solid in milk of magnesia. Is it soluble in water?
Step 1
Answer: No — most hydroxides are insoluble, and magnesium is not a group 1 metal or ammonium. That is why milk of magnesia is a cloudy suspension.
You can now state that most carbonate, phosphate, hydroxide, and sulfide compounds are insoluble except those of group 1 metals and ammonium.
Check your understanding
Which phosphate compound is soluble in water?
AK₃PO₄correct
BAlPO₄
This option is wrong — you marked a typical phosphate soluble — aluminum is not a group 1 metal or ammonium, so AlPO₄ follows the insoluble norm.
CCa₃(PO₄)₂
This option is wrong — you marked a typical phosphate soluble — calcium is not a group 1 metal or ammonium, so this phosphate follows the insoluble norm.
DMg₃(PO₄)₂
This option is wrong — you marked a typical phosphate soluble — magnesium is not a group 1 metal or ammonium, so this phosphate follows the insoluble norm.
Most phosphate compounds are insoluble. The exceptions are compounds of group 1 metals or ammonium. Potassium is a group 1 metal, so K₃PO₄ is soluble.
Check your understanding
Iron(II) sulfide, FeS, gives some mineral specimens their brassy dark color. What do the solubility rules say about FeS in water?
AIt is insoluble.correct
BIt is soluble.
This option is wrong — you missed the family norm — most sulfides are insoluble, and iron is not a group 1 metal or ammonium.
CIt is soluble only in cold water.
This option is wrong — you invented a temperature clause — the sulfide rule has no temperature condition; FeS simply follows the insoluble norm.
DThe rules do not cover sulfides.
This option is wrong — you dropped sulfide from the table — sulfide is one of the four usually-insoluble families, alongside carbonate, phosphate, and hydroxide.
Most carbonate, phosphate, hydroxide, and sulfide compounds are insoluble. Iron is not a group 1 metal or ammonium, so no exception applies. FeS is insoluble.
Check your understanding
Which hydroxide compound is soluble in water?
ALiOHcorrect
BCu(OH)₂
This option is wrong — you marked a typical hydroxide soluble — copper is not a group 1 metal or ammonium, so Cu(OH)₂ follows the insoluble norm.
CFe(OH)₃
This option is wrong — you marked a typical hydroxide soluble — iron is not a group 1 metal or ammonium, so Fe(OH)₃ follows the insoluble norm.
DAl(OH)₃
This option is wrong — you marked a typical hydroxide soluble — aluminum is not a group 1 metal or ammonium, so Al(OH)₃ follows the insoluble norm.
Most hydroxide compounds are insoluble. The exceptions are compounds of group 1 metals or ammonium. Lithium is a group 1 metal, so LiOH is soluble.
Lesson 32 of 45 · SOL-032
Use the solubility rules
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You've now seen all three rule packages: the always-soluble ions, the usually-soluble families with exceptions, and the usually-insoluble families with exceptions. Classifying any ionic compound is a short routine through them.
The idea
First, split the compound's formula into its two ions.
Check the positive ion: a group 1 metal ion or ammonium makes the compound soluble — done.
Check the negative ion: nitrate or acetate also settles it as soluble — done.
The solubility rules at a glance
Rule
Families
Exceptions
Always soluble
group 1 metal ions, NH₄⁺, NO₃⁻, C₂H₃O₂⁻
none
Usually soluble
Cl⁻, Br⁻, I⁻ · SO₄²⁻
halides of Ag and Pb insoluble · sulfates of Ba, Pb, Ca insoluble
Usually insoluble
CO₃²⁻, PO₄³⁻, OH⁻, S²⁻
group 1 or NH₄⁺ compounds soluble
Check the positive ion, then the negative ion, then the family rule and its exceptions.
Otherwise, find the negative ion's family rule — halide, sulfate, carbonate, phosphate, hydroxide, or sulfide — and start from its usual answer.
Last, check whether the positive ion is one of that rule's named exceptions, and flip the answer if it is.
Run PbSO₄ through the routine: no group 1 metal or ammonium, no nitrate or acetate; sulfates are usually soluble, but lead is a named exception — PbSO₄ is insoluble.
Worked examples
Worked example 1. Use the solubility rules to classify silver nitrate, AgNO₃, as soluble or insoluble.
Step 1
Split the formula: Ag⁺ and NO₃⁻.
Step 2
Positive ion check: silver is not a group 1 metal or ammonium.
Step 3
Negative ion check: nitrate is on the always-soluble list — done.
Step 4
AgNO₃ is soluble — the nitrate guarantee settles it before silver's halide exception ever comes up.
Worked example 2. Use the solubility rules to classify copper(II) sulfide, CuS, as soluble or insoluble.
Step 1
Split the formula: Cu²⁺ and S²⁻.
Step 2
Positive ion check: copper is not a group 1 metal or ammonium.
Step 3
Negative ion check: sulfide is not nitrate or acetate.
Step 4
Family rule: sulfides are usually insoluble.
Step 5
Exception check: the soluble sulfides are group 1 or ammonium compounds — copper is neither.
Step 6
CuS is insoluble.
You can now classify an ionic compound as soluble or insoluble in water by checking its ions against the solubility rules.
Check your understanding
Use the solubility rules: which of these compounds is insoluble in water?
APbBr₂correct
BNaBr
This option is wrong — you marked a group 1 compound insoluble — sodium guarantees solubility before any bromide question arises.
CMgBr₂
This option is wrong — you treated magnesium as a bromide exception — the halide exceptions are silver and lead, so MgBr₂ is soluble.
DNH₄Br
This option is wrong — you marked an ammonium compound insoluble — ammonium guarantees solubility with no common exceptions.
Run each formula through the routine: positive ion, negative ion, family rule, exceptions. PbBr₂ has no guaranteed ion; bromides are usually soluble, but lead is a named exception. So PbBr₂ is the insoluble one.
Check your understanding
Use the solubility rules: which of these barium compounds is soluble in water?
ABa(C₂H₃O₂)₂correct
BBaSO₄
This option is wrong — you missed the sulfate exception — barium is one of the three metals whose sulfates are insoluble.
CBaCO₃
This option is wrong — you rescued a carbonate without a rescue ion — barium is not a group 1 metal or ammonium, so BaCO₃ follows the insoluble norm.
DBa₃(PO₄)₂
This option is wrong — you rescued a phosphate without a rescue ion — barium is not a group 1 metal or ammonium, so this phosphate follows the insoluble norm.
Check the negative ion first here: acetate is on the always-soluble list. The acetate guarantee settles Ba(C₂H₃O₂)₂ as soluble, whatever the metal. The sulfate, carbonate, and phosphate partners all leave barium compounds insoluble.
Check your understanding
Use the solubility rules: which of these calcium compounds is insoluble in water?
ACaSO₄correct
BCaCl₂
This option is wrong — you carried calcium's sulfate exception over to chloride — calcium is not a halide exception, so CaCl₂ is soluble.
CCaBr₂
This option is wrong — you carried calcium's sulfate exception over to bromide — the halide exceptions are silver and lead, so CaBr₂ is soluble.
DCa(NO₃)₂
This option is wrong — you overrode the nitrate guarantee — every common nitrate compound is soluble, calcium's included.
Calcium is not a group 1 metal, so the negative ion decides each case. Nitrate guarantees solubility; chloride and bromide are usually soluble with no calcium exception. Sulfate is the family where calcium IS a named exception — CaSO₄ is insoluble.
