Titration Practice Problems with Worked Answers

Titration calculations reward practice more than almost any other topic. The method is always the same, but the questions dress it up in different ways. This set of 20 problems covers every common variation, from single-step questions to multi-step problems of the kind found at the end of exam papers.

Before you start, it’s worth reading titration calculations for the core method:

1. Moles of the known solution = concentration × volume (dm³).
2. Use the balanced equation’s mole ratio.
3. Find what’s asked: concentration, mass, molar mass or purity.

Use Ar values: H 1.0, C 12.0, O 16.0, Na 23.0, Mg 24.3, S 32.1, Cl 35.5, K 39.1, Ca 40.1.

Level 1: one-step, 1 : 1 ratio

1. 25.0 cm³ of NaOH is neutralised by 22.40 cm³ of 0.100 mol/dm³ HCl. Find [NaOH].

2. 20.0 cm³ of HNO₃ is neutralised by 18.60 cm³ of 0.150 mol/dm³ KOH. Find [HNO₃].

3. What volume of 0.250 mol/dm³ HCl neutralises 25.0 cm³ of 0.200 mol/dm³ NaOH?

4. 25.0 cm³ of 0.0800 mol/dm³ NaOH needs 21.25 cm³ of ethanoic acid (CH₃COOH). Find the concentration of the ethanoic acid.

Level 2: other mole ratios

5. 25.0 cm³ of H₂SO₄ is neutralised by 24.20 cm³ of 0.100 mol/dm³ NaOH. Find [H₂SO₄].

6. 20.0 cm³ of 0.0500 mol/dm³ Ca(OH)₂ is neutralised by HCl. The titre is 16.70 cm³. Find [HCl].

7. 25.0 cm³ of 0.0500 mol/dm³ Na₂CO₃ reacts completely with 0.100 mol/dm³ HCl to form CO₂. What titre is expected?

8. What volume of 0.200 mol/dm³ NaOH fully neutralises 10.0 cm³ of 0.150 mol/dm³ H₃PO₄?

Level 3: dilutions and aliquots

9. 25.0 cm³ of a concentrated HCl solution is diluted to 250.0 cm³. A 25.0 cm³ portion of the diluted solution needs 23.60 cm³ of 0.200 mol/dm³ NaOH. Find the concentration of the original HCl.

10. A 10.0 cm³ sample of vinegar is diluted to 100.0 cm³. A 25.0 cm³ portion needs 20.80 cm³ of 0.100 mol/dm³ NaOH. Find the concentration of ethanoic acid in the original vinegar, in mol/dm³ and in g/dm³.

11. 2.12 g of anhydrous sodium carbonate is dissolved and made up to 250.0 cm³. A 25.0 cm³ portion needs 20.00 cm³ of HCl to reach the methyl orange end point. Find [HCl].

Level 4: finding formulas and molar masses

12. 1.575 g of a hydrated diprotic acid, H₂C₂O₄·xH₂O, is made up to 250.0 cm³. 25.0 cm³ portions need 25.00 cm³ of 0.100 mol/dm³ NaOH. Find the molar mass and the value of x.

13. 0.600 g of an unknown monoprotic acid needs 20.00 cm³ of 0.250 mol/dm³ NaOH. Find its molar mass.

14. 4.29 g of hydrated sodium carbonate, Na₂CO₃·xH₂O, is made up to 250.0 cm³. 25.0 cm³ needs 30.00 cm³ of 0.100 mol/dm³ HCl (to CO₂). Find x.

Level 5: back titrations and purity

15. 1.00 g of impure calcium carbonate is added to 50.0 cm³ of 0.500 mol/dm³ HCl (excess). The unreacted acid needs 12.00 cm³ of 0.500 mol/dm³ NaOH. Find the percentage purity of the calcium carbonate.

16. An antacid tablet containing magnesium hydroxide is dissolved in 50.0 cm³ of 0.200 mol/dm³ HCl. The excess acid needs 15.00 cm³ of 0.100 mol/dm³ NaOH. Find the mass of Mg(OH)₂ in the tablet.

