pH and Solubility

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

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.

📘 Definition

pH-Dependent Solubility

A change in the molar solubility of an ionic salt caused by pH-dependent protonation or deprotonation of the salt's anion, which shifts the dissolution equilibrium per Le Chatelier's principle

Example:

Calcium carbonate (the main component of marine shells) solubility increases as ocean pH decreases from ocean acidification

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 $K_{s p}$ . 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)

📐 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, check if the anion is the conjugate base of a weak acid (i.e., if it is basic).
2
(a) CaCO₃: The anion is CO₃²⁻, conjugate base of the weak acid HCO₃⁻. Lower pH increases [H+], which consumes CO₃²⁻ to form H₂CO₃. Equilibrium shifts right, so solubility increases.
3
(b) AgNO₃: The anion is NO₃⁻, conjugate base of strong HNO₃. NO₃⁻ does not react with H+ significantly, so equilibrium does not shift. Solubility stays the same.
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★★★☆☆⏱ 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, $s = [M^{+}] = [A^{-}] + [H A]$ , 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: $K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]}$ . 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 $K_{s p} = [M^{+}] [A^{-}] = s \times (α_{A^{-}} s) = α_{A^{-}} s^{2}$ , which can be solved directly for $s$ . For metal hydroxides at fixed pH, the calculation is even simpler: pH gives [OH⁻] directly, so $s = [M^{n +}] = \frac{K _{s p}}{[ O H ^{-} ] ^{n}}$ .

📐 Worked Example

Calculate the molar solubility of BaSO₃ in a buffered solution at pH = 5.0. Given $K_{s p} (B a S O_{3}) = 8.0 \times 1 0^{- 7}$ , $K_{a} (H S O_{3}^{-}) = 6.3 \times 1 0^{- 8}$

1
Step 1: pH = 5.0, so
2
$[H^+] = 1.0 \times 10^{-5} M$
3
The anion SO₃²⁻ is the conjugate base of HSO₃⁻, so we use Ka for HSO₃⁻ to calculate α.
4
Step 2: Calculate α for SO₃²⁻:
5
$\alpha = \frac{K_a}{K_a + [H^+]} = \frac{6.3 \times 10^{-8}}{6.3 \times 10^{-8} + 1.0 \times 10^{-5}} ≈ 0.00626$
6
Step 3: Relate Ksp to s. For BaSO₃, $s = [B a^{2 +}]$ and $[S O_{3}^{2 -}] = α s$ , so $K_{s p} = [B a^{2 +}] [S O_{3}^{2 -}] = α s^{2}$ .
7
Step 4: Solve for s:
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$s = \sqrt{\frac{K_{sp}}{\alpha}} = \sqrt{\frac{8.0 \times 10^{-7}}{0.00626}} ≈ 0.011 M$
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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. 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}

📐 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 $K_{s p} (C d S) = 8.0 \times 1 0^{- 28}$ , $K_{s p} (Z n S) = 2.0 \times 1 0^{- 20}$ , $K_{a 1} (H_{2} S) = 1.0 \times 1 0^{- 7}$ , $K_{a 2} (H_{2} S) = 1.0 \times 1 0^{- 14}$

1
Step 1: Calculate [S²⁻] at pH 1.0: [H+] = 0.1 M
2
$[S^{2-}] = \frac{0.1(1.0 \times 10^{-7})(1.0 \times 10^{-14})}{(0.1)^2} = 1.0 \times 10^{-20} M$
3
Step 2: Calculate Q for CdS:
4
$Q = [Cd^{2+}][S^{2-}] = (0.01)(1.0 \times 10^{-20}) = 1.0 \times 10^{-22}$
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Step 3: Compare Q to Ksp(CdS): $1.0 \times 1 0^{- 22} > 8.0 \times 1 0^{- 28}$ , so CdS precipitates.
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Step 4: Calculate Q for ZnS:
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$Q = (0.01)(1.0 \times 10^{-20}) = 1.0 \times 10^{-22}$
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Step 5: Compare Q to Ksp(ZnS): $1.0 \times 1 0^{- 22} < 2.0 \times 1 0^{- 20}$ , so ZnS remains dissolved.
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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

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 C₂O₄²⁻

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. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
General pH effect	Solubility of basic anion salts increases as pH decreases	Salts with anions from strong acids have pH-independent solubility
Mass balance (1:1 salt MA)	$s = [M^{+}] = [A^{-}] + [H A]$	Counts all dissolved anion, regardless of protonation state
Fraction of deprotonated anion	$α_{A^{-}} = \frac{K _{a}}{K _{a} + [ H ^{+} ]}$	For monoprotic conjugate bases; for n-charged anions, use Ka(n)
Ksp for 1:1 salt at fixed pH	$K_{s p} = α_{A^{-}} s^{2}$	Works for any fixed pH buffered solution
M(OH)n solubility at fixed pH	$s = \frac{K _{s p}}{[ O H ^{-} ] ^{n}}$	$[O H^{-}] = 1 0^{- (14 - p H)}$ from fixed pH
[S²⁻] in saturated H₂S	$[S^{2 -}] = \frac{0.1 K _{a 1} K _{a 2}}{[ H ^{+} ] ^{2}}$	Only valid for 0.1 M saturated H₂S at 1 atm
Precipitation rule	Precipitate forms if $Q > K_{s p}$	Q = ion product of dissolved ions
Selective separation	More insoluble salts precipitate at lower pH	Used to separate mixtures of metal 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 · AP Chem MCQ
Qualitative pH solubility prediction
2022 · AP Chem FRQ
Solubility calculation at fixed pH

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.