Titration and dilution calculations: Class 10 practice
When an acid exactly neutralises a base, moles of H⁺ = moles of OH⁻, so n₁M₁V₁ = n₂M₂V₂ (n = number of H⁺ or OH⁻ per formula). Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Molarity
The idea behind titration and dilution calculations
When an acid exactly neutralises a base, moles of H⁺ = moles of OH⁻, so n₁M₁V₁ = n₂M₂V₂ (n = number of H⁺ or OH⁻ per formula). Diluting keeps moles constant: M₁V₁ = M₂V₂.
Step-by-step method
- Write basicity/acidity n for each substance (HCl 1, H₂SO₄ 2, NaOH 1, Ca(OH)₂ 2).
- Use n₁M₁V₁ = n₂M₂V₂.
- For dilution, M₁V₁ = M₂V₂.
Worked example 1 (Basic)
How much water must be added to 39 mL of 0.1 M HCl to make it 0.05 M?
- V₂ = 0.1 × 39/0.05 = 78 mL
- Water = 39 mL
- Answer: 39 mL
Worked example 2 (Standard)
What volume of 0.1 M NaOH neutralises 9 mL of 0.1 M HCl?
- 1 × 0.1 × 9 = 1 × 0.1 × V
- V = 9 mL
- Answer: 9 mL
Worked example 3 (Challenge)
What volume of 0.1 M NaOH neutralises 14 mL of 0.5 M H₂SO₄?
- 2 × 0.5 × 14 = 1 × 0.1 × V
- V = 140 mL
- Answer: 140 mL
Common mistakes and misconceptions
- Ignoring that H₂SO₄ gives 2 H⁺
- Adding volumes in the formula
Practice questions
- (Basic) How much water must be added to 26 mL of 0.1 M HCl to make it 0.25 M?
- (Basic) How much water must be added to 28 mL of 0.1 M HCl to make it 0.1 M?
- (Standard) What volume of 0.2 M NaOH neutralises 16 mL of 0.1 M HCl?
- (Standard) What volume of 0.1 M NaOH neutralises 21 mL of 1 M HCl?
- (Challenge) What volume of 0.2 M NaOH neutralises 32 mL of 0.1 M H₂SO₄?
- (Challenge) What volume of 0.25 M NaOH neutralises 50 mL of 1 M H₂SO₄?
Answer key with full working
- Q1: V₂ = 0.1 × 26/0.25 = 10.4 mL → Water = -15.6 mL → Answer: -15.6 mL
- Q2: V₂ = 0.1 × 28/0.1 = 28 mL → Water = 0 mL → Answer: 0 mL
- Q3: 1 × 0.1 × 16 = 1 × 0.2 × V → V = 8 mL → Answer: 8 mL
- Q4: 1 × 1 × 21 = 1 × 0.1 × V → V = 210 mL → Answer: 210 mL
- Q5: 2 × 0.1 × 32 = 1 × 0.2 × V → V = 32 mL → Answer: 32 mL
- Q6: 2 × 1 × 50 = 1 × 0.25 × V → V = 400 mL → Answer: 400 mL
Syllabus note
This skill is listed under "Acids, Bases and Salts" for Class 10 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is titration and dilution calculations in?
"Acids, Bases and Salts", Class 10 Chemistry (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.