Salt Hydrolysis Practice Problems

The pH of a salt solution is a classic place where students discover that “salt” doesn’t mean “neutral”. These 12 problems move from quick classification to full calculations, finishing with the equivalence-point pH in titrations, which is where salt hydrolysis matters most in practice.

Useful relationships at 25 °C:

• Ka × Kb = Kw = 1.0 × 10⁻¹⁴, so pKa + pKb = 14.00 for a conjugate pair
• Weak acid: [H⁺] ≈ √(Ka × c); weak base: [OH⁻] ≈ √(Kb × c)
• Amphiprotic ion: pH ≈ ½(pKa₁ + pKa₂)

Data: Ka(HF) = 6.8 × 10⁻⁴; Ka(HCOOH) = 1.8 × 10⁻⁴; Ka(CH₃COOH) = 1.8 × 10⁻⁵; Ka(HCN) = 6.2 × 10⁻¹⁰; Kb(NH₃) = 1.8 × 10⁻⁵; carbonic acid pKa₁ = 6.35, pKa₂ = 10.33 (Ka₂ = 4.7 × 10⁻¹¹).

For the theory, see salt hydrolysis.

Show your working clearly in every answer, including the constant you used.

Section A: classify

1. Predict whether each 0.1 mol/dm³ solution is acidic, neutral or basic: (a) KNO₃ (b) NH₄Cl (c) Na₂CO₃ (d) CH₃COONa (e) FeCl₃ (f) NaF (g) AlCl₃ (h) K₂S (i) NaHSO₄ (j) CH₃COONH₄

2. Which is more basic at the same concentration: sodium cyanide or sodium fluoride? Explain using Ka values.

Section B: calculate

3. Calculate the pH of 0.20 mol/dm³ NaF.

4. Calculate the pH of 0.050 mol/dm³ NH₄NO₃.

5. Calculate the pH of 0.050 mol/dm³ Na₂CO₃. (Check whether you need the quadratic.)

6. [Al(H₂O)₆]³⁺ has pKa 5.0. Calculate the pH of 0.10 mol/dm³ AlCl₃.

7. Estimate the pH of a solution of sodium hydrogencarbonate.

Section C: working backwards

8. A 0.10 mol/dm³ solution of the sodium salt NaA has pH 9.00. Find Ka of the weak acid HA.

9. Calculate Kb for the ethanoate ion and pKa for the ammonium ion.

Section D: titrations

10. 25.0 cm³ of 0.10 mol/dm³ methanoic acid is titrated with 0.10 mol/dm³ NaOH. Calculate the pH at the equivalence point.

11. 20.0 cm³ of 0.20 mol/dm³ ammonia is titrated with 0.20 mol/dm³ HCl. Calculate the pH at the equivalence point.

12. Explain why ammonium ethanoate solution is almost exactly neutral.

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Answers

1. (a) neutral: strong acid (HNO₃) + strong base (KOH). (b) acidic: NH₄⁺ is the conjugate acid of a weak base. (c) basic: CO₃²⁻ is the conjugate base of the very weak acid HCO₃⁻. (d) basic: ethanoate is the conjugate base of a weak acid. (e) acidic: hydrated Fe³⁺ releases H⁺. (f) slightly basic: F⁻ is the conjugate base of the weak acid HF. (g) acidic: hydrated Al³⁺ releases H⁺. (h) strongly basic: S²⁻ is a strong base in water. (i) acidic: HSO₄⁻ is a fairly strong acid (pKa about 2). (j) neutral (approximately): see question 12.

2. Sodium cyanide. HCN (Ka 6.2 × 10⁻¹⁰) is a much weaker acid than HF (Ka 6.8 × 10⁻⁴), so its conjugate base CN⁻ is a much stronger base than F⁻. Kb(CN⁻) = 1.6 × 10⁻⁵ vs Kb(F⁻) = 1.5 × 10⁻¹¹.

3.

• Kb(F⁻) = 1.0 × 10⁻¹⁴ ÷ 6.8 × 10⁻⁴ = 1.47 × 10⁻¹¹
• [OH⁻] = √(1.47 × 10⁻¹¹ × 0.20) = 1.71 × 10⁻⁶ mol/dm³
• pOH = 5.77, pH = 8.23

4.

• Ka(NH₄⁺) = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰
• [H⁺] = √(5.56 × 10⁻¹⁰ × 0.050) = 5.27 × 10⁻⁶ mol/dm³
• pH = 5.28

5.

• Kb(CO₃²⁻) = 1.0 × 10⁻¹⁴ ÷ 4.7 × 10⁻¹¹ = 2.13 × 10⁻⁴
• Shortcut: √(2.13 × 10⁻⁴ × 0.050) = 3.26 × 10⁻³, which is 6.5% of c, so use the quadratic.
• x² + (2.13 × 10⁻⁴)x − 1.064 × 10⁻⁵ = 0 → x = [OH⁻] = 3.16 × 10⁻³ mol/dm³
• pOH = 2.50, pH = 11.50

6.

