Chemistry of Solutions 202-SN2-RE • CEGEP, Montreal

Revision sheet: solubility and the solubility product (202-SN2-RE)

Solubility is only a special case of equilibrium: the solid is in equilibrium with its ions, and the corresponding constant is called the solubility product, $K_{sp}$ . The whole difficulty lies in going between this constant and the molar solubility, a step that depends entirely on the stoichiometry of the salt.

This sheet gives the three relations to know, the rule for comparing $K_{sp}$ values, the ion product test and the two effects that change a solubility: the common ion and pH. It is written for CEGEP students in Chemistry of Solutions, 202-SN2-RE (formerly 202-NYB).

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The thread of the chapter

A $K_{sp}$ can be compared with another only if the two salts have the SAME stoichiometry. Without that precaution, the least soluble salt is regularly announced as the most soluble, and the error shows nowhere in the calculation.

This chapter is part of 202-SN2-RE, Chemistry of Solutions

The essentials

From $K_{sp}$ to the molar solubility

• For $A_{m}B_{n}(s)\rightleftharpoons mA^{n+}+nB^{m-}$ : $K_{sp}=[A^{n+}]^{m}[B^{m-}]^{n}$ . The SOLID never appears in it.
• Type $AB$ (like $AgCl$ ): $K_{sp}=s^{2}$ , so $s=\sqrt{K_{sp}}$ .
• Type $AB_{2}$ or $A_{2}B$ (like $Ag_{2}CrO_{4}$ ): $K_{sp}=4s^{3}$ , so $s=\sqrt[3]{\dfrac{K_{sp}}{4}}$ .
• Type $A_{3}B_{2}$ (like $Ca_{3}(PO_{4})_{2}$ ): $K_{sp}=108\,s^{5}$ .
• The coefficient enters TWICE: once as a factor in front of $s$ in the concentration, once as an exponent in $K_{sp}$ . Hence the $4$ and the $108$ .

The salt with the SMALLER solubility product is here almost five times MORE SOLUBLE: the stoichiometries differ, so the two constants cannot be compared with each other.

Comparing two $K_{sp}$ values tells you nothing as long as the stoichiometries differ: you must then calculate both solubilities and compare those. It is the most profitable rule of the chapter.

Will it precipitate? The ion product

• $Q$ has exactly the same expression as $K_{sp}$ , but with the ACTUAL concentrations at the instant considered.
• $Q<K_{sp}$ : unsaturated solution, no precipitate. $Q=K_{sp}$ : saturated. $Q>K_{sp}$ : a precipitate forms until $Q$ falls back to $K_{sp}$ .
• When two solutions are MIXED, all the concentrations are first DILUTED by the total volume, before any calculation of $Q$ .
• Selective precipitation: the ion that requires the LOWEST reagent concentration precipitates first.
• A separation is considered quantitative when less than $0.1\%$ of the first ion remains at the moment the second begins to precipitate.

Only

Q

decides: below

K_{sp}

everything stays dissolved, at

K_{sp}

the solution is saturated, above it the solid forms until

Q

falls back to

K_{sp}

The precipitation threshold is always calculated the same way: isolate, in the expression of $K_{sp}$ , the concentration of the ion being added, putting in the actual concentration of the other ion.

The two effects that change a solubility

• COMMON ION: adding one of the two ions shifts the equilibrium toward the solid and makes the solubility DROP, often by several orders of magnitude.
• In practice, hold the concentration of the common ion at its added value and isolate the other: $s=\dfrac{K_{sp}}{[\text{common ion}]}$ for an $AB$ salt.
• pH: a salt whose ANION IS BASIC ( $OH^{-}$ , $F^{-}$ , $CO_{3}^{2-}$ , $S^{2-}$ ) is more soluble the more ACIDIC the medium, because the $H^{+}$ ions consume the anion.
• A salt whose anion comes from a STRONG acid ( $Cl^{-}$ , $NO_{3}^{-}$ , $ClO_{4}^{-}$ ) is insensitive to pH.
• AMPHOTERIC hydroxides: they also redissolve in strongly basic medium, by forming a complex. Their solubility therefore goes through a MINIMUM.

