Solubility equilibria — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: dissolution of sparingly soluble ionic solids, solubility product constant ($K_{sp}$) calculations, molar solubility, common ion effect, precipitation prediction, $Q$ vs $K_{sp}$ comparisons, and pH effects on solubility for CED Unit 7.

You should already know: General equilibrium constant expressions and ICE table methods; acid-base equilibria and pH calculations; ionic compound dissociation rules.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Solubility equilibria?

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, which dissociate completely, sparingly soluble ionic solids only dissolve a small amount, so we use equilibrium tools to quantify their solubility. By convention, $K_{sp}$ (the solubility product constant) is always defined for the dissolution reaction (solid as reactant, ions as products), and pure solids are excluded from the equilibrium expression because their activity is 1.

Per the AP Chemistry CED, solubility equilibria makes up ~15-20% of Unit 7 (Equilibrium), corresponding to ~2-4% of the total AP exam score. It appears in both multiple choice (MCQ) and free response (FRQ) sections, and is frequently combined with acid-base equilibria, thermodynamics, or lab separation techniques in multi-part FRQ questions.

2. $K_{sp}$ and Molar Solubility

Molar solubility ($s$) is defined as the number of moles of a sparingly soluble solid that dissolve to form 1 liter of a saturated solution, with units of mol/L. The solubility product constant $K_{sp}$ is the equilibrium constant for the dissolution reaction. For the general dissociation of a salt $A_x{B}_y$: $$A_x{B}_y(s)\rightleftharpoons x{A}^{y+}(aq)+y{B}^{x-}(aq)$$ The $K_{sp}$ expression omits the solid reactant, so: $$K_{sp}=[A^{y+}{]}^x[B^{x-}{]}^y$$ Relating this to molar solubility, if 1 mole of $A_x{B}_y$ dissolves to give $x$ moles of $A^{y+}$ and $y$ moles of $B^{x-}$, then $[A^{y+}]=xs$ and $[B^{x-}]=ys$. Substituting gives: $$K_{sp}=(xs{)}^x(ys{)}^y=x^x{y}^y{s}^{x+y}$$ A key intuition: smaller $K_{sp}$ 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 $K_{sp}$ of magnesium hydroxide $Mg(OH{)}_2$ is $5.6\times 10^{-12}$ at 25°C. Calculate the molar solubility of $Mg(OH{)}_2$ in pure water.

1. Write the balanced dissolution reaction: $Mg(OH{)}_2(s)\rightleftharpoons M{g}^{2+}(aq)+2O{H}^{-}(aq)$
2. Set up equilibrium concentrations: let $s$ = molar solubility, so $[M{g}^{2+}]=s$, $[O{H}^{-}]=2s$
3. Substitute into the $K_{sp}$ expression: $$K_{sp}=[M{g}^{2+}][O{H}^{-}{]}^2=(s)(2s{)}^2=4s^3$$
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 $K_{sp}$ values for salts with different stoichiometry.

3. Common Ion Effect

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, 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: it is equal to the concentration provided by the soluble second compound. Because $K_{sp}$ 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 simplifies calculations, and 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 $Mg(OH{)}_2$ ($K_{sp}=5.6\times 10^{-12}$) in a 0.020 M solution of magnesium nitrate $Mg(N{O}_3{)}_2$ at 25°C.

1. $Mg(N{O}_3{)}_2$ is fully soluble, so initial $[M{g}^{2+}]=0.020$ M.
2. Let $s$ = molar solubility of $Mg(OH{)}_2$, so equilibrium $[M{g}^{2+}]=0.020+s$ and $[O{H}^{-}]=2s$.
3. Apply the small-s approximation: $s<<0.020$, so $0.020+s\approx 0.020$.
4. Substitute into $K_{sp}$: $$K_{sp}=(0.020)(2s{)}^2=0.080s^2=5.6\times 10^{-12}$$ $$s^2=\frac{5.6\times 10^{-12}}{0.080}=7.0\times 10^{-11}$$ $$s\approx 8.4\times 10^{-6}\text{M}$$
5. Check approximation: $s=8.4\times 10^{-6}$ M is 0.04% of 0.020 M, so the approximation is valid. Compare to $s=1.1\times 10^{-4}$ M in pure water: solubility is ~13x lower, as expected for the common ion effect.