17. 0.250 g of an ammonium salt is warmed with 50.0 cm³ of 0.100 mol/dm³ NaOH (excess) until all the ammonia is driven off. The remaining NaOH needs 17.40 cm³ of 0.100 mol/dm³ HCl. Find the percentage by mass of nitrogen in the salt.

Level 6: stretch

18. 25.0 cm³ of a mixture of HCl and H₂SO₄ is titrated with 0.100 mol/dm³ NaOH; the titre is 30.00 cm³. A separate 25.0 cm³ sample, treated with excess BaCl₂, gives 0.2334 g of BaSO₄ (Mr 233.4). Find the concentrations of HCl and H₂SO₄.

19. A student forgets to rinse the burette with NaOH after washing it with water. Explain the effect on the calculated concentration of the acid in the flask.

20. 25.0 cm³ of 0.100 mol/dm³ ethanoic acid is titrated with 0.100 mol/dm³ NaOH. What is the pH at the half-equivalence point (pKa 4.76), and what volume of NaOH has been added at that point?

---

Answers

1. n(HCl) = 0.100 × 0.02240 = 2.24 × 10⁻³ mol = n(NaOH). [NaOH] = 2.24 × 10⁻³ ÷ 0.0250 = 0.0896 mol/dm³

2. n(KOH) = 0.150 × 0.01860 = 2.79 × 10⁻³ mol. [HNO₃] = 2.79 × 10⁻³ ÷ 0.0200 = 0.140 mol/dm³ (0.1395)

3. n(NaOH) = 0.200 × 0.0250 = 5.00 × 10⁻³ mol = n(HCl). V = 5.00 × 10⁻³ ÷ 0.250 = 0.0200 dm³ = 20.0 cm³

4. n(NaOH) = 0.0800 × 0.0250 = 2.00 × 10⁻³ mol = n(CH₃COOH). c = 2.00 × 10⁻³ ÷ 0.02125 = 0.0941 mol/dm³

5. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. n(NaOH) = 2.42 × 10⁻³ mol, so n(H₂SO₄) = 1.21 × 10⁻³ mol. [H₂SO₄] = 1.21 × 10⁻³ ÷ 0.0250 = 0.0484 mol/dm³

6. Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O. n(Ca(OH)₂) = 1.00 × 10⁻³ mol, so n(HCl) = 2.00 × 10⁻³ mol. [HCl] = 2.00 × 10⁻³ ÷ 0.01670 = 0.120 mol/dm³

7. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. n(Na₂CO₃) = 1.25 × 10⁻³ mol, n(HCl) = 2.50 × 10⁻³ mol. V = 2.50 × 10⁻³ ÷ 0.100 = 25.0 cm³

8. H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O. n(H₃PO₄) = 1.50 × 10⁻³ mol, n(NaOH) = 4.50 × 10⁻³ mol. V = 4.50 × 10⁻³ ÷ 0.200 = 22.5 cm³

9. n(NaOH) = 0.200 × 0.02360 = 4.72 × 10⁻³ mol = n(HCl) in 25.0 cm³ portion. In 250.0 cm³: 4.72 × 10⁻² mol. This came from the original 25.0 cm³: [HCl] = 4.72 × 10⁻² ÷ 0.0250 = 1.89 mol/dm³

10. n(NaOH) = 2.08 × 10⁻³ mol = n(CH₃COOH) in 25.0 cm³. In 100.0 cm³: 8.32 × 10⁻³ mol, from 10.0 cm³ of vinegar. c = 8.32 × 10⁻³ ÷ 0.0100 = 0.832 mol/dm³. Mr(CH₃COOH) = 60.0, so 49.9 g/dm³ (about 5% by mass, typical for vinegar).

