AP Chemistry Solubility Equilibria

AP Chemistry· AP Chemistry CED — Equilibrium· 14 min read

1. Fundamentals of Solubility Equilibrium★★☆☆☆⏱ 3 min

Solubility equilibria describes the dynamic heterogeneous equilibrium that forms when a sparingly soluble ionic compound is added to water: undissolved solid is in equilibrium with its dissolved ions in a saturated solution. Unlike fully soluble compounds that dissociate completely, sparingly soluble ionic solids only dissolve a small amount, so we use equilibrium tools to quantify their solubility. Per AP Chemistry CED, this topic makes up ~15-20% of Unit 7, corresponding to ~2-4% of total AP exam score, appearing in both MCQ and FRQ, often combined with other topics like acid-base equilibria.

📘 Definition

Solubility Product Constant ($K_{sp}$)

K_{sp}

Equilibrium constant for the dissolution of 1 mole of sparingly soluble ionic solid into aqueous ions. Pure solid is excluded from the expression because its activity equals 1.

Example:

For $A g C l (s) ⇌ A g^{+} (a q) + C l^{-} (a q)$ , $K_{s p} = [A g^{+}] [C l^{-}]$

Exam tip:

$K_{s p}$ is always defined for the dissolution reaction (solid as reactant), never for precipitation.

2. $K_{sp}$ and Molar Solubility★★★☆☆⏱ 4 min

📘 Definition

Molar Solubility

s

The number of moles of a sparingly soluble solid that dissolve to form 1 liter of saturated solution, with units of mol/L.

For the general dissociation of a salt $A_{x} B_{y}$ : $A_{x} B_{y} (s) ⇌ x A^{y +} (a q) + y B^{x -} (a q)$ , the $K_{s p}$ expression omits the solid, so relating to molar solubility $s$ :

K_{sp} = (xs)^x (ys)^y = x^x y^y s^{x+y}

A key intuition: smaller $K_{s p}$ means lower solubility only for salts with the same ion stoichiometry. For salts with different numbers of ions, you must calculate molar solubility first to compare.

📐 Worked Example

The $K_{s p}$ of magnesium hydroxide $M g (O H)_{2}$ is $5.6 \times 1 0^{- 12}$ at 25°C. Calculate the molar solubility of $M g (O H)_{2}$ in pure water.

1
Write the balanced dissolution reaction:
$Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$
2
Set up equilibrium concentrations: let $s$ = molar solubility, so $[M g^{2 +}] = s$ , $[O H^{-}] = 2 s$
3
Substitute into the $K_{s p}$ expression:
$K_{sp} = [Mg^{2+}][OH^-]^2 = (s)(2s)^2 = 4s^3$
4
Solve for $s$ :
$s^3 = \frac{5.6 \times 10^{-12}}{4} = 1.4 \times 10^{-12} \\ s = \sqrt[3]{1.4 \times 10^{-12}} \approx 1.1 \times 10^{-4}\ \text{M}$

Exam tip:

When comparing solubility of different salts, always calculate molar solubility explicitly. Do not rely solely on $K_{s p}$ values for salts with different stoichiometry.

3. The Common Ion Effect★★★☆☆⏱ 3 min

The common ion effect describes the reduction in molar solubility of a sparingly soluble salt when one of its constituent ions is already present in solution from a second, fully soluble compound. This follows Le Chatelier’s principle: adding a product ion shifts the dissolution equilibrium back toward the solid reactant, reducing the amount of solid that can dissolve.

The calculation approach is nearly identical to that in pure water, except the initial concentration of the common ion is not zero. Because $K_{s p}$ is very small for most sparingly soluble salts, the additional concentration of the common ion from the dissolving salt is usually negligible compared to the initial concentration from the soluble compound. This allows the small- $s$ approximation, which is acceptable for AP as long as the change is less than 5% of the initial common ion concentration (the 5% rule).

📐 Worked Example

Calculate the molar solubility of $M g (O H)_{2}$ ( $K_{s p} = 5.6 \times 1 0^{- 12}$ ) in a 0.020 M solution of magnesium nitrate $M g (N O_{3})_{2}$ at 25°C.