Lesson 33 of 45 · SOL-033
Predict the precipitate
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You've already seen that mixing two solutions sometimes makes a precipitate — a solid that forms and does not dissolve — and that in a double-replacement reaction the positive ions swap partners. The solubility rules tell you, before you mix anything, whether a solid will appear.
The idea
To predict a precipitate, first write the two products of the double-replacement: the positive ions swap partners, and each new pair gets a correct formula.
Then classify each product with the solubility rules.
An insoluble product forms as a solid — that solid is the precipitate, written with (s).
A soluble product stays dissolved, written with (aq).
If both products are soluble, no solid forms — no precipitate, and the mixture stays clear.
Mix AgNO₃(aq) and NaCl(aq): swapping partners gives AgCl and NaNO₃.
AgCl is insoluble — silver is a halide exception — so a white solid, AgCl(s), appears the instant the solutions meet.
NaNO₃ contains a group 1 metal and nitrate, so it stays dissolved as NaNO₃(aq).
Swapping partners gives AgCl and NaNO₃. AgCl is insoluble, so it appears as a white solid the instant the solutions meet.
Worked examples
Worked example 1. Solutions of lead(II) nitrate, Pb(NO₃)₂, and potassium iodide, KI, are mixed. Predict whether a precipitate forms, and identify it if so.
Step 1
Swap partners: the products are PbI₂ and KNO₃.
Step 2
Classify PbI₂: iodides are usually soluble, but lead is a named exception — insoluble.
Step 3
Classify KNO₃: potassium and nitrate are both always-soluble ions — soluble.
Step 4
A precipitate forms: bright yellow PbI₂(s). The KNO₃ stays dissolved as KNO₃(aq).
Worked example 2. Solutions of sodium carbonate, Na₂CO₃, and calcium chloride, CaCl₂, are mixed. Predict whether a precipitate forms, and identify it if so.
Step 1
Swap partners: the products are CaCO₃ and NaCl.
Step 2
Classify CaCO₃: carbonates are usually insoluble, and calcium is not a rescue ion — insoluble.
Step 3
Classify NaCl: sodium is a group 1 metal — soluble.
Step 4
A precipitate forms: white CaCO₃(s). The NaCl stays dissolved as NaCl(aq).
Worked example 3. Solutions of potassium chloride, KCl, and ammonium nitrate, NH₄NO₃, are mixed. Predict whether a precipitate forms.
Step 1
Swap partners: the products are KNO₃ and NH₄Cl.
Step 2
Classify KNO₃: potassium and nitrate are both always-soluble ions — soluble.
No precipitate forms — both products stay dissolved, and the mixture stays clear.
You can now predict whether mixing two aqueous solutions forms a precipitate by writing the double-replacement products and checking each product against the solubility rules, and identify the precipitate if one forms.
Check your understanding
Solutions of barium chloride, BaCl₂, and potassium sulfate, K₂SO₄, are mixed. Which precipitate, if any, forms?
ABaSO₄correct
BKCl
This option is wrong — you picked the soluble product — potassium guarantees KCl stays dissolved; the solid is the barium sulfate, a named sulfate exception.
CBoth BaSO₄ and KCl
This option is wrong — you sent both products to the bottom — KCl contains a group 1 metal and stays dissolved; only BaSO₄ is insoluble.
DNo precipitate forms
This option is wrong — you missed the sulfate exception — barium is one of the three metals whose sulfates are insoluble, so a solid does form.
Swap partners: the products are BaSO₄ and KCl. BaSO₄ hits the barium sulfate exception — insoluble; KCl is a group 1 compound — soluble. So white BaSO₄(s) is the precipitate, and KCl stays as KCl(aq).
Check your understanding
Solutions of potassium carbonate, K₂CO₃, and magnesium chloride, MgCl₂, are mixed and a white solid appears. Write the formula of the precipitate.
Accepted answer: MgCO₃
Swap partners: the products are MgCO₃ and KCl. Classify each: carbonates are usually insoluble and magnesium is not a rescue ion; KCl is a group 1 compound — soluble. So the white solid is MgCO₃(s).
Check your understanding
Solutions of sodium hydroxide, NaOH, and copper(II) chloride, CuCl₂, are mixed. Which precipitate, if any, forms?
ACu(OH)₂correct
BNaCl
This option is wrong — you picked the soluble product — sodium guarantees NaCl stays dissolved; the solid is the copper(II) hydroxide.
CBoth Cu(OH)₂ and NaCl
This option is wrong — you sent both products to the bottom — NaCl contains a group 1 metal and stays dissolved; only Cu(OH)₂ is insoluble.
DNo precipitate forms
This option is wrong — you rescued a hydroxide without a rescue ion — copper is not a group 1 metal or ammonium, so Cu(OH)₂ forms as a solid.
Swap partners: the products are Cu(OH)₂ and NaCl. Hydroxides are usually insoluble, and copper is not a group 1 metal or ammonium — insoluble. So blue Cu(OH)₂(s) is the precipitate, and NaCl stays as NaCl(aq).
Two pitchers of lemonade look identical. One sip puckers your mouth; the other tastes nearly like plain water. Same ingredients, same amount of drink — so what is actually different inside the two pitchers?
You've already seen that a solution is solute spread evenly through solvent. The difference between the pitchers is how MUCH solute is spread through the same amount of solution.
The idea
Compare two solutions by asking how much solute each holds in the same amount of solution.
The solution holding more solute in the same amount of solution is the stronger one — chemists call it the more 'concentrated' solution.
The solution holding less solute in the same amount of solution is the weaker one — the more 'dilute' solution.
Concentrated and dilute are comparison words: a solution is only more or less concentrated THAN another, and neither word names an exact amount.
A particle diagram shows the comparison directly: with equal volumes drawn, the panel with more solute particles shows the more concentrated solution.
Equal volumes of solution: the panel with more solute particles shows the more concentrated solution.
Worked examples
Worked example 1. Beaker A holds 100 mL of solution made with 2 g of salt. Beaker B holds 100 mL of solution made with 10 g of salt. Which solution is more concentrated?
Step 1
The amounts of solution are the same: 100 mL each.
Step 2
Beaker B holds more solute in that same amount of solution.
Step 3
Beaker B's solution is more concentrated; beaker A's is more dilute.
Worked example 2. Jug 1 holds 500 mL of fruit punch made with 50 mL of punch concentrate. Jug 2 holds 500 mL made with 150 mL of the same concentrate. Which jug holds the more dilute drink?
Step 1
The amounts of solution are the same: 500 mL each.
Step 2
Jug 1 holds less solute in that same amount of solution.
Step 3
Jug 1 holds the more dilute drink; jug 2's is more concentrated.
You can now classify one solution as more concentrated or more dilute than another by comparing how much solute is present in the same amount of solution.
Check your understanding
Four solutions are drawn below with equal volumes. Which particle diagram shows the most concentrated solution?
Acorrect
B
This option is wrong — you picked the panel with the fewest solute particles — with equal volumes, the MOST particles marks the most concentrated solution.
C
This option is wrong — you stopped at a middling count — another panel holds more solute particles in the same volume.
D
This option is wrong — you stopped one short — count again: another panel holds more solute particles in the same volume.
All four panels show the same volume of solution. With equal volumes, the panel with more solute particles shows the more concentrated solution. The 14-particle panel holds the most solute in the same volume, so it is the most concentrated.
Check your understanding
The figure shows particle diagrams of two solutions, P and Q. Which solution is more concentrated?