11. Mr(Na₂CO₃) = 106.0. n = 2.12 ÷ 106.0 = 0.0200 mol in 250.0 cm³, so 2.00 × 10⁻³ mol in 25.0 cm³. n(HCl) = 4.00 × 10⁻³ mol. [HCl] = 4.00 × 10⁻³ ÷ 0.02000 = 0.200 mol/dm³

12. n(NaOH) = 2.50 × 10⁻³ mol, n(acid) in 25.0 cm³ = 1.25 × 10⁻³ mol, in 250.0 cm³ = 1.25 × 10⁻² mol. Mr = 1.575 ÷ 0.0125 = 126. Mr(H₂C₂O₄) = 90.0, so water = 36.0, x = 36.0 ÷ 18.0 = 2.

13. n(NaOH) = 5.00 × 10⁻³ mol = n(acid). Mr = 0.600 ÷ 5.00 × 10⁻³ = 120 g/mol

14. n(HCl) = 3.00 × 10⁻³ mol, n(Na₂CO₃) in 25.0 cm³ = 1.50 × 10⁻³ mol, in 250.0 cm³ = 1.50 × 10⁻² mol. Mr = 4.29 ÷ 0.0150 = 286. Water = 286 − 106 = 180, x = 10.

15. n(HCl) at start = 0.500 × 0.0500 = 0.0250 mol. Excess = n(NaOH) = 0.500 × 0.01200 = 6.00 × 10⁻³ mol. Reacted = 0.0190 mol. n(CaCO₃) = 0.0190 ÷ 2 = 9.50 × 10⁻³ mol. Mass = 9.50 × 10⁻³ × 100.1 = 0.951 g. Purity = 95.1%

16. n(HCl) at start = 0.0100 mol. Excess = 1.50 × 10⁻³ mol. Reacted = 8.50 × 10⁻³ mol. Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O, so n(Mg(OH)₂) = 4.25 × 10⁻³ mol. Mr = 58.3. Mass = 0.248 g

17. n(NaOH) at start = 5.00 × 10⁻³ mol. Excess = n(HCl) = 1.74 × 10⁻³ mol. Reacted with NH₄⁺ = 3.26 × 10⁻³ mol (NH₄⁺ + OH⁻ → NH₃ + H₂O, 1 : 1). n(N) = 3.26 × 10⁻³ mol, mass N = 3.26 × 10⁻³ × 14.0 = 0.0456 g. % N = 0.0456 ÷ 0.250 × 100 = 18.3% (consistent with ammonium sulfate, whose theoretical value is 21.2%, if the sample is about 86% pure).

18. n(BaSO₄) = 0.2334 ÷ 233.4 = 1.000 × 10⁻³ mol = n(H₂SO₄). [H₂SO₄] = 1.000 × 10⁻³ ÷ 0.0250 = 0.0400 mol/dm³. Total n(NaOH) = 3.00 × 10⁻³ mol. H₂SO₄ uses 2.00 × 10⁻³, leaving 1.00 × 10⁻³ for HCl. [HCl] = 1.00 × 10⁻³ ÷ 0.0250 = 0.0400 mol/dm³

19. The water dilutes the NaOH, so more of it is needed to reach the end point. The titre is too high, and the calculated concentration of the acid is too high. See common titration errors.

20. Equivalence is at 25.0 cm³, so half-equivalence is at 12.5 cm³. There, [CH₃COOH] = [CH₃COO⁻], so pH = pKa = 4.76. See titration curves.

Where people lose marks

• Forgetting to divide cm³ by 1000.
• Using the wrong mole ratio for H₂SO₄, Ca(OH)₂, Na₂CO₃ and H₃PO₄.
• Forgetting to scale up from a 25.0 cm³ portion to the full volumetric flask.
• In back titrations, forgetting to subtract the excess.
• Giving too many or too few significant figures (three is usually right).

Key takeaways

• Every titration problem starts with moles of the solution you fully know.
• The balanced equation gives the ratio; dilution and aliquot steps need careful scaling.
• Back titrations work by subtraction: acid added minus acid left over equals acid used.
• Always sense-check purity (not above 100%) and formula answers (whole numbers of water).
• Check concentrations with the molarity calculator.