• Ka = 10⁻⁵·⁰ = 1.0 × 10⁻⁵
• [H⁺] = √(1.0 × 10⁻⁵ × 0.10) = 1.0 × 10⁻³ mol/dm³
• pH = 3.00

7. HCO₃⁻ is amphiprotic, with pKa values of 6.35 (for H₂CO₃) and 10.33 (for HCO₃⁻). pH ≈ ½(6.35 + 10.33) = 8.34, roughly independent of concentration.

8.

• pOH = 14.00 − 9.00 = 5.00, so [OH⁻] = 1.0 × 10⁻⁵ mol/dm³
• Kb(A⁻) = (1.0 × 10⁻⁵)² ÷ 0.10 = 1.0 × 10⁻⁹
• Ka(HA) = 1.0 × 10⁻¹⁴ ÷ 1.0 × 10⁻⁹ = 1.0 × 10⁻⁵ (pKa 5.00)

9.

• Kb(CH₃COO⁻) = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰
• pKa(NH₄⁺) = 14.00 − pKb(NH₃) = 14.00 − 4.74 = 9.26 (often quoted as 9.25)

10.

• At equivalence, all methanoic acid is converted to methanoate: 2.5 × 10⁻³ mol in 50.0 cm³, so [HCOO⁻] = 0.050 mol/dm³.
• Kb = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁴ = 5.56 × 10⁻¹¹
• [OH⁻] = √(5.56 × 10⁻¹¹ × 0.050) = 1.67 × 10⁻⁶ mol/dm³
• pOH = 5.78, pH = 8.22

11.

• At equivalence, [NH₄⁺] = 4.0 × 10⁻³ mol ÷ 0.0400 dm³ = 0.10 mol/dm³.
• Ka(NH₄⁺) = 5.56 × 10⁻¹⁰
• [H⁺] = √(5.56 × 10⁻¹¹) = 7.45 × 10⁻⁶ mol/dm³
• pH = 5.13

12. Both ions hydrolyse: NH₄⁺ as a weak acid (Ka = 5.6 × 10⁻¹⁰) and CH₃COO⁻ as a weak base (Kb = 5.6 × 10⁻¹⁰). Because these two constants are almost exactly equal, the H⁺ produced by one is balanced by the OH⁻ produced by the other, so the solution stays close to pH 7.

Why these answers matter

Questions 10 and 11 show why indicator choice depends on the titration. A weak acid–strong base equivalence point near pH 8.2 needs phenolphthalein; a strong acid–weak base equivalence point near pH 5.1 needs methyl orange or methyl red. See how to choose an indicator and titration curves.

Salt hydrolysis in everyday life

These calculations aren’t just exam exercises. The same reasoning explains why baking soda neutralises indigestion (hydrogencarbonate is mildly basic), why ammonium fertilisers gradually acidify soil (the ammonium ion is a weak acid), why aluminium sulfate turns hydrangeas blue (hydrated aluminium ions are acidic), and why soap feels slippery (soap is the salt of a weak fatty acid and a strong base, so its solutions are alkaline). Whenever a label lists a “salt” as an acidity regulator, as many food and cosmetic products do, salt hydrolysis is the chemistry behind it.

Strategy summary

1. Split the salt into its ions.
2. Discard spectators: Group 1 cations, heavier Group 2 cations, and anions of strong acids (Cl⁻, Br⁻, I⁻, NO₃⁻, ClO₄⁻).
3. Identify what’s left: a weak base (anion of a weak acid), a weak acid (NH₄⁺, hydrated small highly charged metal ions) or both.
4. Find the right constant using Ka × Kb = Kw if needed.
5. Calculate as for any weak acid or weak base, checking the 5% rule.

Common mistakes

• Using the parent acid’s Ka for the anion without converting to Kb.
• Forgetting that the concentration after a titration is diluted by the combined volume.
• Assuming every salt with a metal is neutral (Fe³⁺ and Al³⁺ aren’t).
• Treating NaHSO₄ like NaHCO₃: one is acidic, the other slightly basic.
• Forgetting the Group 2 exception: magnesium and beryllium salts can be slightly acidic, because their small, doubly charged ions hydrolyse a little, whereas calcium, strontium and barium salts are neutral.
• Rounding intermediate Kb values too early, which can shift the final pH by a few hundredths.

Key takeaways

• Anions of weak acids make salts basic; cations of weak bases and small, highly charged metal ions make them acidic.
• Use Ka × Kb = Kw to convert between a parent acid’s Ka and its anion’s Kb.
• Treat the hydrolysing ion as an ordinary weak acid or base in calculations.
• Equivalence-point pH in a titration is simply the pH of the salt solution formed.
• Check results with the pH calculator.