Exam tip: Never apply the stoichiometric coefficient of the dissolving salt to the initial concentration of the common ion. For example, do not write $[M{g}^{2+}]=2\times 0.020$ M in the example above; the coefficient only applies to the change in concentration from the dissolving sparingly soluble salt.

4. Predicting Precipitation with $Q$ vs $K_{sp}$

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 $K_{sp}$ but uses initial concentrations of the ions (immediately after mixing, before equilibrium is established). The rules for prediction are straightforward:

• If $Q>K_{sp}$: The solution is supersaturated, precipitation occurs until $Q=K_{sp}$
• If $Q=K_{sp}$: The solution is saturated, at equilibrium, no precipitate forms
• If $Q<K_{sp}$: The solution is unsaturated, no precipitate forms, all ions remain 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 a mixture of ions based on their different $K_{sp}$ values.

Worked Example

A chemist mixes 100.0 mL of 0.0040 M calcium chloride $CaC{l}_2$ with 100.0 mL of 0.0030 M sodium sulfate $N{a}_{2}S{O}_4$. The $K_{sp}$ of calcium sulfate $CaS{O}_4$ is $4.9\times 10^{-5}$. Will a precipitate form?

1. Calculate final concentrations after mixing: total volume = 100.0 + 100.0 = 200.0 mL. $$[C{a}^{2+}]=\frac{(0.0040\text{M})(0.1000\text{L})}{0.2000\text{L}}=0.0020\text{M}$$ $$[S{O}_4^{2-}]=\frac{(0.0030\text{M})(0.1000\text{L})}{0.2000\text{L}}=0.0015\text{M}$$
2. Calculate $Q$ using the same form as $K_{sp}$: $$Q=[C{a}^{2+}][S{O}_4^{2-}]=(0.0020)(0.0015)=3.0\times 10^{-6}$$
3. Compare $Q$ to $K_{sp}$: $3.0\times 10^{-6}<4.9\times 10^{-5}$, so $Q<K_{sp}$. 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 MCQs and FRQs.

5. pH Effects on Solubility

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 salts containing hydroxide anions, increasing pH (higher [OH-]) decreases solubility via the common ion effect, while decreasing pH increases solubility.

Worked Example

What is the molar solubility of $Mg(OH{)}_2$ ($K_{sp}=5.6\times 10^{-12}$) at pH 10.00 at 25°C?

1. Calculate [OH-] from pH: $pOH=14.00-10.00=4.00$, so $[O{H}^{-}]=1.0\times 10^{-4}$ M.
2. Let $s$ = molar solubility of $Mg(OH{)}_2$, so $[M{g}^{2+}]=s$.
3. Substitute into $K_{sp}$: $$K_{sp}=[M{g}^{2+}][O{H}^{-}{]}^2=s(1.0\times 10^{-4}{)}^2=1.0\times 10^{-8}s=5.6\times 10^{-12}$$
4. Solve for $s$: $s=5.6\times 10^{-4}M$. Compare to $s=1.1\times 10^{-4}$ M in pure water (pH ~10.3 for saturated $Mg(OH{)}_2$), so at lower pH 10, solubility is higher, as expected.

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; AgCl, AgBr, and other halide salts have pH-independent solubility.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Including the concentration of the solid salt in the $K_{sp}$ 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 $K_{sp}$.
• Wrong move: Comparing $K_{sp}$ values directly to rank solubility of salts with different ion stoichiometries. Why: Students associate lower $K_{sp}$ 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 (OH-, F-, CO3^2-, C2O4^2-) cause pH-dependent solubility.
• Wrong move: Defining $K_{sp}$ 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, $K_{sp}$ 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. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

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 CaF2 C) AgCl D) Lead(II) sulfate PbSO4

Worked Solution: To solve this, we recall that only salts with anions that are conjugate bases of weak acids have pH-dependent solubility, and solubility increases at lower pH. Br-, Cl-, and SO4^2- are all conjugate bases of strong acids (HBr, HCl, H2SO4), so their salts have pH-independent solubility. F- is the conjugate base of HF, a weak acid. At low pH, F- reacts with H+ to form HF, lowering [F-], shifting dissolution equilibrium right and increasing solubility. The correct answer is B.