ASolution Qcorrect
BSolution P
This option is wrong — you picked the panel with fewer solute particles — with equal volumes drawn, more particles means more concentrated.
CThe two solutions are equally concentrated
This option is wrong — you treated the panels as identical — count the solute dots: Q holds more in the same volume.
DThe diagrams cannot settle the comparison
This option is wrong — you gave up on readable diagrams — equal volumes are drawn, so the particle counts settle the comparison directly.
Both panels show the same volume of solution. Solution Q's panel holds more solute particles in that same volume. So solution Q is more concentrated, and P is more dilute.
Check your understanding
Bottle 1 holds 300 mL of sports drink made with 15 g of drink powder. Bottle 2 holds 300 mL made with 33 g of the same powder. Which drink is more dilute?
ABottle 1's drinkcorrect
BBottle 2's drink
This option is wrong — you flipped the comparison words — bottle 2 holds MORE powder in the same amount of drink, which makes it the more concentrated one.
CNeither — the drinks are equally strong because the volumes match
This option is wrong — you compared only the volumes — equal volumes holding different amounts of solute are not equally concentrated.
DNeither — dilute only describes solutions made with water alone
This option is wrong — you restricted dilute to plain water — any solution can be the more dilute of a pair, whatever its solvent.
The amounts of solution are the same: 300 mL each. Bottle 1 holds less solute in that same amount of solution. Less solute in the same amount of solution means more dilute.
Lesson 35 of 45 · SOL-035
More solute, stronger solution
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Did You Know?
You've already seen that the more concentrated of two solutions holds more solute in the same amount of solution. Now change a single solution: what happens when you add MORE solute to it?
The idea
Stir a second spoonful of drink mix into the same pitcher of water, and the drink turns stronger.
The amount of solution has barely changed, but more solute is now dissolved in it.
More solute in the same amount of solution is exactly what more concentrated means.
Same amount of solution, more dissolved solute — the solution is now more concentrated.
So adding more solute to a fixed volume of solution makes the solution more concentrated — as long as the added solute all dissolves.
Worked examples
Worked example 1. A chef stirs an extra pinch of salt into a bowl of broth, and it dissolves completely. What happens to the broth's concentration?
Step 1
The amount of broth is unchanged.
Step 2
More solute is now dissolved in that same amount of solution.
Step 3
The broth becomes more concentrated.
Worked example 2. A student adds 5.0 g more copper(II) sulfate to a beaker of copper(II) sulfate solution, and it all dissolves. The volume stays essentially the same. What happens to the solution's concentration, and what would you see?
Step 1
The amount of solution is essentially unchanged.
Step 2
More solute is now dissolved in that same amount of solution, so the solution is more concentrated.
Step 3
The solution becomes more concentrated — its blue color deepens.
You can now predict that adding more solute to a fixed volume of solution makes the solution more concentrated.
Check your understanding
An aquarist stirs an extra measure of aquarium salt into a tank, and it dissolves completely. The water level is unchanged. What happens to the solution's concentration?
AIt increases.correct
BIt decreases.
This option is wrong — you ran the change backwards — more solute dissolved in the same amount of solution means MORE concentrated, not less.
CIt stays the same.
This option is wrong — you let the unchanged water level freeze the concentration — the solution amount is the same, but the solute in it grew, so the concentration rose.
DIt cannot change unless water is added or removed.
This option is wrong — you tied concentration to the water alone — concentration compares solute to solution, and the solute side just changed.
The amount of solution is unchanged. More solute is now dissolved in that same amount of solution. More solute in the same amount of solution means the solution is more concentrated.
Check your understanding
A camper stirs a second scoop of lemonade powder into the same jug of lemonade, and it dissolves fully. Compared with before, the lemonade is now…
Amore concentratedcorrect
Bmore dilute
This option is wrong — you ran the change backwards — the jug now holds more solute in the same amount of drink, which is what more concentrated means.
Cequally concentrated
This option is wrong — you froze the concentration because the jug looks the same — the drink's volume barely changed, but the dissolved solute doubled.
Dno longer a solution
This option is wrong — you retired the word solution too early — the powder dissolved fully, so the drink is still a solution, just a stronger one.
The amount of lemonade is essentially unchanged. More solute is now dissolved in that same amount of solution. So the lemonade is more concentrated — a stronger-tasting drink.
Check your understanding
A gardener's hummingbird feeder holds sugar water. She empties it, mixes a fresh batch in the same bottle with the same amount of water but an extra spoonful of sugar, and the sugar dissolves fully. Why is the new batch more concentrated?
AIt holds more dissolved solute in the same amount of solution.correct
BIt was made more recently, and fresh solutions are stronger.
This option is wrong — you made age the cause — a solution's strength comes from its solute-to-solution comparison, not from when it was mixed.
CThe extra sugar makes the batch heavier, and heavier solutions are more concentrated.
This option is wrong — you reached the right direction through weight — concentration compares dissolved solute to the amount of solution, not the bottle's total weight.
DThe sugar settles at the bottom, where it strengthens the solution.
This option is wrong — you left the sugar undissolved — the sugar dissolved fully and spread evenly, and the strength rises because more solute is in the same amount of solution.
Same amount of water, so essentially the same amount of solution. More solute is dissolved in it. More solute in the same amount of solution is exactly what more concentrated means.
Lesson 36 of 45 · SOL-036
More volume, weaker solution
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Did You Know?
You've already seen that adding more solute to a fixed volume makes a solution more concentrated. The volume can change too — and it works the other way.
The idea
Keep the amount of solute fixed, and give it a larger volume of solution to occupy.
Spreading a fixed amount of solute through a larger volume of solution makes the solution less concentrated — more dilute.
The same number of solute particles now share more space, so each liter of solution holds fewer of them.
One spoonful of drink mix stirred into a small pitcher of water makes a strong-tasting drink.
The same amount of solute in a larger volume of solution: fewer solute particles in each liter — more dilute.
The same one spoonful stirred into a larger pitcher of water makes a weaker-tasting drink — the drink is more dilute.
This mirrors the last lesson: more solute in the same volume strengthens a solution, and more volume for the same solute weakens it.
Worked examples
Worked example 1. Two sugar cubes are dissolved to make a 250 mL bottle of iced tea. Two identical sugar cubes are dissolved to make a 1 L bottle of iced tea. Which bottle holds the more dilute tea?
Step 1
The amount of sugar is the same in both bottles.
Step 2
The 1 L bottle spreads that sugar through a larger volume, so it is less concentrated.
Step 3
The 1 L bottle holds the more dilute tea.
Worked example 2. The same 30 g of aquarium salt is dissolved in a 20 L tank and in a 60 L tank. Which tank's water is more dilute?
Step 1
The amount of salt is fixed at 30 g in each tank.
Step 2
The 60 L tank spreads that salt through a larger volume.
Step 3
The 60 L tank's water is more dilute.
You can now predict that spreading a fixed amount of solute through a larger volume of solution makes the solution less concentrated.
Check your understanding
One packet of powdered lemonade is dissolved to make 500 mL of drink. An identical packet is dissolved to make 2 L of drink. Which statement compares the two drinks correctly?
AThe 2 L drink is more dilute.correct
BThe 500 mL drink is more dilute.
This option is wrong — you flipped the direction — the larger volume spreads the same powder thinner, so the 2 L drink is the more dilute one.
CThe two drinks are equally concentrated, because each holds one packet.
This option is wrong — you judged concentration by the amount of solute alone — concentration compares the solute with the volume it occupies.