An amphoteric hydroxide dissolves at BOTH ends of the scale: in acidic medium because the

H^{+}

ions consume the hydroxide, in strongly basic medium because it forms a complex.

The common-ion effect and the pH effect are two direct applications of Le Chatelier's principle: you add a species of the equilibrium to shift it toward the solid, or you remove one to shift it toward dissolution.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Comparing two $K_{sp}$ values of different stoichiometries

the whole question, 5 points, with the conclusion reversed

What not to write

“ $K_{sp}(Ag_{2}CrO_{4})=1.1\times 10^{-12}$ is smaller than $K_{sp}(AgCl)=1.8\times 10^{-10}$ , so silver chromate is less soluble.”

What to write

“The stoichiometries differ: $s(AgCl)=\sqrt{K_{sp}}=1.34\times 10^{-5}$ and $s(Ag_{2}CrO_{4})=\sqrt[3]{\dfrac{K_{sp}}{4}}=6.50\times 10^{-5}$ mol/L. The chromate is in fact almost FIVE TIMES more soluble.”

Why: A $K_{sp}$ is a product of different powers depending on the salt: it is not the same physical quantity from one type to another. Only the molar solubilities are directly comparable.

2. Writing $K_{sp}=s^{2}$ for every salt

the whole question, 6 points, with a factor of

60

on the result

What not to write

“ $Ag_{2}CrO_{4}$ : $K_{sp}=s^{2}$ , so $s=\sqrt{1.1\times 10^{-12}}=1.05\times 10^{-6}$ mol/L.”

What to write

“Two silver ions per formula unit: $[Ag^{+}]=2s$ and $[CrO_{4}^{2-}]=s$ , so $K_{sp}=(2s)^{2}(s)=4s^{3}$ and $s=6.50\times 10^{-5}$ mol/L.”

Two silver ions for one chromate ion:

[Ag^{+}]=2s

, and the

2

comes back squared in

K_{sp}

. Hence

4s^{3}

, and never

s^{2}

Why: The coefficient $2$ enters twice: as a factor in the silver ion concentration and as an exponent in $K_{sp}$ . That is where the $4$ comes from, and it is exactly what the error removes.

3. Putting the solid in the expression of the solubility product

3 points, and a wrong expression for the constant

What not to write

“ $K_{sp}=\dfrac{[Ag^{+}][Cl^{-}]}{[AgCl]}$ .”

What to write

“ $K_{sp}=[Ag^{+}][Cl^{-}]$ : the pure solid never appears, its concentration being a constant already absorbed into $K_{sp}$ .”

Why: It is the same rule as for any heterogeneous equilibrium. It also explains why adding solid changes nothing: the solution stays saturated at the same concentration.

4. Forgetting the dilution before calculating the ion product

3 points, and a conclusion that is sometimes reversed

What not to write

“ $50.0$ mL of $Ag^{+}$ at $0.0020$ mol/L and $50.0$ mL of $Cl^{-}$ at $0.0020$ mol/L are mixed: $Q=(0.0020)^{2}=4.0\times 10^{-6}$ .”

What to write

“The volume doubles, so each concentration is DIVIDED BY TWO: $Q=(0.0010)^{2}=1.0\times 10^{-6}$ . A precipitate does form, but the number was off by a factor of $4$ .”

Mixing doubles the volume: each ion is at

0.0010

mol/L, and it is with these DILUTED concentrations that

Q

is calculated.

Why: Mixing always dilutes. When $Q$ lands near $K_{sp}$ , this dilution alone decides whether or not there is a precipitate, and neglecting it leads you to announce a precipitate that does not exist.

5. Ignoring the common-ion effect

the whole question, 5 points

What not to write

“The solubility of silver chloride in a $0.100$ mol/L sodium chloride solution is $1.34\times 10^{-5}$ mol/L, as in pure water.”