Question 2 (Free Response)

Silver sulfate $A{g}_{2}S{O}_4$ dissociates as $A{g}_{2}S{O}_4(s)\rightleftharpoons 2A{g}^{+}(aq)+S{O}_4^{2-}(aq)$, with $K_{sp}=1.2\times 10^{-5}$ at 25°C. (a) Calculate the molar solubility of $A{g}_{2}S{O}_4$ in pure water at 25°C. (b) Calculate the molar solubility of $A{g}_{2}S{O}_4$ in 0.15 M sodium sulfate $N{a}_{2}S{O}_4$ at 25°C. (c) A solution contains 0.010 M Cl- and 0.010 M SO4^2-. Solid silver nitrate is added slowly (no volume change). Which precipitate forms first: AgCl ($K_{sp}=1.8\times 10^{-10}$) or $A{g}_{2}S{O}_4$? Justify your answer.

Worked Solution: (a) Let $s$ = molar solubility, so $[A{g}^{+}]=2s$, $[S{O}_4^{2-}]=s$. $$K_{sp}=(2s{)}^2(s)=4s^3=1.2\times 10^{-5}$$ $$s^3=3.0\times 10^{-6}⟹s=\sqrt[3]{3.0\times 10^{-6}}\approx 1.4\times 10^{-2}\text{M}$$

(b) $N{a}_{2}S{O}_4$ is soluble, so initial $[S{O}_4^{2-}]=0.15$ M. Let $s$ = molar solubility, so $[S{O}_4^{2-}]=0.15+s\approx 0.15$ (approximation valid since $K_{sp}$ is small). $$K_{sp}=(2s{)}^2(0.15)=0.60s^2=1.2\times 10^{-5}$$ $$s^2=2.0\times 10^{-5}⟹s\approx 4.5\times 10^{-3}\text{M}$$

(c) Calculate the minimum $[A{g}^{+}]$ required for precipitation: For AgCl: $[A{g}^{+}{]}_{min}=\frac{K_{sp}}{[C{l}^{-}]}=\frac{1.8\times 10^{-10}}{0.010}=1.8\times 10^{-8}$ M. For $A{g}_{2}S{O}_4$: $[A{g}^{+}{]}_{min}^2=\frac{K_{sp}}{[S{O}_4^{2-}]}=\frac{1.2\times 10^{-5}}{0.010}=1.2\times 10^{-3}⟹[A{g}^{+}{]}_{min}=0.035$ M. AgCl requires a much lower $[A{g}^{+}]$ to precipitate, so AgCl precipitates first.

Question 3 (Application / Real-World Style)

Water treatment plants use addition of phosphate to remove lead ions from contaminated drinking water. Lead(II) phosphate $P{b}_3(P{O}_4{)}_2$ has a very low $K_{sp}$ of $3.0\times 10^{-44}$ at 25°C. If the maximum allowable concentration of lead in drinking water is 15 ppb (which equals $7.2\times 10^{-8}$ M $P{b}^{2+}$), what minimum concentration of phosphate must be maintained in the water to meet the standard?

Worked Solution:

1. Write the dissolution reaction: $P{b}_3(P{O}_4{)}_2(s)\rightleftharpoons 3P{b}^{2+}(aq)+2P{O}_4^{3-}(aq)$, so $K_{sp}=[P{b}^{2+}{]}^3[P{O}_4^{3-}{]}^2$.
2. Rearrange to solve for $[P{O}_4^{3-}]$ at maximum allowable $[P{b}^{2+}]$: $$[P{O}_4^{3-}{]}^2=\frac{K_{sp}}{[P{b}^{2+}{]}^3}=\frac{3.0\times 10^{-44}}{(7.2\times 10^{-8}{)}^3}=\frac{3.0\times 10^{-44}}{3.73\times 10^{-22}}=8.0\times 10^{-23}$$ $$[P{O}_4^{3-}]=\sqrt{8.0\times 10^{-23}}\approx 9.0\times 10^{-12}\text{M}$$ This very low minimum concentration means that even a small amount of added phosphate is sufficient to reduce lead levels below the allowable standard in drinking water.

8. Quick Reference Cheatsheet

9. 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 $K_{sp}$, 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 chapter, 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.

Free energy and equilibrium Acid-base equilibria Thermodynamics of dissolution Qualitative analysis separations