Ka and pKa Practice Problems

Ka and pKa connect almost every part of acid–base chemistry: pH calculations, buffers, titration curves, indicators and even drug design. These 15 problems build from simple conversions to multi-step questions. Work through them in order, and check each answer before moving on.

Key relationships (at 25 °C):

• Ka = [H⁺][A⁻] ÷ [HA]
• pKa = −log Ka and Ka = 10⁻ᵖᴷᵃ
• For a weak acid: [H⁺] ≈ √(Ka × c), valid when less than about 5% ionises
• pKa + pKb = 14.00 for a conjugate pair
• pH = pKa when the acid is half ionised

Revise the theory in Ka and pKa and pH of weak acids if needed.

Section A: conversions and comparisons

1. An acid has Ka = 3.5 × 10⁻⁴ mol/dm³. What is its pKa?

2. Hydrocyanic acid has pKa = 9.21. What is its Ka?

3. Rank these acids from strongest to weakest:

• Acid P: Ka = 1.8 × 10⁻⁵
• Acid Q: pKa = 3.2
• Acid R: Ka = 6.3 × 10⁻⁵
• Acid S: pKa = 4.9

4. How many times stronger is an acid with pKa 2.5 than one with pKa 5.5?

Section B: from experimental data

5. A 0.100 mol/dm³ solution of a weak monoprotic acid has pH 2.88. Calculate Ka and pKa.

6. A 0.0250 mol/dm³ solution of a weak acid is 3.0% ionised. Calculate Ka and pKa.

7. In a titration of a weak acid with NaOH, the pH at the half-equivalence point is 3.85. What are the pKa and Ka of the acid?

Section C: calculating pH and ionisation

8. Calculate the pH of 0.0500 mol/dm³ benzoic acid (pKa 4.20). Check the approximation.

9. Calculate the percent ionisation of 0.200 mol/dm³ of an acid with Ka = 1.0 × 10⁻⁵.

10. Calculate the pH of 0.0100 mol/dm³ chloroethanoic acid (Ka 1.35 × 10⁻³). Is the square-root approximation valid?

Section D: conjugates and species

11. An acid has pKa 4.20. What are the pKb and Kb of its conjugate base?

12. Hypochlorous acid has pKa 7.53. What fraction is ionised in a solution at pH 7.53? At pH 8.53?

13. Which phosphate species dominates at pH 10? (pKa values of H₃PO₄: 2.15, 7.20, 12.35)

14. An acid HA has pKa 5.0. In a solution buffered at pH 3.0, what is the ratio [A⁻] : [HA]?

15. Aspirin has pKa 3.5. What percentage is in the un-ionised form in the stomach (pH 1.5)?

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Answers

1. pKa = −log(3.5 × 10⁻⁴) = 3.46

2. Ka = 10⁻⁹·²¹ = 6.2 × 10⁻¹⁰ mol/dm³

3. Convert all to pKa: P = 4.74, Q = 3.2, R = 4.20, S = 4.9. Lower pKa = stronger: Q (3.2) > R (4.20) > P (4.74) > S (4.9)

4. The difference is 3 pKa units, and each unit is a factor of 10: 10³ = 1000 times stronger.

5.

• [H⁺] = 10⁻²·⁸⁸ = 1.32 × 10⁻³ mol/dm³ = [A⁻]
• [HA] = 0.100 − 0.00132 = 0.0987 mol/dm³
• Ka = (1.32 × 10⁻³)² ÷ 0.0987 = 1.76 × 10⁻⁵ mol/dm³
• pKa = 4.75 (consistent with ethanoic acid)

6.

• [H⁺] = [A⁻] = 0.030 × 0.0250 = 7.5 × 10⁻⁴ mol/dm³
• [HA] = 0.0250 − 0.00075 = 0.02425 mol/dm³
• Ka = (7.5 × 10⁻⁴)² ÷ 0.02425 = 2.3 × 10⁻⁵ mol/dm³
• pKa = 4.63

7. At half-equivalence, pH = pKa, so pKa = 3.85 and Ka = 10⁻³·⁸⁵ = 1.4 × 10⁻⁴ mol/dm³.

8.

• Ka = 10⁻⁴·²⁰ = 6.31 × 10⁻⁵
• [H⁺] ≈ √(6.31 × 10⁻⁵ × 0.0500) = √(3.16 × 10⁻⁶) = 1.78 × 10⁻³ mol/dm³
• pH = 2.75
• Check: 1.78 × 10⁻³ ÷ 0.0500 = 3.6%, below 5%, so the approximation is valid.

