pH and solubility — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: The effect of pH on solubility of ionic hydroxides, sulfides, and salts of weak conjugate bases, calculation of molar solubility at fixed pH, qualitative prediction of precipitation, and application of simultaneous Ksp and Ka equilibrium relationships.

You should already know: How to calculate Ksp from molar solubility and vice versa for pure water. How to relate pH, [H+], [OH-], and Ka/Kb for weak acids and bases. How to set up ICE tables for simultaneous equilibria.

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 pH and solubility?

pH and solubility is the study of how hydronium ion concentration alters the solubility of ionic compounds whose anions act as weak Brønsted-Lowry bases. When a sparingly soluble ionic salt dissolves, it releases free anions into solution. If these anions can react with H+ (in acidic solution) or OH- (in basic solution), this reaction pulls or pushes the dissolution equilibrium according to Le Chatelier's principle, changing the total amount of salt that can dissolve.

This topic is part of Unit 7: Equilibrium on the AP Chemistry CED, making up approximately 6% of the unit's exam weight, translating to 2-4% of the total AP exam score. pH-solubility relationships appear in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ typically tests qualitative predictions, while FRQ combines pH-solubility calculations with precipitation separation or real-world context. This topic is critical for understanding phenomena like tooth decay, geochemical mineral formation, and drinking water treatment.

2. Qualitative Effect of pH on Solubility

The core relationship between pH and solubility relies on a key observation: 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: $$\text{MA}(s)\rightleftharpoons {\text{M}}^{+}(aq)+{\text{A}}^{-}(aq)K_{sp}=[{\text{M}}^{+}][A^{-}]$$ In acidic solution (low pH, high [H+]), A- reacts with H+ via ${\text{A}}^{-}(aq)+{\text{H}}^{+}(aq)\rightleftharpoons \text{HA}(aq)$. This reaction 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.

In general: solubility of salts with basic anions increases as pH decreases (the solution becomes more acidic). For salts with anions from strong acids (e.g. Cl-, NO3-, Br-), the anions are negligible bases, do not react with H+, so solubility is unaffected by pH. For metal hydroxides (e.g. Mg(OH)2), which contain OH- as the anion, solubility always increases in acidic solution (H+ reacts with OH- to form water, consuming product) and decreases in basic solution (excess OH- acts as a common ion, shifting equilibrium left).

Worked Example

Predict whether the solubility of each salt increases, decreases, or stays the same when pH is lowered from 7 to 3: (a) CaCO3, (b) AgNO3, (c) Fe(OH)3.

1. Step 1: For each salt, identify if the anion is the conjugate base of a weak acid (i.e., if it is basic).
2. Step 2: (a) CaCO3: The anion is CO3^2-, conjugate base of HCO3-, which is a weak acid. Lower pH = higher [H+] that consumes CO3^2- to form H2CO3. Equilibrium shifts right, so solubility increases.
3. Step 3: (b) AgNO3: The anion is NO3-, conjugate base of strong HNO3. NO3- does not react with H+ to a significant extent, so equilibrium does not shift. Solubility stays the same.
4. Step 4: (c) Fe(OH)3: The anion is OH-, a strong base. Lower pH means excess H+ reacts with OH- to form water, consuming product. Equilibrium shifts right, so solubility increases.

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

When pH is fixed (usually in a buffered solution), we know [H+] directly, which simplifies calculations significantly. The key rule is that the molar solubility s of the salt equals the total concentration of the anion (in all protonation forms) in solution, per the 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.

We relate [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 we can solve directly for s. For metal hydroxides at fixed pH, the calculation is even simpler: pH gives [OH-] directly, so $s=[M^{n+}]=\frac{K_{sp}}{[O{H}^{-}{]}^n}$.

Worked Example

Calculate the molar solubility of BaSO3 in a buffered solution at pH = 5.0. Given $K_{sp}(BaSO3)=8.0\times 10^{-7}$, $K_a(HSO{3}^{-})=6.3\times 10^{-8}$.

1. Step 1: pH = 5.0, so $[H^{+}]=1.0\times 10^{-5}M$. The anion SO3^2- is the conjugate base of HSO3-, so we use Ka for HSO3- to calculate α.
2. Step 2: Calculate α for SO3^2-: $\alpha =\frac{K_a}{K_a+[H^{+}]}=\frac{6.3\times 10^{-8}}{6.3\times 10^{-8}+1.0\times 10^{-5}}\approx 0.00626$.
3. Step 3: Relate Ksp to s. For BaSO3, $s=[B{a}^{2+}]$, $[S{O}_3^{2-}]=\alpha s$, so $K_{sp}=[B{a}^{2+}][S{O}_3^{2-}]=\alpha s^2$.
4. Step 4: Solve for s: $s=\sqrt{\frac{K_{sp}}{\alpha }}=\sqrt{\frac{8.0\times 10^{-7}}{0.00626}}\approx 0.011M$. For comparison, solubility in pure water at pH 7 is ~0.0009 M, which matches the expected increase in acidic solution.