1
$M g (N O_{3})_{2}$ is fully soluble, so initial $[M g^{2 +}] = 0.020$ M.
2
Let $s$ = molar solubility of $M g (O H)_{2}$ , so equilibrium $[M g^{2 +}] = 0.020 + s$ and $[O H^{-}] = 2 s$ .
3
Apply the small- $s$ approximation: $s << 0.020$ , so $0.020 + s \approx 0.020$ .
4
Substitute into $K_{s p}$ :
$K_{sp} = (0.020)(2s)^2 = 0.080 s^2 = 5.6 \times 10^{-12}$
5
Solve for $s$ and check approximation:
$s^2 = \frac{5.6 \times 10^{-12}}{0.080} = 7.0 \times 10^{-11} \\ s \approx 8.4 \times 10^{-6}\ \text{M}$
6
The approximation is valid (s is 0.04% of initial 0.020 M), and solubility is ~13x lower than in pure water, as expected.

Exam tip:

Never apply the stoichiometric coefficient of the dissolving salt to the initial concentration of the common ion. The coefficient only applies to the change in concentration from the dissolving salt.

4. Predicting Precipitation with $Q$ vs $K_{sp}$★★★☆☆⏱ 3 min

To predict whether a precipitate will form when two solutions containing ions of a potential sparingly soluble salt are mixed, we use the reaction quotient $Q$ , which has the same mathematical form as $K_{s p}$ but uses initial concentrations of the ions (immediately after mixing, before equilibrium is established). The prediction rules are:

If $Q > K_{s p}$ : Solution is supersaturated, precipitation occurs until $Q = K_{s p}$
If $Q = K_{s p}$ : Solution is saturated at equilibrium, no net precipitation forms
If $Q < K_{s p}$ : Solution is unsaturated, no precipitate forms, all ions stay dissolved

This framework is also used for selective precipitation, a separation technique where you add a precipitating ion slowly to precipitate one ion at a time, separating mixtures of ions based on different $K_{s p}$ values.

📐 Worked Example

A chemist mixes 100.0 mL of 0.0040 M calcium chloride $C a C l_{2}$ with 100.0 mL of 0.0030 M sodium sulfate $N a_{2} S O_{4}$ . The $K_{s p}$ of calcium sulfate $C a S O_{4}$ is $4.9 \times 1 0^{- 5}$ . Will a precipitate form?

1
Calculate final concentrations after mixing, total volume = 200.0 mL:
$[Ca^{2+}] = \frac{(0.0040\ \text{M})(0.1000\ \text{L})}{0.2000\ \text{L}} = 0.0020\ \text{M} \\ [SO_4^{2-}] = \frac{(0.0030\ \text{M})(0.1000\ \text{L})}{0.2000\ \text{L}} = 0.0015\ \text{M}$
2
Calculate $Q$ using the same form as $K_{s p}$ :
$Q = [Ca^{2+}][SO_4^{2-}] = (0.0020)(0.0015) = 3.0 \times 10^{-6}$
3
Compare $Q$ to $K_{s p}$ : $3.0 \times 1 0^{- 6} < 4.9 \times 1 0^{- 5}$ , so $Q < K_{s p}$ . No calcium sulfate precipitate will form.

Exam tip:

Always remember to dilute ion concentrations when mixing solutions. This is the most frequently missed step in precipitation prediction problems.

5. pH Effects on Solubility★★★★☆⏱ 3 min

The solubility of a sparingly soluble salt depends on pH if the salt's anion is the conjugate base of a weak acid. This is because the anion will react with $H^{+}$ at low pH, removing it from solution, shifting the dissolution equilibrium right and increasing solubility. If the anion is the conjugate base of a strong acid, it does not react with $H^{+}$ , so solubility is independent of pH. For hydroxide-containing salts, higher pH (higher $[O H^{-}]$ ) decreases solubility via the common ion effect, while lower pH increases solubility.

📐 Worked Example

What is the molar solubility of $M g (O H)_{2}$ ( $K_{s p} = 5.6 \times 1 0^{- 12}$ ) at pH 10.00 at 25°C?

1
Calculate $[O H^{-}]$ from pH: $pO H = 14.00 - 10.00 = 4.00$ , so $[O H^{-}] = 1.0 \times 1 0^{- 4}$ M.
2
Let $s$ = molar solubility of $M g (O H)_{2}$ , so $[M g^{2 +}] = s$ .
3
Substitute into $K_{s p}$ :
$K_{sp} = [Mg^{2+}][OH^-]^2 = s(1.0 \times 10^{-4})^2 = 1.0 \times 10^{-8} s = 5.6 \times 10^{-12}$
4
Solve for $s$ :
$s = 5.6 \times 10^{-4}\ \text{M}$
5
Compared to $s = 1.1 \times 1 0^{- 4}$ M in pure water, solubility is higher at lower pH, as expected.