What to write

“The chloride ion is already present: $s=\dfrac{K_{sp}}{[Cl^{-}]}=\dfrac{1.8\times 10^{-10}}{0.100}=1.8\times 10^{-9}$ mol/L, that is $7500$ times lower.”

On the logarithmic scale, the common ion pushes the solubility back by nearly four powers of ten: from

1.34\times 10^{-5}

1.8\times 10^{-9}

mol/L.

Why: The common ion shifts the equilibrium toward the solid. The effect is huge because the solubility in pure water is already minuscule next to the concentration added: this is what makes it possible to wash a precipitate without losing it.

6. Believing that pH has no effect on a solubility

4 points on the reasoning question

What not to write

“Calcium carbonate is a salt: its solubility does not depend on pH.”

What to write

“The carbonate ion is a BASE: in acidic medium, it is consumed by $H^{+}$ ions, the equilibrium shifts toward dissolution and the solubility INCREASES.”

The solubility of calcium carbonate climbs as the pH drops, because

H^{+}

consumes the carbonate ion; that of silver chloride does not move, since

Cl^{-}

comes from a strong acid.

Why: Removing a product from an equilibrium shifts it to the right. This is what explains the dissolution of limestone by acid rain and the attack on tooth enamel by fizzy drinks.

7. Isolating the wrong ion when calculating a precipitation threshold

3 points

What not to write

“To find when silver chloride precipitates in a solution that is $0.100$ mol/L in chloride, I calculate $[Cl^{-}]=\dfrac{K_{sp}}{[Ag^{+}]}$ .”

What to write

“It is SILVER that is being added, so it is silver that gets isolated: $[Ag^{+}]=\dfrac{K_{sp}}{[Cl^{-}]}=\dfrac{1.8\times 10^{-10}}{0.100}=1.8\times 10^{-9}$ mol/L.”

Silver is poured in, so the thresholds are read in

[Ag^{+}]

: the chloride precipitates from

1.8\times 10^{-9}

, the chromate only nearly six thousand times further along.

Why: The threshold always concerns the ion being poured in, the other being fixed by the problem. Isolating the wrong one gives a number that does not answer the question asked.

8. Confusing molar solubility and mass solubility

2 points, and a comparison with the tables that becomes impossible

What not to write

“The solubility of silver chloride is $1.34\times 10^{-5}$ , so $1.34\times 10^{-5}$ g/L.”

What to write

“ $1.34\times 10^{-5}$ mol/L multiplied by $M=143.3$ g/mol gives $1.92\times 10^{-3}$ g/L, that is $1.92$ mg/L.”

Why: Solubility tables are often in grams per litre, while $K_{sp}$ calculations are always in moles per litre. Going from one to the other is a simple multiplication by the molar mass, but it gets forgotten.

9. Believing that adding solid raises the concentration of a saturated solution

3 points on the reasoning question

What not to write

“I add more silver chloride to the bottom of the beaker, so the solution becomes more concentrated in silver ions.”

What to write

“The solution is already SATURATED: $[Ag^{+}][Cl^{-}]=K_{sp}$ and nothing changes. The added solid stays at the bottom.”

Little or a lot of solid at the bottom, the saturated solution has the SAME concentration: the solid does not enter

K_{sp}

Why: The solid does not enter the expression of $K_{sp}$ , so its amount has no effect on the position of the equilibrium. This is exactly why a saturated solution has a reproducible concentration.

10. Forgetting that a complex redissolves a precipitate

4 points on the reasoning question

What not to write

“Silver chloride is insoluble, so it stays at the bottom whatever is added.”

What to write

“In the presence of ammonia, the complex $Ag(NH_{3})_{2}^{+}$ forms: the free silver is consumed, the equilibrium shifts toward dissolution, and the precipitate DISAPPEARS.”

Ammonia captures the silver as

Ag(NH_{3})_{2}^{+}

: the dissolution equilibrium shifts to the right and the precipitate disappears.

Why: A complex removes the ion from solution as surely as an acid removes a basic anion. This is the principle of photographic fixer, and the example expected in the theory question.

