pH and Solubility — AP Chemistry
1. What is pH-Dependent Solubility? ★★☆☆☆ ⏱ 3 min
pH and solubility describes how hydronium ion concentration changes the solubility of ionic compounds whose anions act as weak Brønsted-Lowry bases. When a sparingly soluble salt dissolves, it releases free anions into solution. If these anions react with H+ (in acidic solution) or OH- (in basic solution), Le Chatelier's principle shifts the dissolution equilibrium, changing how much salt can dissolve.
This topic makes up ~2-4% of the total AP Chemistry exam score, appearing in both multiple-choice and free-response questions, and has real-world applications in environmental chemistry, geochemistry, and drinking water treatment.
2. Qualitative Effect of pH on Solubility ★★☆☆☆ ⏱ 4 min
Only salts with anions that are conjugate bases of weak acids have pH-dependent solubility. For a general 1:1 sparingly soluble salt MA where A⁻ is the conjugate base of weak acid HA, the dissolution equilibrium is:
\text{MA}(s) \rightleftharpoons \text{M}^+(aq) + \text{A}^-(aq) \quad K_{sp} = [\text{M}^+][A^-]
In acidic solution (low pH, high [H+]), A⁻ reacts with H+ to form HA. This consumes free A⁻, lowering [A⁻] below the value required for $K_{sp}$. To restore equilibrium, more solid MA dissolves to replenish A⁻, increasing total solubility.
- Solubility of salts with basic anions **increases as pH decreases** (more acidic solution)
- For salts with anions from strong acids (e.g. Cl⁻, NO₃⁻, Br⁻), anions are negligible bases that do not react with H+, so solubility is unaffected by pH
- For metal hydroxides (e.g. Mg(OH)₂), solubility always increases in acidic solution (H+ reacts with OH⁻ to form water) and decreases in basic solution (excess OH⁻ acts as a common ion shifting equilibrium left)
Exam tip: On AP MCQ, the most common distractor is a salt with an anion from a strong acid that is claimed to have pH-dependent solubility. Always check the parent acid of the anion first to eliminate wrong options quickly.
3. Calculating Molar Solubility at Fixed pH ★★★☆☆ ⏱ 4 min
When pH is fixed (usually in a buffered solution), [H+] is known directly, which simplifies calculations. The key rule is that molar solubility $s$ equals the total concentration of the anion (in all protonation forms) in solution, per dissolution stoichiometry.
For a 1:1 salt MA, $s = [M^+] = [A^-] + [HA]$, because every mole of dissolved MA produces one mole of A in some form. Relating [A⁻] to total dissolved A using the acid dissociation constant for HA: $K_a = \frac{[H^+][A^-]}{[HA]}$. Rearranging gives the fraction of A in the free (deprotonated) form:
\alpha_{A^-} = \frac{[A^-]}{[A^-] + [HA]} = \frac{K_a}{K_a + [H^+]}
Substituting into the Ksp expression gives $K_{sp} = [M^+][A^-] = s \times (\alpha_{A^-} s) = \alpha_{A^-} s^2$, which can be solved directly for $s$. For metal hydroxides at fixed pH, the calculation is even simpler: pH gives [OH⁻] directly, so $s = [M^{n+}] = \frac{K_{sp}}{[OH^-]^n}$.
Exam tip: For polyprotic acids, always match the Ka to the charge of the anion. A dianion (e.g. C₂O₄²⁻) requires Ka2, a trianion (e.g. PO₄³⁻) requires Ka3. Using the wrong Ka is the most common error in these calculations.
4. Selective Precipitation of Ions by pH Adjustment ★★★★☆ ⏱ 3 min
A common AP Chemistry application of pH-solubility relationships is the separation of mixed metal ions by selective precipitation of their insoluble sulfide or hydroxide salts. Many metal sulfides have very different Ksp values: the least soluble sulfides precipitate at low pH (where [S²⁻] is very low), while more soluble sulfides remain dissolved, allowing clean separation.
- Calculate [S²⁻] (or [OH⁻]) at the given pH using the Ka of the parent acid
- Calculate the ion product Q for each metal salt using the given initial metal concentration
- If Q > Ksp, the salt precipitates; if Q < Ksp, it remains dissolved
For saturated H₂S (a common source of sulfide), [H₂S] = 0.1 M at 1 atm, so [S²⁻] can be calculated directly from pH:
K_{a1}K_{a2} = \frac{[H^+]^2[S^{2-}]}{[H_2S]} = \frac{[H^+]^2[S^{2-}]}{0.1} \implies [S^{2-}] = \frac{0.1 K_{a1}K_{a2}}{[H^+]^2}
Exam tip: Do not set the metal ion concentration equal to solubility in selective precipitation problems. The initial metal concentration is given, so use that value directly to calculate Q.
Common Pitfalls
Why: Students memorize 'lower pH increases solubility' without checking if the anion is basic. AgCl's Cl⁻ is the conjugate base of a strong acid, so it does not react with H+
Why: Students are used to pure water Ksp calculations where no protonation occurs, so they forget that most A⁻ is protonated in acidic solution, meaning [A⁻] < s
Why: Students confuse the order of deprotonation steps and do not connect anion charge to dissociation number
Why: Students forget that buffered solutions fix pH, so small amounts of OH⁻ from dissolution do not change [OH⁻]
Why: Students mix up the direction of equilibrium shift: Q < Ksp means the reaction shifts right to dissolve more solid
Why: Students confuse consumption of the anion with addition of a common ion