✓ Quick check

Test your understanding of pH effects:

Which of the following sparingly soluble salts will have increased molar solubility when the pH of the saturated solution is lowered from 7 to 3?
- A) AgBr
- B) Calcium fluoride $C a F_{2}$
- C) AgCl
- D) Lead(II) sulfate $P b S O_{4}$
Reveal answer
B) Calcium fluoride $CaF_2$ —
Only salts with anions that are conjugate bases of weak acids have pH-dependent solubility. $B r^{-}$ , $C l^{-}$ , and $S O_{4}^{2 -}$ are conjugate bases of strong acids, so their solubility does not change with pH. $F^{-}$ is the conjugate base of weak acid $H F$ , so solubility increases at lower pH.

Exam tip:

A common MCQ trick asks which salt's solubility changes with pH. Remember: only salts with basic anions (conjugate of weak acid) are pH-dependent; halide salts like AgCl have pH-independent solubility.

6. Common Pitfalls

Wrong move:

Including the concentration of the solid salt in the $K_{s p}$ expression.

Why:

Students habitually include all reactants and products in equilibrium expressions, forgetting pure solids have an activity of 1.

Correct move:

Always omit pure solids (and pure liquids) from any equilibrium expression, including $K_{s p}$ .

Wrong move:

Comparing $K_{s p}$ values directly to rank solubility of salts with different ion stoichiometries.

Why:

Students associate lower $K_{s p}$ with lower solubility, and incorrectly generalize this to all salts.

Correct move:

Always calculate molar solubility for each salt first, then compare the resulting $s$ values.

Wrong move:

Forgetting to dilute ion concentrations when mixing solutions to calculate $Q$ .

Why:

Students reuse the original concentrations from before mixing, ignoring the increased total volume.

Correct move:

Always calculate moles of each ion, then divide by the final total volume to get concentrations for $Q$ .

Wrong move:

Assuming all salts have pH-dependent solubility.

Why:

Students see pH effects in hydroxide examples and generalize to all sparingly soluble salts.

Correct move:

Check the anion: only anions that are conjugate bases of weak acids cause pH-dependent solubility.

Wrong move:

Defining $K_{s p}$ for a precipitation reaction (ions as reactants, solid as product) instead of dissolution.

Why:

Students reverse the reaction when answering precipitation questions, leading to an inverted K value.

Correct move:

By definition, $K_{s p}$ is always for the dissolution of 1 mole of solid into ions, so solid is always the reactant.

Wrong move:

Applying the stoichiometric coefficient of the dissolving salt to the initial common ion concentration.

Why:

Students confuse the stoichiometry of the dissolution reaction with the source of the common ion.

Correct move:

Only add the stoichiometric multiple of $s$ to the initial common ion concentration; do not multiply the initial concentration by the coefficient.

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
General $K_{s p}$ for $A_{x} B_{y}$	$K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}$	Pure solid is omitted; $K_{s p}$ only changes with temperature
Molar solubility (pure water)	$K_{s p} = x^{x} y^{y} s^{x + y}$	$s$ = moles of solid dissolved per liter of saturated solution
Ion product $Q$	$Q = [A^{y +}]^{x} [B^{x -}]^{y}$	Uses initial concentrations after mixing, not equilibrium
Precipitation rule	$Q > K_{s p}$ : precipitate; $Q = K_{s p}$ : saturated; $Q < K_{s p}$ : no precipitate	Used for all precipitation prediction problems
Common ion effect	$s_{co mm o ni o n} < s_{p u r e w a t er}$	Le Chatelier shift left from added product ion
5% approximation rule	Approximation valid if $s < 5%$ of initial common ion concentration	If invalid, solve the full polynomial equation
pH effect rule	Solubility increases at lower pH for salts with basic anions	Basic anions = conjugate bases of weak acids; no effect for strong acid anions
Selective precipitation rule	Salt requiring lower precipitating ion concentration precipitates first	Used for separating mixtures of ions

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Identify pH-dependent solubility
2022 · FRQ
Selective precipitation calculation
2021 · FRQ
Common ion effect solubility

What's Next

Solubility equilibria is a core application of general equilibrium principles, and it underpins many common AP Chemistry topics that build on equilibrium fundamentals. Mastery of $K_{s p}$ , common ion effects, and precipitation prediction is required for lab-based FRQs focused on qualitative separation of ions, and for mixed problems that combine solubility with acid-base equilibria. Without mastering the concepts in this module, you will struggle to solve multi-concept problems that connect equilibrium to other units, like thermodynamics of dissolution or acid-base titrations of sparingly soluble bases. Next, you will extend solubility concepts to complex ion formation, which explains the increased solubility of some salts in the presence of excess ligands.