Exam tip: For polyprotic acids, always match the Ka to the charge of the anion. A dianion (e.g. C2O4^2-) requires Ka2, a trianion (e.g. PO4^3-) requires Ka3. Using the wrong Ka is the most common error in these calculations.

4. Selective Precipitation of Ions by pH Adjustment

A common AP Chemistry application of pH-solubility relationships is the separation of mixed metal ions by selective precipitation of their insoluble salts (usually sulfides or hydroxides). Many metal sulfides have very different Ksp values: the least soluble sulfides will precipitate at low pH (where [S2-] is very low), while more soluble sulfides remain dissolved, allowing clean separation of the two metal ions.

The step-by-step method for this problem is: 1) Calculate [S2-] (or [OH-]) at the given pH using the Ka of the parent acid; 2) Calculate the ion product Q for each metal sulfide using the given initial metal concentration; 3) If Q > Ksp, the salt precipitates; if Q < Ksp, the salt remains dissolved. For saturated H2S (a common source of sulfide), [H2S] = 0.1 M at 1 atm, so we can calculate [S2-] directly from pH: $$K_{a1}{K}_{a2}=\frac{[H^{+}{]}^2[S^{2-}]}{[H_{2}S]}=\frac{[H^{+}{]}^2[S^{2-}]}{0.1}⟹[S^{2-}]=\frac{0.1K_{a1}{K}_{a2}}{[H^{+}{]}^2}$$

Worked Example

A solution contains 0.01 M Cd2+ and 0.01 M Zn2+. The solution is saturated with H2S (0.1 M) and adjusted to pH = 1.0. Will CdS precipitate? Will ZnS precipitate? Given $K_{sp}(CdS)=8.0\times 10^{-28}$, $K_{sp}(ZnS)=2.0\times 10^{-20}$, $K_{a1}(H2S)=1.0\times 10^{-7}$, $K_{a2}(H2S)=1.0\times 10^{-14}$.

1. Step 1: Calculate [S2-] at pH 1.0: $[H^{+}]=0.1M$, so $[S^{2-}]=\frac{0.1(1.0\times 10^{-7})(1.0\times 10^{-14})}{(0.1{)}^2}=1.0\times 10^{-20}M$.
2. Step 2: Calculate Q for CdS: $Q=[C{d}^{2+}][S^{2-}]=(0.01)(1.0\times 10^{-20})=1.0\times 10^{-22}$.
3. Step 3: Compare Q to Ksp(CdS): $1.0\times 10^{-22}>8.0\times 10^{-28}$, so CdS precipitates.
4. Step 4: Calculate Q for ZnS: $Q=(0.01)(1.0\times 10^{-20})=1.0\times 10^{-22}<2.0\times 10^{-20}$, so Q < Ksp, so ZnS remains dissolved.
5. Step 5: Conclusion: CdS precipitates at pH 1.0, ZnS does not, so the two ions can be separated.

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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming that salts like AgCl or Ba(NO3)2 become more soluble in acidic solution. 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+. Correct move: Always check the parent acid of the anion first: if the parent acid is strong, pH has no effect on solubility.
• Wrong move: When calculating solubility at fixed pH, set [A-] = s directly instead of accounting for protonation of A-. 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. Correct move: Always write a mass balance that counts all protonation forms of the anion, then relate [A-] to s via α.
• Wrong move: For polyprotic acids, use Ka1 instead of Ka2 for a dianion like C2O4^2-. Why: Students confuse the order of deprotonation steps and do not connect anion charge to dissociation number. Correct move: For an anion with charge -n, always use the nth acid dissociation constant Ka(n) to calculate α.
• Wrong move: For metal hydroxide solubility at fixed pH, include OH- from the salt in an ICE table to calculate [OH-]. Why: Students forget that buffered solutions fix pH, so small amounts of OH- from dissolution do not change [OH-]. Correct move: If pH is fixed, calculate [OH-] directly from pH, do not adjust it for OH- from the dissolving salt.
• Wrong move: Claim that precipitation occurs when Q < Ksp. Why: Students mix up the direction of equilibrium shift: Q < Ksp means the reaction shifts right to dissolve more solid. Correct move: Always remember: solid precipitate forms only when Q > Ksp; Q < Ksp means all ions stay dissolved.
• Wrong move: Double-count the common ion effect when pH changes, adding H+ as a common ion to the dissolution equilibrium. Why: Students confuse consumption of the anion with addition of a common ion. Correct move: Treat protonation as consumption of free anion, not addition of common ion, and only add common ion if it is explicitly added from another source.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following salts will experience the largest increase in molar solubility when the pH of a solution is decreased from 7.0 to 3.0? A) AgBr B) Ca3(PO4)2 C) NaNO3 D) AgCl

