pH Calculations Practice Problems with Answers

The only way to get fast and accurate at pH calculations is to do lots of them. This set starts gently and builds up to the kind of multi-step questions that appear at the end of exam papers.

Try each question before reading the answer. Assume a temperature of 25 °C (so Kw = 1.0 × 10⁻¹⁴ and pH + pOH = 14.00) unless a question says otherwise. If you need a refresher on the methods, see how to calculate pH.

A suggestion before you start: write every step down, even the ones that feel obvious. In exams, method marks are awarded for showing the concentration you used and the equation you applied, so a slip with the calculator doesn’t have to cost you the whole question. Give pH values to two decimal places unless told otherwise, and keep units on every concentration.

Section A: strong acids

1. Calculate the pH of 0.0010 mol/dm³ hydrochloric acid.

2. Calculate the pH of 0.20 mol/dm³ nitric acid.

3. Calculate the pH of 0.045 mol/dm³ hydrochloric acid.

4. A solution of HCl has a pH of 3.60. What is its concentration?

5. How many times greater is [H⁺] in a solution of pH 2 than in a solution of pH 5?

Section B: strong bases

6. Calculate the pH of 0.015 mol/dm³ potassium hydroxide.

7. Calculate the pH of 0.0025 mol/dm³ calcium hydroxide.

8. A sodium hydroxide solution has a pH of 11.40. What is [OH⁻]?

9. 2.00 g of sodium hydroxide is dissolved in water and made up to 500 cm³. Calculate the pH. (Mr of NaOH = 40.0)

Section C: from mass and by dilution

10. 0.365 g of hydrogen chloride gas is dissolved in water to make 250 cm³ of solution. Calculate the pH. (Mr of HCl = 36.5)

11. 10.0 cm³ of 0.50 mol/dm³ HCl is diluted with water to a total volume of 250 cm³. What is the pH of the diluted solution?

12. A strong acid has a pH of 1.0. It is diluted by adding water until its volume is 1000 times larger. What is the new pH?

Section D: mixing acids and bases

13. 30.0 cm³ of 0.200 mol/dm³ HCl is mixed with 25.0 cm³ of 0.200 mol/dm³ NaOH. Calculate the pH of the mixture.

14. 20.0 cm³ of 0.100 mol/dm³ HNO₃ is mixed with 25.0 cm³ of 0.100 mol/dm³ KOH. Calculate the pH.

15. What volume of 0.100 mol/dm³ NaOH must be added to 25.0 cm³ of 0.100 mol/dm³ HCl to reach pH 7?

Section E: stretch questions

16. Calculate the pH of 1.0 × 10⁻⁸ mol/dm³ HCl.

17. A weak acid HA has Ka = 1.0 × 10⁻⁵. Calculate the pH of a 0.050 mol/dm³ solution.

18. At 50 °C, Kw = 5.5 × 10⁻¹⁴. Calculate the pH of pure water at this temperature. Is it acidic, neutral or alkaline?

19. Two solutions have pH values of 3.0 and 4.0. Equal volumes are mixed. Estimate the pH of the mixture.

20. 50.0 cm³ of an HCl solution of pH 1.30 is added to 50.0 cm³ of an NaOH solution of pH 12.70. What is the pH of the result?

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Answers with working

1. [H⁺] = 0.0010 → pH = −log(0.0010) = 3.00

2. pH = −log(0.20) = 0.70. pH values below 1 are perfectly possible for concentrated strong acids.

3. pH = −log(0.045) = 1.35

4. [H⁺] = 10⁻³·⁶⁰ = 2.5 × 10⁻⁴ mol/dm³. For a strong monoprotic acid, that’s also the acid concentration.

5. The difference is 3 pH units, and each unit is a factor of 10, so 10³ = 1000 times greater.

6. [OH⁻] = 0.015 → pOH = −log(0.015) = 1.82 → pH = 14.00 − 1.82 = 12.18

7. Ca(OH)₂ releases two OH⁻ per formula unit: [OH⁻] = 0.0050 → pOH = 2.30 → pH = 11.70

8. pOH = 14.00 − 11.40 = 2.60 → [OH⁻] = 10⁻²·⁶⁰ = 2.5 × 10⁻³ mol/dm³

9. Moles NaOH = 2.00 ÷ 40.0 = 0.0500 mol. Concentration = 0.0500 ÷ 0.500 = 0.100 mol/dm³. pOH = 1.00 → pH = 13.00

10. Moles HCl = 0.365 ÷ 36.5 = 0.0100 mol. Concentration = 0.0100 ÷ 0.250 = 0.0400 mol/dm³. pH = −log(0.0400) = 1.40

11. Moles H⁺ = 0.50 × 0.0100 = 0.0050 mol. New concentration = 0.0050 ÷ 0.250 = 0.020 mol/dm³. pH = 1.70

12. Diluting a strong acid by 10 raises the pH by 1. By 1000 = 10³, the pH rises by 3: pH 4.0. (This shortcut works because the result is still well away from pH 7.)