DThe drinks cannot be compared without knowing the mass of powder in a packet.
This option is wrong — you asked for information the comparison does not need — the packets are identical, so only the volumes differ.
The amount of powder is the same in both drinks. Spreading a fixed amount of solute through a larger volume of solution makes the solution less concentrated. So the 2 L drink is more dilute.
Check your understanding
A chemist dissolves 0.10 mol of copper(II) sulfate to make 100 mL of solution, and another 0.10 mol to make 400 mL of solution. Which solution is less concentrated?
AThe 400 mL solution.correct
BThe 100 mL solution.
This option is wrong — you flipped the direction — the same solute in a larger volume is the weaker solution, and 400 mL is the larger volume.
CNeither — the two solutions are equally concentrated.
This option is wrong — you compared only the amounts of solute — the amounts match, but the volumes they occupy do not.
DNeither — a solution's concentration is a fixed property of copper(II) sulfate.
This option is wrong — you treated concentration as belonging to the substance — concentration describes a particular solution, and it changes with the volume.
Both solutions hold 0.10 mol of copper(II) sulfate. Spreading a fixed amount of solute through a larger volume of solution makes the solution less concentrated. The 400 mL solution spreads the same solute through four times the volume, so it is less concentrated.
Check your understanding
The same three drops of blue food coloring are stirred into glass A, holding 100 mL of water, and into glass B, holding 500 mL of water. Which glass holds the more dilute mixture?
AGlass B.correct
BGlass A.
This option is wrong — you flipped the direction — the same dye in the larger volume is spread thinner, and that is glass B.
CNeither — both glasses received the same three drops.
This option is wrong — you judged by the amount of dye alone — the dye amounts match, but glass B spreads its dye through five times the volume.
DNeither — food coloring keeps the same strength in any amount of water.
This option is wrong — you treated the dye's concentration as fixed — spreading the same dye through more water leaves fewer dye particles in each milliliter.
Both glasses hold the same amount of dye. Spreading a fixed amount of solute through a larger volume of solution makes the solution less concentrated. Glass B has the larger volume, so its mixture is more dilute — and it looks paler for exactly that reason.
Lesson 37 of 45 · SOL-037
Molarity
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Did You Know?
Wonder this:
Two bottles of salt water sit on a shelf, one labeled 'concentrated' and one labeled 'dilute'. A procedure asks for exactly 0.10 mol of salt — which bottle, and how much of it? The labels cannot say. Chemists need concentration as a number.
You've already compared concentrations qualitatively — more solute or more volume, stronger or weaker. Here is the number chemists put on it.
The equation
Chemists measure concentration by counting the moles of solute dissolved in each liter of solution.
This measure is called 'molarity', written with the symbol M.
M = n ÷ V
n is the moles of solute, and V is the volume of the solution in liters.
M
molarity (mol/L)
n
moles of solute (mol)
V
volume of solution (L)
This M is the molarity in mol/L — not the molar mass from Unit 7. When a problem needs both, this course writes 'molar mass' out in words.
Dissolve 1.0 mol of NaCl in enough water to make 1.0 L of solution, and the result is a 1.0 M solution.
Read '1.0 M' aloud as '1.0 molar'.
V counts the volume of the finished solution, not the volume of water you poured in.
A higher molarity means a more concentrated solution, and a lower molarity means a more dilute one.
Worked examples
Worked example 1. A bottle is labeled 3.0 M KBr. What does the label say about each liter of the solution?
Step 1
Answer: each liter of the solution contains 3.0 mol of dissolved KBr.
Worked example 2. In the equation M = n ÷ V, what does each symbol stand for, and what unit does it carry?
Step 1
Write down the values in the question
M is the molarity, in moles per liter (mol/L).
n is the moles of solute, in mol.
V is the volume of the solution, in liters.
Step 2
Write down the equation
M = n ÷ V
Step 3
Substitute in the values, and calculate
You can now state that molarity (symbol M) measures concentration as moles of solute per liter of solution, M = n ÷ V, where n is the moles of solute and V is the volume of solution in liters.
Check your understanding
A label reads 2.0 M NaNO₃. What does the label mean?
AEach liter of the solution contains 2.0 mol of dissolved NaNO₃.correct
BEach liter of water added contains 2.0 mol of dissolved NaNO₃.
This option is wrong — you measured against the water poured in — molarity counts moles per liter of finished SOLUTION.
CThe bottle contains 2.0 mol of NaNO₃ in total, whatever its size.
This option is wrong — you read the label as a total amount — molarity is an amount PER LITER, so the total depends on the bottle's volume.
DEach liter of the solution contains 2.0 g of dissolved NaNO₃.
This option is wrong — you swapped grams in for moles — the n in molarity is a mole count, not a mass.
Molarity counts the moles of solute dissolved in each liter of solution. 2.0 M means 2.0 mol of NaNO₃ in every liter of the solution. The count is per liter of solution — not per liter of added water, not grams, and not a bottle total.
Check your understanding
In the equation M = n ÷ V, what does V stand for?
AThe volume of the solution, in liters.correct
BThe volume of water added, in liters.
This option is wrong — you counted the water instead of the solution — V is the volume of the finished solution, which includes the dissolved solute.
CThe volume of the solute before it dissolved.
This option is wrong — you attached V to the solute — the solute contributes moles through n, and V measures the whole solution.
DThe volume of the solution, in milliliters.
This option is wrong — you kept the right quantity but the wrong unit — molarity is moles per LITER, so V must be in liters.
In M = n ÷ V, V is the volume of the finished solution, measured in liters. n is the moles of solute, and M comes out in moles per liter.
Check your understanding
Which preparation produces a 1.0 M solution of KCl?
ADissolving 1.0 mol of KCl in enough water to make 1.0 L of solution.correct
BAdding 1.0 mol of KCl to 1.0 L of water.
This option is wrong — you measured the water instead of the solution — dissolved KCl adds volume, so this mixture ends up holding slightly MORE than 1.0 L and reads slightly below 1.0 M.
CDissolving 1.0 g of KCl in enough water to make 1.0 L of solution.
This option is wrong — you swapped grams in for moles — molarity counts moles of solute, and 1.0 g of KCl is far less than 1.0 mol.
DDissolving 1.0 mol of KCl in enough water to make 1.0 mL of solution.
This option is wrong — you used the wrong volume unit — molarity is moles per LITER, and 1.0 mol in a single milliliter would be enormously concentrated.
Molarity is moles of solute per liter of SOLUTION. 1.0 mol of KCl made up to a total solution volume of 1.0 L gives 1.0 M. Water-added, grams, and milliliters each break one piece of the definition.
Lesson 38 of 45 · SOL-038
Calculate molarity
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Did You Know?
You've already seen that molarity counts the moles of solute in each liter of solution: M = n ÷ V. Now run the equation on real numbers.
The equation
To calculate a molarity, divide the moles of solute by the volume of the solution in liters.
M = n ÷ V
The moles n carry mol, the volume V carries liters, and the answer M carries moles per liter — written mol/L, or M on labels.
If the volume is given in milliliters, divide it by 1000 to get liters before substituting.
Sanity check: more moles in the same volume must give a bigger M, and the same moles in more volume must give a smaller M.
Worked examples
Worked example 1. What is the molarity of a solution made by dissolving 0.50 mol of KCl in enough water to make 2.0 L of solution?
