Chemistry · Lesson 20

Practical 2: acid–base titration calculations

WAECNECO

What the syllabus asks

WAEC: acid–base titrations; calculations of concentration, mass concentration, molar mass, basicity, dilution, gas volumes, back titration. Links c-units on the mole and acids/bases.

NECO: acid–base titrations; calculations of concentration, mass concentration, molar mass, basicity, dilution, gas volumes, back titration. Links c-units on the mole and acids/bases.

What you will learn

Write balanced equations for common acid–base titrations.
Apply the mole-ratio formula correctly.
Convert between mass and molar concentrations.
Answer a complete WAEC-style Question 1.
Calculate moles, masses and gas volumes from titration data.
Solve dilution problems.
Determine molar mass, relative atomic mass and basicity.
Carry out back-titration calculations.

Key points

Equation first; ratio second.
C_A V_A/C_B V_B = n_A/n_B.
g per V cm³ × 1000/V = g dm⁻³.
mol dm⁻³ = g dm⁻³ ÷ molar mass.
C₁V₁ = C₂V₂.
Basicity = mol base ÷ mol acid.
3 s.f. and units every time.

Start here: The mole-ratio formula

Every acid–base titration calculation rests on one idea: at the end point the acid and the base have reacted exactly in the ratio of the balanced equation. If the equation is a acid + b base → products, then

(C_A × V_A) ÷ (C_B × V_B) = n_A ÷ n_B = a ÷ b

where C is molar concentration in mol dm⁻³, V is volume in cm³ (the 1000s cancel because both volumes are in the same unit), and n_A : n_B is the mole ratio from the equation. WAEC schemes award separate marks for writing the mole ratio, making the unknown the subject, substituting correctly and evaluating to three significant figures with the unit, so show every one of those lines.

The alternative, step-by-step mole method

Some students prefer to find moles directly: moles of known = C × V/1000; use the ratio to find moles of unknown; divide by the unknown’s volume in dm³. Both methods earn full marks when the working is clear. The step method is safer for unusual questions (back titration, mixtures), and the formula method is faster for standard ones.

Worked example 20.1: both methods

25.0 cm³ of 0.100 mol dm⁻³ NaOH needed 24.87 cm³ of H₂SO₄. Find the concentration of the acid.

Formula. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so n_A/n_B = 1/2. C_A = (C_BV_B × 1) ÷ (V_A × 2) = (0.100 × 25.0) ÷ (24.87 × 2) = 0.0503 mol dm⁻³.

Moles. Moles NaOH = 0.100 × 25.0/1000 = 0.00250 mol. Moles H₂SO₄ = 0.00250 ÷ 2 = 0.00125 mol in 24.87 cm³. C_A = 0.00125 × 1000/24.87 = 0.0503 mol dm⁻³.

Notice that the two volumes must be matched to the right solutions: V_A is the volume of acid (here the titre from the burette), V_B is the volume of base (here the pipette volume). Swapping them is the commonest single error in the whole paper.

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Practice questions

1. The mole ratio of H₂SO₄ to NaOH in their reaction is

1 : 2
2 : 1
1 : 1
1 : 3

Show answer

A: 1 : 2
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

2. The mole ratio of HCl to Na₂CO₃ is

1 : 2
1 : 1
3 : 1
2 : 1

Show answer

D: 2 : 1
2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂.

3. A solution contains 4.00 g of NaOH per dm³. Its concentration is [Na = 23, O = 16, H = 1]

4.00 mol dm⁻³
0.0100 mol dm⁻³
0.100 mol dm⁻³
0.400 mol dm⁻³

Show answer

C: 0.100 mol dm⁻³
4.00 ÷ 40 = 0.100.

4. 25.0 cm³ of 0.100 mol dm⁻³ NaOH needed 24.87 cm³ of H₂SO₄. The acid concentration is

0.0251 mol dm⁻³
0.0503 mol dm⁻³
0.101 mol dm⁻³
0.201 mol dm⁻³

Show answer

B: 0.0503 mol dm⁻³
C = 0.100 × 25.0 ÷ (2 × 24.87).

5. 0.0500 mol dm⁻³ H₂C₂O₄ (23.72 cm³) neutralised 25.0 cm³ of NaOH. The NaOH concentration is

0.0949 mol dm⁻³
0.0474 mol dm⁻³
0.0237 mol dm⁻³
0.190 mol dm⁻³

Show answer

A: 0.0949 mol dm⁻³
Ratio 1 : 2: C = 2 × 0.0500 × 23.72 ÷ 25.0.

Key words

Mole ratio: Ratio of moles of reactants from the balanced equation.
Molar concentration: Moles of solute per dm³ of solution.
Mass concentration: Grams of solute per dm³ of solution.
Basicity: Number of replaceable hydrogen ions per molecule of acid.
Dilution factor: Final volume ÷ initial volume.
Back titration: Titration of the excess of a reagent added to a sample.
s.t.p.: Standard temperature and pressure: 0 °C, 1 atm; molar volume 22.4 dm³.

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