13.

• H⁺: 0.200 × 0.0300 = 0.00600 mol
• OH⁻: 0.200 × 0.0250 = 0.00500 mol
• Excess H⁺ = 0.00100 mol in a total of 55.0 cm³ (0.0550 dm³)
• [H⁺] = 0.00100 ÷ 0.0550 = 0.0182 mol/dm³
• pH = 1.74

14.

• H⁺: 0.100 × 0.0200 = 0.00200 mol
• OH⁻: 0.100 × 0.0250 = 0.00250 mol
• Excess OH⁻ = 0.00050 mol in 0.0450 dm³ → [OH⁻] = 0.0111 mol/dm³
• pOH = 1.95 → pH = 12.05

15. At pH 7, moles of OH⁻ added = moles of H⁺ present = 0.100 × 0.0250 = 0.00250 mol. Volume of NaOH = 0.00250 ÷ 0.100 = 0.0250 dm³ = 25.0 cm³. (Strong acid + strong base in equal moles gives exactly pH 7 at 25 °C.)

16. You can’t ignore water here. Let [H⁺] = x. The acid supplies 1.0 × 10⁻⁸ and water supplies the rest, with x × [OH⁻] = 1.0 × 10⁻¹⁴ and [OH⁻] = x − 1.0 × 10⁻⁸. Solving the quadratic x² − (1.0 × 10⁻⁸)x − 1.0 × 10⁻¹⁴ = 0 gives x = 1.05 × 10⁻⁷ mol/dm³. pH = 6.98. Slightly acidic, not pH 8.

17. Using [H⁺] ≈ √(Ka × c) = √(1.0 × 10⁻⁵ × 0.050) = √(5.0 × 10⁻⁷) = 7.1 × 10⁻⁴ mol/dm³. pH = 3.15. Check the approximation: 7.1 × 10⁻⁴ is about 1.4% of 0.050, so ignoring the small amount of HA that ionised is fine.

18. In pure water [H⁺] = [OH⁻], so [H⁺] = √(5.5 × 10⁻¹⁴) = 2.35 × 10⁻⁷ mol/dm³. pH = 6.63. The water is still neutral, because [H⁺] still equals [OH⁻]. Neutral only means pH 7 at 25 °C.

19. [H⁺] values are 1.0 × 10⁻³ and 1.0 × 10⁻⁴. Mixing equal volumes halves each concentration: total [H⁺] = (1.0 × 10⁻³ + 1.0 × 10⁻⁴) ÷ 2 = 5.5 × 10⁻⁴ mol/dm³. pH = 3.26. Notice it’s not the average of 3 and 4. Because pH is logarithmic, the more acidic solution dominates. (This assumes both are strong acids.)

20. pH 1.30 → [H⁺] = 10⁻¹·³⁰ = 0.0501 mol/dm³. pH 12.70 → pOH = 1.30 → [OH⁻] = 0.0501 mol/dm³. Equal volumes and equal concentrations mean equal moles, so they neutralise exactly. pH = 7.00.

How did you do?

• 1–9 correct: you’ve got the core skills. Focus on mixing problems next.
• 10–15 correct: solid. The stretch questions test real understanding of what pH means.
• 16–20 correct: excellent. You understand the limits of the simple formulas, which is exactly what top-grade questions test.

The five mistakes that cost the most marks

1. Forgetting the ×2 for Ca(OH)₂, Ba(OH)₂ and (fully dissociated) H₂SO₄.
2. Dividing by one volume instead of the total after mixing.
3. Averaging pH values instead of averaging concentrations.
4. Stopping at pOH and giving it as the pH.
5. Claiming a very dilute acid has pH above 7.

Keep practising

• Check any answer with the pH calculator.
• Revise the theory in the pH scale explained and strong vs weak acids.
• For titration-style questions, try molarity explained and the molarity calculator.