Worked Solution: To answer this, we first recall that only salts with basic anions (conjugate bases of weak acids) have pH-dependent solubility. Lowering pH increases [H+], which consumes the basic anion, shifts dissolution equilibrium right, and increases solubility. AgBr (A) has Br-, the conjugate base of strong HBr; NaNO3 (C) has NO3-, conjugate base of strong HNO3; AgCl (D) has Cl-, conjugate base of strong HCl. None of these anions react with H+ significantly, so their solubility does not change. Ca3(PO4)2 (B) has PO4^3-, the conjugate base of the weak acid HPO4^2-, which reacts with H+ in acidic solution. The correct answer is B.

Question 2 (Free Response)

A student studies the solubility of magnesium hydroxide, Mg(OH)2. (a) Write the balanced dissolution equilibrium for Mg(OH)2(s) and the expression for Ksp. (b) Ksp of Mg(OH)2 is 5.6 × 10-12. Calculate the molar solubility of Mg(OH)2 in a buffered solution at pH = 10.0. (c) A student claims that adding a small amount of 0.1 M HCl to the solution will increase the molar solubility of Mg(OH)2. Is the student correct? Justify your answer.

Worked Solution: (a) Balanced equilibrium: $${\text{Mg(OH)}}_2(s)\rightleftharpoons {\text{Mg}}^{2+}(aq)+2{\text{OH}}^{-}(aq)$$ Ksp expression: $$K_{sp}=[{\text{Mg}}^{2+}][{\text{OH}}^{-}{]}^2$$

(b) pH = 10.0, so pOH = 14 - 10.0 = 4.0, so [OH-] = 1.0 × 10-4 M. Since pH is buffered, [OH-] is constant. Molar solubility s = [Mg2+], so rearrange Ksp: $$s=\frac{K_{sp}}{[O{H}^{-}{]}^2}=\frac{5.6\times 10^{-12}}{(1.0\times 10^{-4}{)}^2}=5.6\times 10^{-4}M$$

(c) The student is correct. Adding HCl lowers the pH of the solution, decreasing [OH-]. This causes the ion product Q = [Mg2+][OH-]^2 to drop below Ksp. To re-establish equilibrium, more solid Mg(OH)2 dissolves, increasing molar solubility.

Question 3 (Application / Real-World Style)

Ocean acidification lowers the average pH of surface ocean water from 8.2 to 8.1 due to increased atmospheric CO2. The shells of marine organisms are primarily made of calcium carbonate, CaCO3, with Ksp = 4.5 × 10-7. Ka of HCO3- is 4.7 × 10-11. Calculate the molar solubility of CaCO3 at pH 8.2 and at pH 8.1, and explain what the result means for marine shells.

Worked Solution: For pH 8.2: [H+] = 10-8.2 = 6.3 × 10-9 M. α for CO3^2- = $\frac{K_a}{K_a+[H^{+}]}=\frac{4.7\times 10^{-11}}{4.7\times 10^{-11}+6.3\times 10^{-9}}\approx 0.0074$. $s=\sqrt{K_{sp}/\alpha }=\sqrt{4.5\times 10^{-7}/0.0074}\approx 0.0078M$. For pH 8.1: [H+] = 10-8.1 = 7.9 × 10-9 M. α = $\frac{4.7\times 10^{-11}}{4.7\times 10^{-11}+7.9\times 10^{-9}}\approx 0.0059$. $s=\sqrt{4.5\times 10^{-7}/0.0059}\approx 0.0087M$. The solubility of CaCO3 increases by ~12% when pH drops from 8.2 to 8.1. This means ocean acidification causes the calcium carbonate shells of marine organisms to dissolve at a faster rate, threatening marine organism survival.

7. Quick Reference Cheatsheet

8. What's Next

pH and solubility is the capstone application of multiple simultaneous equilibria in Unit 7: Equilibrium, and it is a direct prerequisite for the next major topic: solubility and complex ion formation. In that topic, you will extend the same simultaneous equilibrium approach you learned here to account for metal ions reacting with ligands to form soluble complex ions, which further alters the solubility of sparingly soluble salts. Without mastering how pH changes alter solubility via Le Chatelier's principle and mass balance of protonated anions, you will struggle to set up correct equilibrium expressions for complex ion problems, which are frequently tested in AP free response. This topic also provides foundational context for understanding environmental chemistry concepts that are common scenario-based problems on the exam.

Next topics to study: Solubility and Complex Ion Equilibria Le Chatelier's Principle Ksp Calculations Acid-Base Equilibria