9.

• [H⁺] ≈ √(1.0 × 10⁻⁵ × 0.200) = √(2.0 × 10⁻⁶) = 1.41 × 10⁻³ mol/dm³
• % ionised = 1.41 × 10⁻³ ÷ 0.200 × 100 = 0.71%

10.

• Shortcut: √(1.35 × 10⁻³ × 0.0100) = √(1.35 × 10⁻⁵) = 3.67 × 10⁻³, which is 37% of c. Not valid.
• Solve x² + (1.35 × 10⁻³)x − 1.35 × 10⁻⁵ = 0: x = [−1.35 × 10⁻³ + √((1.35 × 10⁻³)² + 5.4 × 10⁻⁵)] ÷ 2 = 3.06 × 10⁻³ mol/dm³
• pH = 2.51

11. pKb = 14.00 − 4.20 = 9.80; Kb = 10⁻⁹·⁸⁰ = 1.6 × 10⁻¹⁰

12.

• At pH 7.53 = pKa: [A⁻] = [HA], so 50% ionised.
• At pH 8.53 (one unit above pKa): [A⁻] : [HA] = 10 : 1, so 10 ÷ 11 = 91% ionised.

13. pH 10 lies between pKa₂ (7.20) and pKa₃ (12.35), so HPO₄²⁻ dominates.

14. log([A⁻] ÷ [HA]) = 3.0 − 5.0 = −2.0 → ratio = 1 : 100 (mostly HA).

15. log([A⁻] ÷ [HA]) = 1.5 − 3.5 = −2 → [A⁻] : [HA] = 1 : 100. Un-ionised fraction = 100 ÷ 101 = 99%.

A strategy for “find Ka from data” questions

Questions 5 and 6 are the most common type in exams, and they always follow the same four steps. First, turn whatever you’re given (a pH, a percentage ionised, or a conductivity reading) into the equilibrium concentration of H⁺. Second, recognise that for a monoprotic weak acid with no added salt, [A⁻] equals [H⁺], because every molecule that ionises produces one of each. Third, find the equilibrium concentration of HA by subtracting the ionised amount from the starting concentration. Many textbooks allow you to skip this subtraction when ionisation is small, but doing it never costs marks. Finally, substitute into Ka = [H⁺][A⁻] ÷ [HA], and convert to pKa if asked.

Writing these four steps as a mini ICE table makes the working easy for an examiner to follow, and it catches the most common mistake: forgetting that [A⁻] and [H⁺] are equal.

Why pKa matters beyond exams

pKa values are used every day in real chemistry. Pharmaceutical chemists use them to predict how much of a drug is ionised in the stomach, the blood and inside cells, which controls how well it’s absorbed and where it goes. Biochemists choose buffers for enzyme experiments by matching pKa to the pH they need. Environmental chemists use the pKa of carbonic acid to understand ocean chemistry, and the pKa of hypochlorous acid to explain why swimming pool pH matters for disinfection. Question 12 above is exactly the calculation a pool engineer would use. And question 15 is the reasoning pharmacologists use to explain why aspirin is absorbed partly through the stomach lining.

Tips for Ka and pKa questions

• Always check the 5% rule after using the square-root shortcut. Relatively strong weak acids (pKa below about 3) often fail it at low concentrations.
• Keep an extra significant figure in intermediate steps; logs amplify rounding errors.
• Link pKa to pH. If pH = pKa, it’s half ionised; one unit either side is 10 : 1.
• Watch units. Ka has units of mol/dm³; pKa has no units.
• Sense-check your answer. Most weak acids in school questions have pKa values between about 2 and 10; a result outside that range usually signals a slip.

Common mistakes

1. Thinking a higher pKa means a stronger acid (it’s the reverse).
2. Forgetting to subtract the ionised amount from [HA] when calculating Ka from pH.
3. Using Ka directly as [H⁺].
4. Confusing the pKa of an acid with the pKb of its conjugate base.

Key takeaways

• pKa = −log Ka; lower pKa means a stronger acid, by a factor of 10 per unit.
• Ka can be found from a measured pH or percent ionisation of a known concentration.
• pH = pKa at half-equivalence in a titration.
• The ratio [A⁻] : [HA] = 10^(pH − pKa) tells you which form dominates at any pH.
• Check answers with the pH calculator.