Salt Hydrolysis & Polyprotic Acids
- Predict whether a salt solution is acidic, basic, or neutral from the origin of its ions
- Calculate the pH of a salt solution using Kb = Kw/Ka and an ICE table
- Explain stepwise dissociation of a polyprotic acid where Ka1 ≫ Ka2
A dissolved salt is not automatically neutral
When a salt dissolves, its ions can react with water — hydrolysis — and shift the pH. The trick is to trace each ion back to the acid and base that formed the salt. An ion that is the conjugate base of a weak acid (like CH₃COO⁻ from acetic acid) acts as a weak base and raises pH. An ion that is the conjugate acid of a weak base (like NH₄⁺ from ammonia) acts as a weak acid and lowers pH. Ions from strong acids or bases (Na⁺, K⁺, Cl⁻, NO₃⁻) are spectators — they do not hydrolyze and leave pH unchanged.
The four cases
Combine the two ions to predict the solution. Strong acid + strong base (NaCl) → both ions are spectators → neutral. Weak acid + strong base (CH₃COONa) → the anion is a weak base → basic. Strong acid + weak base (NH₄Cl) → the cation is a weak acid → acidic. Weak acid + weak base (NH₄CH₃COO) → compare Ka of the cation with Kb of the anion; the larger one wins. Always ask: which ion came from the weak partner? That ion controls the pH.
Polyprotic acids give up protons one at a time
A polyprotic acid (H₂CO₃, H₃PO₄, H₂SO₃) has more than one ionizable proton, and each ionization has its own constant: Ka1, Ka2, Ka3. Pulling a positive H⁺ off an already-negative ion is progressively harder, so Ka1 ≫ Ka2 ≫ Ka3 — typically by factors of 10⁴ to 10⁵. The practical consequence: for pH calculations you treat the acid as if only the first ionization matters, because the second step contributes a negligible amount of additional H⁺. Their titration curves show one steep jump per proton.
Find the pH of 0.10 M sodium acetate, CH₃COONa. Acetic acid has Ka = 1.8 × 10⁻⁵.
- 1.Na⁺ is a spectator; acetate CH₃COO⁻ is the conjugate base of a weak acid, so it hydrolyzes: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻.
- 2.Get Kb from Kw/Ka: Kb = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰.
- 3.ICE with x = [OH⁻] gives Kb = x² / (0.10 − x). Since Kb is tiny, approximate 0.10 − x ≈ 0.10.
- 4.x² = Kb · C = (5.6 × 10⁻¹⁰)(0.10) = 5.6 × 10⁻¹¹.
- 5.x = √(5.6 × 10⁻¹¹) = 7.5 × 10⁻⁶ M = [OH⁻].
- 6.pOH = −log(7.5 × 10⁻⁶) = 5.13, so pH = 14 − 5.13 = 8.87.
A salt-hydrolysis problem is just a weak-acid or weak-base equilibrium in disguise. The only extra step is getting the right K from Kb = Kw/Ka (for an anion) or Ka = Kw/Kb (for a cation). After that, it is the same ICE table you already know.
Is an aqueous solution of ammonium chloride, NH₄Cl, acidic, basic, or neutral?
The fluoride ion F⁻ is the conjugate base of HF (Ka = 6.8 × 10⁻⁴). What is Kb for F⁻?
For carbonic acid, Ka1 = 4.3 × 10⁻⁷ and Ka2 = 4.7 × 10⁻¹¹. Which statement about a solution of H₂CO₃ is correct?
On the exam, first sort each salt ion into "spectator" or "hydrolyzes." Group 1/Group 2 cations and the anions of strong acids (Cl⁻, Br⁻, I⁻, NO₃⁻, ClO₄⁻) are spectators. Any leftover ion traced to a weak acid or weak base is the one that sets the pH.
Answer the 3 checkpoints as you read.
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