Step 1
Write down the values in the question
n = 0.50 mol
V = 2.0 L
Step 2
Write down the equation
M = n ÷ V
Step 3
Substitute in the values, and calculate
M = 0.50 ÷ 2.0
M = 0.25 M
Worked example 2. What is the molarity of a solution containing 0.30 mol of sucrose in 600.0 mL of solution?
Step 1
Write down the values in the question
n = 0.30 mol
V = 600.0 mL = 0.6000 L
Step 2
Write down the equation
M = n ÷ V
Step 3
Substitute in the values, and calculate
M = 0.30 ÷ 0.6000
M = 0.50 M
You can now calculate the molarity of a solution from the moles of solute and the volume of solution in liters, using M = n ÷ V.
Check your understanding
A solution contains 1.2 mol of NaNO₃ in 3.0 L of solution. What is its molarity, in mol/L?
Answer: 0.40mol/L(tolerance ±0.005)
Write down the values in the question: n = 1.2 mol V = 3.0 L Write down the equation: M = n ÷ V Substitute in the values, and calculate: M = 1.2 ÷ 3.0 M = 0.40 M
Check your understanding
A solution contains 0.60 mol of CaCl₂ in 0.80 L of solution. What is its molarity, in mol/L?
Answer: 0.75mol/L(tolerance ±0.005)
Write down the values in the question: n = 0.60 mol V = 0.80 L Write down the equation: M = n ÷ V Substitute in the values, and calculate: M = 0.60 ÷ 0.80 M = 0.75 M
Check your understanding
A flask holds 0.15 mol of glucose dissolved in 250.0 mL of solution. What is the solution's molarity, in mol/L?
Answer: 0.60mol/L(tolerance ±0.005)
Write down the values in the question: n = 0.15 mol V = 250.0 mL = 0.2500 L Write down the equation: M = n ÷ V Substitute in the values, and calculate: M = 0.15 ÷ 0.2500 M = 0.60 M
Lesson 39 of 45 · SOL-039
Moles from molarity
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Did You Know?
You've already calculated a molarity from moles and volume. Labels usually run the other way: the bottle states its molarity, and you need the moles in the volume you pour.
The equation
The label gives M, you measure out V, and the unknown is the moles n.
Start from the one molarity equation: M = n ÷ V.
Make n the subject by multiplying both sides by V, giving n = M × V.
It is the same equation rearranged, not a new equation to memorize.
V must be in liters — divide milliliters by 1000 first.
Sanity check: pouring less than a liter of a 1.0 M solution must give less than 1.0 mol.
Worked examples
Worked example 1. How many moles of NaCl are in 0.50 L of a 0.20 M NaCl solution?
Step 1
Write down the values in the question
M = 0.20 M
V = 0.50 L
Step 2
Write down the equation
M = n ÷ V
Step 3
Make the unknown the subject
n = M × V
Step 4
Substitute in the values, and calculate
n = 0.20 × 0.50
n = 0.10 mol
Worked example 2. How many moles of KNO₃ are in 250.0 mL of a 0.60 M KNO₃ solution?
Step 1
Write down the values in the question
M = 0.60 M
V = 250.0 mL = 0.2500 L
Step 2
Write down the equation
M = n ÷ V
Step 3
Make the unknown the subject
n = M × V
Step 4
Substitute in the values, and calculate
n = 0.60 × 0.2500
n = 0.15 mol
You can now calculate the moles of solute in a given volume of solution by rearranging M = n ÷ V to n = M × V.
Check your understanding
How many moles of NaBr are in 3.0 L of a 0.25 M NaBr solution?
Answer: 0.75mol(tolerance ±0.005)
Write down the values in the question: M = 0.25 M V = 3.0 L Write down the equation: M = n ÷ V Make n the subject: n = M × V Substitute in the values, and calculate: n = 0.25 × 3.0 n = 0.75 mol
Check your understanding
How many moles of CaCl₂ are in 0.40 L of a 0.50 M CaCl₂ solution?
Answer: 0.20mol(tolerance ±0.005)
Write down the values in the question: M = 0.50 M V = 0.40 L Write down the equation: M = n ÷ V Make n the subject: n = M × V Substitute in the values, and calculate: n = 0.50 × 0.40 n = 0.20 mol
Check your understanding
How many moles of glucose are in 500.0 mL of a 0.30 M glucose solution?
Answer: 0.15mol(tolerance ±0.005)
Write down the values in the question: M = 0.30 M V = 500.0 mL = 0.5000 L Write down the equation: M = n ÷ V Make n the subject: n = M × V Substitute in the values, and calculate: n = 0.30 × 0.5000 n = 0.15 mol
Lesson 40 of 45 · SOL-040
Molarity from grams
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Did You Know?
You can already turn grams into moles with the molar mass, and moles plus volume into molarity. Real bottles are made up from weighed grams, so the two routines chain together.
The equation
A balance reads grams, but molarity counts moles — so the mass must become moles first.
Step 1: moles of solute = mass ÷ molar mass.
Step 2: molarity = moles ÷ volume of solution in liters — the same M = n ÷ V as before.
Work the two steps in order, writing the intermediate moles on its own line.
The mass is in grams, the molar mass in g/mol, the volume in liters, and the answer in M.
Take 4.0 g of NaOH (molar mass 40.0 g/mol) dissolved in enough water to make 0.50 L of solution.
Step 1 gives n = 4.0 ÷ 40.0 = 0.10 mol.
Step 2 gives M = 0.10 ÷ 0.50 = 0.20 M.
Sanity check: 4.0 g is a tenth of NaOH's 40.0 g/mol, so the moles must come out at a tenth of a mole.
Worked examples
Worked example 1. 17.0 g of NaNO₃ (molar mass 85.0 g/mol) is dissolved in enough water to make 2.00 L of solution. What is the molarity?
Step 1
Write down the values in the question
m = 17.0 g
molar mass = 85.0 g/mol
V = 2.00 L
Find the moles of solute first:
n = m ÷ molar mass
n = 17.0 ÷ 85.0
n = 0.200 mol
Step 2
Write down the equation
M = n ÷ V
Step 3
Substitute in the values, and calculate
M = 0.200 ÷ 2.00
M = 0.100 M
Worked example 2. 23.8 g of KBr (molar mass 119.0 g/mol) is dissolved in enough water to make 500.0 mL of solution. What is the molarity?
Step 1
Write down the values in the question
m = 23.8 g
molar mass = 119.0 g/mol
V = 500.0 mL = 0.5000 L
Find the moles of solute first:
n = m ÷ molar mass
n = 23.8 ÷ 119.0
n = 0.200 mol
Step 2
Write down the equation
M = n ÷ V
Step 3
Substitute in the values, and calculate
M = 0.200 ÷ 0.5000
M = 0.400 M
You can now calculate the molarity of a solution from the mass of solute in grams by first converting the mass to moles using the molar mass, then dividing by the volume of solution in liters.
Check your understanding
10.6 g of Na₂CO₃ (molar mass 106.0 g/mol) is dissolved in enough water to make 0.500 L of solution. What is the solution's molarity, in mol/L?
Answer: 0.200mol/L(tolerance ±0.002)
Write down the values in the question: m = 10.6 g molar mass = 106.0 g/mol V = 0.500 L Find the moles of solute first: n = m ÷ molar mass n = 10.6 ÷ 106.0 n = 0.100 mol Write down the equation: M = n ÷ V Substitute in the values, and calculate: M = 0.100 ÷ 0.500 M = 0.200 M
Check your understanding
90.0 g of glucose (molar mass 180.0 g/mol) is dissolved in enough water to make 2.00 L of solution. What is the solution's molarity, in mol/L?