Which method to choose

Which relation between $K_{sp}$ and $s$

The stoichiometry of the salt, read from its formula

The coefficient of each ion is used twice, as a factor in front of

s

and as an exponent: that is the whole origin of the

4

, the

27

and the

108

If $AB$ , one cation for one anion → $K_{sp}=s^{2}$

Example: $AgCl$ , $BaSO_{4}$ , $CaCO_{3}$
If $AB_{2}$ or $A_{2}B$ → $K_{sp}=4s^{3}$

Example: $Ag_{2}CrO_{4}$ , $PbI_{2}$ , $Mg(OH)_{2}$
If $AB_{3}$ or $A_{3}B$ → $K_{sp}=27\,s^{4}$

Example: $Al(OH)_{3}$ , $Fe(OH)_{3}$
If $A_{3}B_{2}$ → $K_{sp}=108\,s^{5}$

Example: $Ca_{3}(PO_{4})_{2}$
If an ion is ALREADY present in solution → do not set $s$ for it: fix its concentration and isolate the other

this is the common-ion effect, and it bypasses the relations above entirely

The numerical coefficient is found again in one line from the ICE table: $(2s)^{2}(s)=4s^{3}$ , $(3s)^{3}(2s)^{2}=108s^{5}$ . Memorizing it is useful, knowing how to rederive it is more so.

What question is the solubility problem asking

The verb of the problem

If “calculate the solubility” → start from $K_{sp}$ and the stoichiometry, or fix the common ion if there is one

Example: $s=\sqrt{K_{sp}}$ for an $AB$ salt in pure water
If “will a precipitate form” → dilute, calculate $Q$ , compare with $K_{sp}$

Example: $Q=1.0\times 10^{-6}$ against $K_{sp}=1.8\times 10^{-10}$ : yes
If “from what concentration does it precipitate” → isolate in $K_{sp}$ the ion being added, the other being fixed by the problem

Example: $[Ag^{+}]=\dfrac{K_{sp}}{[Cl^{-}]}$
If “which precipitates first” → calculate both thresholds and keep the SMALLER

Example: $1.8\times 10^{-9}$ against $1.05\times 10^{-5}$ : the chloride first
If “is the separation complete” → calculate what remains of the first ion at the threshold of the second, and compare it with $0.1\%$

below this threshold, the separation is considered quantitative

The question “which precipitates first” is never answered by comparing the $K_{sp}$ values: it is answered by comparing the THRESHOLDS, which also depend on the concentrations present.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Handling a selective precipitation

When to use it: A solution contains two anions and a cation that precipitates both is added gradually.

1 Write the two dissolution equilibria with their $K_{sp}$ , stating the stoichiometry of each.
2 For each salt, isolate the concentration of the ADDED ion needed to reach $K_{sp}$ , using the actual concentration of the other ion.
3 Compare the two thresholds: the smaller one designates the salt that precipitates first, and say so explicitly.
4 Calculate what remains of the first ion at the moment the second begins to precipitate, reusing the $K_{sp}$ of the first salt.
5 Express this remainder as a percentage of the starting amount and conclude on whether the separation is quantitative.

Concluding sentence

“The chloride precipitates from $[Ag^{+}]=1.8\times 10^{-9}$ mol/L, the chromate only at $1.05\times 10^{-5}$ mol/L: the chloride goes first, and when the chromate begins, only $1.7\times 10^{-5}$ mol/L of chloride remains, that is $0.017\%$ of the start: the separation is quantitative.”

The trap: Comparing the $K_{sp}$ values instead of the thresholds. The initial concentrations of the two anions enter the thresholds, and they can reverse the ranking suggested by the constants alone.

Marking: Typically 2 points for the two $K_{sp}$ expressions, 2 for the two thresholds, 1 for the conclusion on the order, 2 for the calculated remainder, 1 for the percentage.

Calculating a solubility with a common ion

When to use it: The problem asks for the solubility of a salt in a solution that already contains one of its ions.

