ScienceAges 16-18

Acid-base titration stoichiometry

Use balanced diprotic neutralisation stoichiometry (H₂SO₄ + 2NaOH) to calculate an unknown alkali concentration from titration volumes. Review suitability for your learners before teaching.

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Worksheet StudioPractice pack · Ages 16-18

Acid-base titration stoichiometry

Take your time. Show your thinking.

1. In an acid-base titration, 25 cm³ of 1.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 25 cm³ × 1.6 M · 2NaOH: 40 cm³ × ? M
2. In an acid-base titration, 25 cm³ of 1.4 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 25 cm³ × 1.4 M · 2NaOH: 20 cm³ × ? M
3. In an acid-base titration, 40 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 40 cm³ × 0.8 M · 2NaOH: 40 cm³ × ? M
4. In an acid-base titration, 20 cm³ of 3.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 20 cm³ × 3.6 M · 2NaOH: 40 cm³ × ? M
5. In an acid-base titration, 20 cm³ of 4 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 20 cm³ × 4 M · 2NaOH: 40 cm³ × ? M
6. In an acid-base titration, 40 cm³ of 2.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 40 cm³ × 2.5 M · 2NaOH: 50 cm³ × ? M
7. In an acid-base titration, 20 cm³ of 4.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 20 cm³ × 4.5 M · 2NaOH: 50 cm³ × ? M
8. In an acid-base titration, 20 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 20 cm³ × 0.8 M · 2NaOH: 20 cm³ × ? M
9. In an acid-base titration, 30 cm³ of 4.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 30 cm³ × 4.8 M · 2NaOH: 40 cm³ × ? M
10. In an acid-base titration, 50 cm³ of 3.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 50 cm³ × 3.6 M · 2NaOH: 40 cm³ × ? M
11. In an acid-base titration, 50 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 50 cm³ × 0.8 M · 2NaOH: 20 cm³ × ? M
12. In an acid-base titration, 25 cm³ of 1.2 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 25 cm³ × 1.2 M · 2NaOH: 20 cm³ × ? M

12 questions · Seed 44 · Age guidance, not curriculum certification

View the answer key

Worksheet StudioAnswer key · Ages 16-18

Acid-base titration stoichiometry

Answers and reasoning. Review before teaching.

1. In an acid-base titration, 25 cm³ of 1.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 1.6 × 25) ÷ 40 = 2 mol/dm³.
2. In an acid-base titration, 25 cm³ of 1.4 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3.5 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 1.4 × 25) ÷ 20 = 3.5 mol/dm³.
3. In an acid-base titration, 40 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
1.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.8 × 40) ÷ 40 = 1.6 mol/dm³.
4. In an acid-base titration, 20 cm³ of 3.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 3.6 × 20) ÷ 40 = 3.6 mol/dm³.
5. In an acid-base titration, 20 cm³ of 4 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
4 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 4 × 20) ÷ 40 = 4 mol/dm³.
6. In an acid-base titration, 40 cm³ of 2.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
4 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 2.5 × 40) ÷ 50 = 4 mol/dm³.
7. In an acid-base titration, 20 cm³ of 4.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 4.5 × 20) ÷ 50 = 3.6 mol/dm³.
8. In an acid-base titration, 20 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
1.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.8 × 20) ÷ 20 = 1.6 mol/dm³.
9. In an acid-base titration, 30 cm³ of 4.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
7.2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 4.8 × 30) ÷ 40 = 7.2 mol/dm³.
10. In an acid-base titration, 50 cm³ of 3.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
9 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 3.6 × 50) ÷ 40 = 9 mol/dm³.
11. In an acid-base titration, 50 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
4 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.8 × 50) ÷ 20 = 4 mol/dm³.
12. In an acid-base titration, 25 cm³ of 1.2 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 1.2 × 25) ÷ 20 = 3 mol/dm³.

12 questions · Seed 44 · Age guidance, not curriculum certification

Worksheet StudioPractice pack · Ages 16-18

Acid-base titration stoichiometry

Take your time. Show your thinking.

1. In an acid-base titration, 25 cm³ of 1.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 25 cm³ × 1.6 M · 2NaOH: 40 cm³ × ? M
2. In an acid-base titration, 25 cm³ of 1.4 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 25 cm³ × 1.4 M · 2NaOH: 20 cm³ × ? M
3. In an acid-base titration, 40 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 40 cm³ × 0.8 M · 2NaOH: 40 cm³ × ? M
4. In an acid-base titration, 20 cm³ of 3.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 20 cm³ × 3.6 M · 2NaOH: 40 cm³ × ? M
5. In an acid-base titration, 20 cm³ of 4 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 20 cm³ × 4 M · 2NaOH: 40 cm³ × ? M
6. In an acid-base titration, 40 cm³ of 2.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 40 cm³ × 2.5 M · 2NaOH: 50 cm³ × ? M
7. In an acid-base titration, 20 cm³ of 4.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 20 cm³ × 4.5 M · 2NaOH: 50 cm³ × ? M
8. In an acid-base titration, 20 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 20 cm³ × 0.8 M · 2NaOH: 20 cm³ × ? M
9. In an acid-base titration, 30 cm³ of 4.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 30 cm³ × 4.8 M · 2NaOH: 40 cm³ × ? M
10. In an acid-base titration, 50 cm³ of 3.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 50 cm³ × 3.6 M · 2NaOH: 40 cm³ × ? M
11. In an acid-base titration, 50 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 50 cm³ × 0.8 M · 2NaOH: 20 cm³ × ? M
12. In an acid-base titration, 25 cm³ of 1.2 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 25 cm³ × 1.2 M · 2NaOH: 20 cm³ × ? M

12 questions · Seed 44 · Age guidance, not curriculum certification

Worksheet StudioAnswer key · Ages 16-18

Acid-base titration stoichiometry

Answers and reasoning. Review before teaching.

1. In an acid-base titration, 25 cm³ of 1.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 1.6 × 25) ÷ 40 = 2 mol/dm³.
2. In an acid-base titration, 25 cm³ of 1.4 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3.5 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 1.4 × 25) ÷ 20 = 3.5 mol/dm³.
3. In an acid-base titration, 40 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
1.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.8 × 40) ÷ 40 = 1.6 mol/dm³.
4. In an acid-base titration, 20 cm³ of 3.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 3.6 × 20) ÷ 40 = 3.6 mol/dm³.
5. In an acid-base titration, 20 cm³ of 4 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
4 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 4 × 20) ÷ 40 = 4 mol/dm³.
6. In an acid-base titration, 40 cm³ of 2.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
4 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 2.5 × 40) ÷ 50 = 4 mol/dm³.
7. In an acid-base titration, 20 cm³ of 4.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 4.5 × 20) ÷ 50 = 3.6 mol/dm³.
8. In an acid-base titration, 20 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
1.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.8 × 20) ÷ 20 = 1.6 mol/dm³.
9. In an acid-base titration, 30 cm³ of 4.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
7.2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 4.8 × 30) ÷ 40 = 7.2 mol/dm³.
10. In an acid-base titration, 50 cm³ of 3.6 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
9 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 3.6 × 50) ÷ 40 = 9 mol/dm³.
11. In an acid-base titration, 50 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
4 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.8 × 50) ÷ 20 = 4 mol/dm³.
12. In an acid-base titration, 25 cm³ of 1.2 mol/dm³ H₂SO₄ completely neutralises 20 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 1.2 × 25) ÷ 20 = 3 mol/dm³.

12 questions · Seed 44 · Age guidance, not curriculum certification