Answer: 0.250mol/L(tolerance ±0.002)
Write down the values in the question: m = 90.0 g molar mass = 180.0 g/mol V = 2.00 L Find the moles of solute first: n = m ÷ molar mass n = 90.0 ÷ 180.0 n = 0.500 mol Write down the equation: M = n ÷ V Substitute in the values, and calculate: M = 0.500 ÷ 2.00 M = 0.250 M
Check your understanding
68.46 g of sucrose (molar mass 342.3 g/mol) is dissolved in enough water to make 400.0 mL of solution. What is the solution's molarity, in mol/L?
Answer: 0.500mol/L(tolerance ±0.002)
Write down the values in the question: m = 68.46 g molar mass = 342.3 g/mol V = 400.0 mL = 0.4000 L Find the moles of solute first: n = m ÷ molar mass n = 68.46 ÷ 342.3 n = 0.2000 mol Write down the equation: M = n ÷ V Substitute in the values, and calculate: M = 0.2000 ÷ 0.4000 M = 0.500 M
Lesson 41 of 45 · SOL-041
Why diluting weakens a solution
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Have You Ever Wondered?
Wonder this:
Lemonade mixed too strong gets fixed with more water — and it works: the drink tastes weaker. But nothing left the glass. Where did the strength go?
You've already predicted that a fixed amount of solute in a larger volume makes a more dilute solution. This lesson names that move and pins down what changes — and what does not.
The idea
Adding more solvent to a solution is called 'diluting' it.
Diluting lowers the concentration, because the same amount of solute is spread through a larger volume.
The amount of solute does not change — every solute particle that was there before is still there.
Diluting: the same amount of solute is spread through a larger volume — fewer solute particles in each liter.
Only the volume grows, so each liter of solution now holds fewer moles of solute.
That is the lemonade answer: the same sugar and lemon are spread through a larger volume, so every sip carries less of them — yet the glass holds all the lemonade it ever did.
Concentration and amount are different questions: diluting changes how crowded the solute is, not how much of it there is.
Worked examples
Worked example 1. A beaker holds 0.20 mol of dissolved CuSO₄. Water is added, doubling the solution's volume. What happens to (a) the amount of dissolved CuSO₄ and (b) the concentration?
Step 1
No CuSO₄ was added or removed, so the amount is unchanged.
Step 2
The concentration falls, because the same amount of solute is spread through a larger volume.
Step 3
(a) still 0.20 mol; (b) the concentration is lower.
Worked example 2. Why does a pot of soup taste less salty after water is added, even though all the salt is still in the pot?
Step 1
The amount of salt is unchanged — adding water removes nothing.
Step 2
The taste weakens because the same amount of solute is spread through a larger volume.
Step 3
The salt is spread through a larger volume, so each spoonful carries less of it.
You can now explain why adding solvent to a solution — called diluting it — lowers the concentration but leaves the amount of solute unchanged, because the same amount of solute is spread through a larger volume.
Check your understanding
Water is added to a glass of sweet tea. Which pair of statements is correct?
AThe concentration decreases, and the amount of dissolved sugar stays the same.correct
BThe concentration decreases, and the amount of dissolved sugar decreases.
This option is wrong — you let the sugar disappear — diluting adds only solvent, so every sugar particle is still in the glass.
CThe concentration stays the same, and the amount of dissolved sugar stays the same.
This option is wrong — you froze the concentration along with the amount — the same sugar in a larger volume means fewer sugar particles per liter.
DThe concentration increases, and the amount of dissolved sugar stays the same.
This option is wrong — you ran the concentration the wrong way — spreading the same sugar through more liquid weakens the tea.
Diluting adds only solvent, so the amount of sugar is unchanged. The concentration falls, because the same amount of solute is spread through a larger volume. Amount unchanged, concentration down — that pair is the signature of dilution.
Check your understanding
Why does diluting a solution lower its concentration?
ABecause the same amount of solute is spread through a larger volume.correct
BBecause the added solvent destroys part of the dissolved solute.
This option is wrong — you removed solute that never leaves — diluting changes the volume, not the amount of solute.
CBecause the added solvent pushes some solute out of solution as a solid.
This option is wrong — you invented a crystallization — adding more solvent gives the solute MORE room to stay dissolved, not less.
DBecause the solute particles shrink when more solvent surrounds them.
This option is wrong — you changed the particles instead of their crowding — the particles are untouched; there are simply fewer of them in each liter.
Diluting lowers the concentration, because the same amount of solute is spread through a larger volume. No solute is destroyed, expelled, or changed. Only the crowding changes: the same particles, more space.
Check your understanding
A 100 mL sample of yellow food-coloring solution is poured into a jug, and water is added until the jug holds 1 L. How does the number of dye particles in the jug compare with the number that was in the 100 mL sample?
AThe jug holds exactly the same number of dye particles.correct
BThe jug holds one-tenth as many dye particles.
This option is wrong — you read the paler color as lost dye — the color fades because the same particles are spread through ten times the volume, not because any left.
CThe jug holds ten times as many dye particles.
This option is wrong — you let the added water bring dye with it — pure water adds solvent only.
DThe comparison cannot be made without weighing the dye.
This option is wrong — you asked for information the comparison does not need — no dye entered or left, so the count is unchanged.
Adding water adds solvent only — every dye particle from the sample is still in the jug. The mixture looks paler because the same amount of solute is spread through a larger volume. Concentration changed; amount did not.
Lesson 42 of 45 · SOL-042
The dilution equation
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Did You Know?
You've already seen that diluting lowers the concentration but leaves the moles of solute unchanged, and that n = M × V counts the moles in any solution. Put the two facts together and an equation falls out.
The equation
Count the moles of solute before dilution with n = M × V: they come to M₁ × V₁.
Count the moles after dilution the same way: they come to M₂ × V₂.
Diluting adds only solvent, so the moles of solute after dilution equal the moles before.
Setting the two counts equal gives the 'dilution equation': M₁V₁ = M₂V₂.
The subscript 1 marks the solution before dilution, and the subscript 2 marks the solution after.
M₁
molarity before dilution (mol/L)
V₁
volume before dilution (L)
M₂
molarity after dilution (mol/L)
V₂
total volume after dilution (L)
V₂ is the total volume of the diluted solution, not the volume of solvent added.
After a dilution M₂ is smaller and V₂ is larger, because the same amount of solute is spread through a larger volume.
Worked examples
Worked example 1. A KNO₃ solution is diluted. In the dilution equation, what does the product M₂V₂ count?
Step 1
Answer: the moles of KNO₃ dissolved after the dilution — the same moles the solution held before.
Worked example 2. 50.0 mL of a CuSO₄ solution is diluted to a total volume of 400.0 mL. Which symbols do the two volumes match in M₁V₁ = M₂V₂?
Step 1
Write down the values in the question
50.0 mL is the volume before dilution, so it is V₁.
400.0 mL is the total volume after dilution, so it is V₂.
Step 2
Write down the equation
M₁V₁ = M₂V₂
Step 3
Substitute in the values, and calculate
V₁ = 50.0 mL and V₂ = 400.0 mL.
You can now state the dilution equation M₁V₁ = M₂V₂, which works because the moles of solute before and after dilution are equal, and n = M × V counts those moles on each side.
Check your understanding
Write the dilution equation, using subscript 1 for the solution before dilution and subscript 2 for the solution after.