1 Write the equilibrium and the expression of $K_{sp}$ , then set up an ICE table starting from the concentration ALREADY PRESENT for the common ion.
2 Let $s$ be the change, and note that $s$ will be negligible next to the concentration of the common ion.
3 Substitute, keeping the concentration of the common ion at its initial value, and isolate $s$ .
4 Check afterwards that $s$ is indeed negligible, with the same $5\%$ criterion as for the other equilibria.
5 Compare with the solubility in pure water and comment on the factor obtained.

Concluding sentence

“In a $0.100$ mol/L sodium chloride solution, $[Cl^{-}]\approx 0.100$ and $s=\dfrac{K_{sp}}{0.100}=1.8\times 10^{-9}$ mol/L, about $7500$ times lower than in pure water.”

The trap: Adding the solubility to the concentration of the common ion without checking that it is negligible. It almost always is here, but the check is worth the mark for rigour.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The $K_{sp}$ recalculated from the solubility

Substitute the ion concentrations found into the expression of $K_{sp}$ : you must get back the value given in the problem. This is the check that validates the stoichiometry.

A recalculated $K_{sp}$ four times too small for $Ag_{2}CrO_{4}$ : the factor $2$ was left out of the silver concentration.
The order of magnitude of a solubility

A salt called insoluble has a solubility of the order of $10^{-5}$ mol/L, that is a few milligrams per litre. A value above $10^{-2}$ contradicts the problem.

A solubility of $0.1$ mol/L for a salt described as insoluble: the root was taken on the wrong exponent.
The common-ion effect in the right direction

The solubility with a common ion must always be SMALLER than in pure water. A larger result signals an error in isolating the unknown.

A solubility of $1.8\times 10^{-9}$ against $1.34\times 10^{-5}$ : consistent, the drop is on the right side.
The dilution applied before any calculation of $Q$

Reread each concentration entering $Q$ and check that it is referred to the TOTAL volume of the mixture.

Two equal volumes mixed without dividing by two: $Q$ is four times too large for an $AB$ salt.
The direction of the pH effect

Check whether the anion is basic. If so, acidifying must INCREASE the solubility; if the anion comes from a strong acid, pH should change nothing.

A silver chloride solubility announced as depending on pH: the chloride ion is not basic.

The typical problem, taken apart

Separating chloride from chromate by selective precipitation

A solution contains chloride ions at $0.100$ mol/L and chromate ions at $0.0100$ mol/L. A silver nitrate solution is added dropwise, with no appreciable change in volume.

First calculate the molar solubilities of silver chloride and silver chromate in pure water, and compare them with the solubility products. Then determine which of the two salts precipitates first, and what fraction of the first ion remains in solution when the second begins to precipitate. Finally, calculate the solubility of silver chloride in a $0.100$ mol/L sodium chloride solution.

Data: $K_{sp}(AgCl)=1.8\times 10^{-10}$ and $K_{sp}(Ag_{2}CrO_{4})=1.1\times 10^{-12}$ .

The situation of the example: two anions in the same beaker, and the silver being added dropwise without changing the volume.

Step 1

$AgCl(s)\rightleftharpoons Ag^{+}+Cl^{-}$ gives $K_{sp}=s^{2}$ , so $s=\sqrt{1.8\times 10^{-10}}=1.34\times 10^{-5}$ mol/L.

Why

One cation for one anion: this is the only case where the square root is enough. Writing the equilibrium before the formula avoids applying this root to a salt that does not suit it.

Step 2

$Ag_{2}CrO_{4}(s)\rightleftharpoons 2Ag^{+}+CrO_{4}^{2-}$ gives $[Ag^{+}]=2s$ and $[CrO_{4}^{2-}]=s$ , so $K_{sp}=(2s)^{2}(s)=4s^{3}$ and $s=\sqrt[3]{\dfrac{1.1\times 10^{-12}}{4}}=6.50\times 10^{-5}$ mol/L.

Why

The coefficient $2$ appears twice, as a factor and as an exponent: this double use produces the $4$ , and it is exactly what the formula $s=\sqrt{K_{sp}}$ would make disappear.