Accepted answer: M₁V₁ = M₂V₂
Each side counts the moles of solute with n = M × V. Before: M₁V₁. After: M₂V₂. Diluting adds only solvent, so the counts are equal: M₁V₁ = M₂V₂.
Check your understanding
In the dilution equation M₁V₁ = M₂V₂, what does each side of the equation count?
AThe moles of solute — the same before and after the dilution.correct
BThe concentration of the solution before and after.
This option is wrong — you read each side as a bare molarity — multiplying by the volume turns the molarity into a mole count, and it is the MOLES that match.
CThe volume of solvent added during the dilution.
This option is wrong — you counted the added solvent — the equation never mentions it; each side is molarity times volume, which counts dissolved moles.
DThe mass of the solution before and after.
This option is wrong — you swapped mass in for moles — M × V carries mol/L times L, which is mol.
n = M × V counts the moles of solute in any solution. M₁V₁ counts them before dilution, and M₂V₂ counts them after. Diluting adds only solvent, so the two counts are equal.
Check your understanding
Why are the two sides of M₁V₁ = M₂V₂ equal?
ABecause the moles of solute stay the same, and M × V counts those moles on each side.correct
BBecause diluting leaves the concentration unchanged on both sides.
This option is wrong — you conserved the wrong quantity — dilution LOWERS the concentration; it is the moles of solute that stay fixed.
CBecause the volume of the solution stays the same during a dilution.
This option is wrong — you froze the volume — dilution grows the volume from V₁ to a larger V₂; that growth is the whole point.
DBecause a solution always contains equal amounts of solute and solvent.
This option is wrong — you invented a solute-solvent balance — solutions hold far more solvent than solute, and the equation compares before with after, not solute with solvent.
Diluting adds only solvent, so the moles of solute after dilution equal the moles before. n = M × V counts those moles: M₁V₁ before, M₂V₂ after. Equal counts give the equation M₁V₁ = M₂V₂.
Lesson 43 of 45 · SOL-043
New concentration after dilution
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Did You Know?
You've already seen the dilution equation M₁V₁ = M₂V₂ and why its two sides are equal. When a solution is diluted, the equation predicts the new concentration.
The equation
List what the problem gives: the starting molarity M₁, the starting volume V₁, and the total volume after dilution V₂.
Write down the dilution equation: M₁V₁ = M₂V₂.
Make M₂ the subject by dividing both sides by V₂, giving M₂ = M₁V₁ ÷ V₂.
The two volumes must be in the same unit — both in mL or both in L — because they divide and the unit cancels.
V₂ is the total volume of the diluted solution, not the volume of water added.
Sanity check: a diluted solution is weaker, so M₂ must come out smaller than M₁.
Worked examples
Worked example 1. 50.0 mL of a 6.0 M NaCl solution is diluted with water to a total volume of 300.0 mL. What is the molarity of the diluted solution?
Step 1
Write down the values in the question
M₁ = 6.0 M
V₁ = 50.0 mL
V₂ = 300.0 mL
Step 2
Write down the equation
M₁V₁ = M₂V₂
Step 3
Make the unknown the subject
M₂ = M₁V₁ ÷ V₂
Step 4
Substitute in the values, and calculate
M₂ = (6.0 × 50.0) ÷ 300.0
M₂ = 1.0 M
Worked example 2. 0.200 L of a 2.4 M glucose solution is diluted to a total volume of 1.20 L. What is the molarity of the diluted solution?
Step 1
Write down the values in the question
M₁ = 2.4 M
V₁ = 0.200 L
V₂ = 1.20 L
Step 2
Write down the equation
M₁V₁ = M₂V₂
Step 3
Make the unknown the subject
M₂ = M₁V₁ ÷ V₂
Step 4
Substitute in the values, and calculate
M₂ = (2.4 × 0.200) ÷ 1.20
M₂ = 0.40 M
You can now calculate the concentration of a diluted solution by rearranging M₁V₁ = M₂V₂ to M₂ = M₁V₁ ÷ V₂.
Check your understanding
10.0 mL of a 5.0 M NaNO₃ solution is diluted to a total volume of 250.0 mL. What is the molarity of the diluted solution, in mol/L?
Answer: 0.20mol/L(tolerance ±0.005)
Write down the values in the question: M₁ = 5.0 M V₁ = 10.0 mL V₂ = 250.0 mL Write down the equation: M₁V₁ = M₂V₂ Make M₂ the subject: M₂ = M₁V₁ ÷ V₂ Substitute in the values, and calculate: M₂ = (5.0 × 10.0) ÷ 250.0 M₂ = 0.20 M
Check your understanding
40.0 mL of a 1.5 M KBr solution is diluted to a total volume of 120.0 mL. What is the molarity of the diluted solution, in mol/L?
Answer: 0.50mol/L(tolerance ±0.005)
Write down the values in the question: M₁ = 1.5 M V₁ = 40.0 mL V₂ = 120.0 mL Write down the equation: M₁V₁ = M₂V₂ Make M₂ the subject: M₂ = M₁V₁ ÷ V₂ Substitute in the values, and calculate: M₂ = (1.5 × 40.0) ÷ 120.0 M₂ = 0.50 M
Check your understanding
0.250 L of a 1.2 M sucrose solution is diluted to a total volume of 2.00 L. What is the molarity of the diluted solution, in mol/L?
Answer: 0.15mol/L(tolerance ±0.005)
Write down the values in the question: M₁ = 1.2 M V₁ = 0.250 L V₂ = 2.00 L Write down the equation: M₁V₁ = M₂V₂ Make M₂ the subject: M₂ = M₁V₁ ÷ V₂ Substitute in the values, and calculate: M₂ = (1.2 × 0.250) ÷ 2.00 M₂ = 0.15 M
Lesson 44 of 45 · SOL-044
How much stock solution to use
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Have You Ever Wondered?
Wonder this:
A lab keeps one strong bottle of each solution and makes everything weaker from it — less shelf space, fewer bottles. Today's experiment needs 100.0 mL of 0.50 M NaCl, and the shelf bottle reads 2.0 M. How much of the strong bottle do you pour?
You've already used M₁V₁ = M₂V₂ to find a diluted solution's new concentration. Labs run the equation the other way: the target is fixed, and the unknown is how much of the strong solution to start from.
The equation
A concentrated solution kept for making weaker ones is called a 'stock solution'.
The recipe question is: what volume V₁ of the stock delivers the moles the target solution needs?
Write down the dilution equation: M₁V₁ = M₂V₂.
Make V₁ the subject by dividing both sides by M₁, giving V₁ = M₂V₂ ÷ M₁.
M₁ is the stock's molarity; M₂ and V₂ describe the target solution you want.
V₁ comes out in whatever unit V₂ uses, because the two molarities divide and their unit cancels.
To finish the preparation, measure out V₁ of stock and add water until the total volume reaches V₂.
For the shelf problem: V₁ = (0.50 × 100.0) ÷ 2.0 = 25.0 mL of stock, topped up with water to 100.0 mL.
Sanity check: the stock is stronger than the target, so V₁ must come out smaller than V₂.
Worked examples
Worked example 1. What volume of a 4.0 M KBr stock solution is needed to prepare 600.0 mL of a 0.30 M KBr solution?
Step 1
Write down the values in the question
M₁ = 4.0 M
M₂ = 0.30 M
V₂ = 600.0 mL
Step 2
Write down the equation
M₁V₁ = M₂V₂
Step 3
Make the unknown the subject
V₁ = M₂V₂ ÷ M₁
Step 4
Substitute in the values, and calculate
V₁ = (0.30 × 600.0) ÷ 4.0
V₁ = 45.0 mL
Worked example 2. What volume of a 6.00 M CaCl₂ stock solution is needed to prepare 2.00 L of a 0.150 M CaCl₂ solution?
Step 1
Write down the values in the question
M₁ = 6.00 M
M₂ = 0.150 M
V₂ = 2.00 L
Step 2
Write down the equation
M₁V₁ = M₂V₂
Step 3
Make the unknown the subject
V₁ = M₂V₂ ÷ M₁
Step 4
Substitute in the values, and calculate
V₁ = (0.150 × 2.00) ÷ 6.00
V₁ = 0.0500 L, which is 50.0 mL
You can now calculate the volume of a concentrated starting solution (a stock solution) needed to prepare a target dilution by rearranging M₁V₁ = M₂V₂ to V₁ = M₂V₂ ÷ M₁.
Check your understanding
What volume of a 2.0 M NaNO₃ stock solution is needed to prepare 250.0 mL of a 0.40 M NaNO₃ solution? Give your answer in mL.
Answer: 50.0mL(tolerance ±0.5)
Write down the values in the question: M₁ = 2.0 M M₂ = 0.40 M V₂ = 250.0 mL Write down the equation: M₁V₁ = M₂V₂ Make V₁ the subject: V₁ = M₂V₂ ÷ M₁ Substitute in the values, and calculate: V₁ = (0.40 × 250.0) ÷ 2.0 V₁ = 50.0 mL
Check your understanding
What volume of a 1.2 M CuSO₄ stock solution is needed to prepare 400.0 mL of a 0.30 M CuSO₄ solution? Give your answer in mL.
Answer: 100.0mL(tolerance ±0.5)
Write down the values in the question: M₁ = 1.2 M M₂ = 0.30 M V₂ = 400.0 mL Write down the equation: M₁V₁ = M₂V₂ Make V₁ the subject: V₁ = M₂V₂ ÷ M₁ Substitute in the values, and calculate: V₁ = (0.30 × 400.0) ÷ 1.2 V₁ = 100.0 mL
Check your understanding
What volume of a 3.0 M K₂CO₃ stock solution is needed to prepare 1.00 L of a 0.060 M K₂CO₃ solution? Give your answer in liters.
Answer: 0.020L(tolerance ±0.0005)
Write down the values in the question: M₁ = 3.0 M M₂ = 0.060 M V₂ = 1.00 L Write down the equation: M₁V₁ = M₂V₂ Make V₁ the subject: V₁ = M₂V₂ ÷ M₁ Substitute in the values, and calculate: V₁ = (0.060 × 1.00) ÷ 3.0 V₁ = 0.020 L
Lesson 45 of 45 · SOL-045
Stoichiometry with solutions
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Did You Know?
You've already predicted precipitates from mixing solutions, used mole ratios from balanced equations, and counted a solution's moles with n = M × V. This lesson chains them: solution in, moles or grams of product out.
The equation
A dissolved reactant's moles come straight from its bottle: make n the subject of M = n ÷ V, giving n = M × V.
From there the routine is the one you know: the balanced equation's mole ratio converts those moles into moles of the substance asked about.
If the question asks for a mass, finish with mass = moles × molar mass.
Work the chain in order, one quantity per line: moles of the dissolved reactant, mole ratio, then the answer.
'Excess' beside the other reactant means it never runs out — the dissolved reactant you measured sets every amount.
Try it: 0.200 L of 0.10 M AgNO₃ is mixed with excess NaCl, and AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).
Moles of AgNO₃: n = 0.10 × 0.200 = 0.020 mol.
The equation shows 1 mol of AgCl for every 1 mol of AgNO₃, so 0.020 mol of AgCl forms.
Worked examples
Worked example 1. 0.100 L of a 0.20 M KI solution reacts completely with excess Pb(NO₃)₂: Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq). How many moles of PbI₂ form?
Step 1
Write down the values in the question
M = 0.20 M
V = 0.100 L
Find the moles of the dissolved reactant first:
M = n ÷ V
Make n the subject:
n = M × V
n(KI) = 0.20 × 0.100
n(KI) = 0.020 mol
Use the mole ratio from the balanced equation:
2 mol KI → 1 mol PbI₂
Step 2
Write down the equation
n(PbI₂) = n(KI) ÷ 2
Step 3
Substitute in the values, and calculate
n(PbI₂) = 0.020 ÷ 2
n(PbI₂) = 0.010 mol
Worked example 2. 0.500 L of a 0.10 M Ba(NO₃)₂ solution reacts completely with excess K₂SO₄: Ba(NO₃)₂(aq) + K₂SO₄(aq) → BaSO₄(s) + 2 KNO₃(aq). What mass of BaSO₄ forms? (Molar mass of BaSO₄ = 233.4 g/mol.)
You can now calculate the moles or mass of another substance in a reaction from the volume and molarity of a reactant solution, by finding the moles of the dissolved reactant with n = M × V and then using the mole ratio from the balanced equation.
Check your understanding
0.250 L of a 0.20 M K₂CO₃ solution reacts completely with excess CaCl₂: CaCl₂(aq) + K₂CO₃(aq) → CaCO₃(s) + 2 KCl(aq). How many moles of CaCO₃ form?
Answer: 0.050mol(tolerance ±0.0005)
Write down the values in the question: M = 0.20 M V = 0.250 L Find the moles of the dissolved reactant first: M = n ÷ V Make n the subject: n = M × V n(K₂CO₃) = 0.20 × 0.250 n(K₂CO₃) = 0.050 mol Use the mole ratio from the balanced equation: 1 mol K₂CO₃ → 1 mol CaCO₃ n(CaCO₃) = 0.050 mol
Check your understanding
0.300 L of a 0.20 M NaOH solution reacts completely with excess MgSO₄: 2 NaOH(aq) + MgSO₄(aq) → Mg(OH)₂(s) + Na₂SO₄(aq). How many moles of Mg(OH)₂ form?
Answer: 0.030mol(tolerance ±0.0005)
Write down the values in the question: M = 0.20 M V = 0.300 L Find the moles of the dissolved reactant first: M = n ÷ V Make n the subject: n = M × V n(NaOH) = 0.20 × 0.300 n(NaOH) = 0.060 mol Use the mole ratio from the balanced equation: 2 mol NaOH → 1 mol Mg(OH)₂ n(Mg(OH)₂) = 0.060 ÷ 2 n(Mg(OH)₂) = 0.030 mol
Check your understanding
0.200 L of a 0.50 M MgCl₂ solution reacts completely with excess K₂CO₃: MgCl₂(aq) + K₂CO₃(aq) → MgCO₃(s) + 2 KCl(aq). What mass of MgCO₃ forms, in grams? (Molar mass of MgCO₃ = 84.3 g/mol.)
Answer: 8.43g(tolerance ±0.05)
Write down the values in the question: M = 0.50 M V = 0.200 L Find the moles of the dissolved reactant first: M = n ÷ V Make n the subject: n = M × V n(MgCl₂) = 0.50 × 0.200 n(MgCl₂) = 0.100 mol Use the mole ratio from the balanced equation: 1 mol MgCl₂ → 1 mol MgCO₃, so n(MgCO₃) = 0.100 mol Write down the equation for the mass: mass = moles × molar mass Substitute in the values, and calculate: mass = 0.100 × 84.3 mass = 8.43 g
Summary video — Molarity, dilution, and stoichiometry with solutions