Step 3

Comparison: the chromate has the SMALLER $K_{sp}$ and yet the LARGER solubility, almost five times that of the chloride.

Why

This is the conclusion that the first part of the question is aiming at, and it is counterintuitive: two constants of different dimensions cannot be compared directly.

Step 4

Chloride threshold: $[Ag^{+}]=\dfrac{K_{sp}}{[Cl^{-}]}=\dfrac{1.8\times 10^{-10}}{0.100}=1.8\times 10^{-9}$ mol/L.

Why

You isolate the ion being POURED IN, the other being fixed by the problem. This reading direction is what makes the question solvable, and reversing it gives a number unrelated to the experiment.

Step 5

Chromate threshold: $[Ag^{+}]^{2}=\dfrac{K_{sp}}{[CrO_{4}^{2-}]}=\dfrac{1.1\times 10^{-12}}{0.0100}=1.1\times 10^{-10}$ , so $[Ag^{+}]=1.05\times 10^{-5}$ mol/L.

Why

The exponent $2$ requires a square root at the end: it is the only difference in treatment between the two salts, and it again comes from stoichiometry.

Step 6

$1.8\times 10^{-9}<1.05\times 10^{-5}$ : the CHLORIDE precipitates first, and nearly six thousand times more silver is needed to bring out the chromate.

Why

The order is read from the thresholds, not from the constants. The ratio between the two thresholds directly measures the margin available for stopping the addition at the right moment.

Step 7

Chloride remaining at the chromate threshold: $[Cl^{-}]=\dfrac{1.8\times 10^{-10}}{1.05\times 10^{-5}}=1.72\times 10^{-5}$ mol/L, that is $\dfrac{1.72\times 10^{-5}}{0.100}\times 100=0.017\%$ of the starting chloride.

Why

This percentage, and it alone, answers the question about the separation. Below $0.1\%$ , the separation is called quantitative, which is very much the case here.

Step 8

Common-ion effect: in $0.100$ mol/L $NaCl$ , $s=\dfrac{K_{sp}}{0.100}=1.8\times 10^{-9}$ mol/L, against $1.34\times 10^{-5}$ in pure water, that is $7500$ times lower.

Why

You do not set $s$ again for the common ion: its concentration is imposed. The solubility found is indeed negligible next to $0.100$ , which validates the approximation.

Step 9

Check: $[Ag^{+}][Cl^{-}]=(1.8\times 10^{-9})(0.100)=1.8\times 10^{-10}$ , equal to the $K_{sp}$ in the problem.

Why

Recalculating the constant from the concentrations obtained tests the stoichiometry, the isolation and the rounding all at once, without redoing any reasoning.

The conclusion, written out

“Despite a smaller solubility product, silver chromate is almost five times more soluble than the chloride. The chloride precipitates first, from $1.8\times 10^{-9}$ mol/L of silver, and only $0.017\%$ of it remains when the chromate begins to precipitate: the separation is quantitative. In a $0.100$ mol/L sodium chloride solution, the solubility of silver chloride drops by a factor of $7500$ .”

The classic mistake on this problem: Concluding from $1.1\times 10^{-12}<1.8\times 10^{-10}$ that the chromate is the less soluble. The comparison is illegitimate, the two constants not varying as the same power of the solubility, and the conclusion is exactly the opposite of reality.

Learn by heart

• $K_{sp}$ never contains the solid, and two $K_{sp}$ values can be compared only for the SAME stoichiometry.
• $AB$ : $K_{sp}=s^{2}$ . $AB_{2}$ or $A_{2}B$ : $K_{sp}=4s^{3}$ . $AB_{3}$ : $27s^{4}$ . $A_{3}B_{2}$ : $108s^{5}$ .
• The coefficient counts TWICE: as a factor in front of $s$ and as an exponent.
• $Q<K_{sp}$ : unsaturated. $Q>K_{sp}$ : a precipitate forms. Dilute BEFORE calculating $Q$ .
• Common-ion effect: fix the concentration of the ion present and isolate the other; the solubility drops.
• Precipitation threshold: isolate the ion you ADD, the other being given.
• Basic anion ( $OH^{-}$ , $CO_{3}^{2-}$ , $F^{-}$ , $S^{2-}$ ): the solubility INCREASES in acidic medium.
• A complex, like that of silver with ammonia, redissolves a precipitate.

Frequently asked questions

Can two solubility products be compared to tell which salt is less soluble?

Only if the two salts have the same stoichiometry. Otherwise the two constants do not even vary as the same power of the solubility, and the comparison makes no sense. Silver chromate has a solubility product about 160 times smaller than silver chloride, and yet it is almost five times more soluble. You have to calculate both molar solubilities.

Why does the coefficient appear twice in the relation between Ksp and solubility?

Because it first acts as a factor: if the salt releases two cations, the concentration of that cation is twice the solubility. It then acts as an exponent in the expression of the solubility product. It is the combination of the two that gives the factor four for a two-to-one salt.

How do you know whether a precipitate will form?

You calculate the ion product, which has the same expression as the solubility product but with the actual concentrations, after allowing for the dilution caused by mixing. If this quotient exceeds the solubility product, precipitation occurs until the two are equal again. If it is lower, the solution stays clear.

Why is a salt less soluble in the presence of one of its ions?

Because the dissolution equilibrium shifts toward the solid, as Le Chatelier's principle predicts. Adding chloride forces silver chloride to precipitate further. The effect is spectacular: in a solution with one tenth of a mole of chloride per litre, the solubility drops by a factor of several thousand.

Does pH affect the solubility of a salt?

Yes, as soon as the anion of the salt is basic, like hydroxide, carbonate, fluoride or sulfide. In acidic medium, hydronium ions consume this anion, the equilibrium shifts toward dissolution and the solubility increases. An anion that comes from a strong acid, like chloride or nitrate, is insensitive to it.

Which salt precipitates first when a common reagent is added?

The one that requires the lowest concentration of the added reagent. You calculate this threshold for each salt by isolating the poured-in ion in its solubility product, with the actual concentration of the other ion. It is therefore not only a matter of constants: the initial concentrations also come in and can reverse the ranking.

Practise it

Corrected exercises: Solubility and the solubility product, 202-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: solubility and the solubility product (202-SN2-RE)

The essentials

From KspK_{sp}Ksp​ to the molar solubility

Will it precipitate? The ion product

The two effects that change a solubility

The mistakes that cost marks

1. Comparing two KspK_{sp}Ksp​ values of different stoichiometries

2. Writing Ksp=s2K_{sp}=s^{2}Ksp​=s2 for every salt

3. Putting the solid in the expression of the solubility product

4. Forgetting the dilution before calculating the ion product

5. Ignoring the common-ion effect

6. Believing that pH has no effect on a solubility

7. Isolating the wrong ion when calculating a precipitation threshold

8. Confusing molar solubility and mass solubility

9. Believing that adding solid raises the concentration of a saturated solution

10. Forgetting that a complex redissolves a precipitate

Which method to choose

Which relation between KspK_{sp}Ksp​ and sss

What question is the solubility problem asking

How the answer is expected to be written

Handling a selective precipitation

Calculating a solubility with a common ion

Check before you hand in

The typical problem, taken apart

Separating chloride from chromate by selective precipitation

Learn by heart

Frequently asked questions

Can two solubility products be compared to tell which salt is less soluble?

Why does the coefficient appear twice in the relation between Ksp and solubility?

How do you know whether a precipitate will form?

Why is a salt less soluble in the presence of one of its ions?

Does pH affect the solubility of a salt?

Which salt precipitates first when a common reagent is added?

Corrected exercises: Solubility and the solubility product, 202-SN2-RE

See also

Looking for a 202-SN2-RE tutor in Montreal?

From $K_{sp}$ to the molar solubility

1. Comparing two $K_{sp}$ values of different stoichiometries

2. Writing $K_{sp}=s^{2}$ for every salt

Which relation between $K_{sp}$ and $s$