If 20 cm ^3 of sodium hydroxide was neutralized by 20 cm ^3 of 0.01 mol dm ^3 tetraoxosulphate(VI) acid, what…

Step 1: Analyze the question We have a neutralization (titration) between NaOH (a base) and H₂SO₄ (tetraoxosulphate(VI) acid). We need the concentration of NaOH.

Step 2: Balanced equation 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

The mole ratio is NaOH : H₂SO₄ = 2 : 1. This ratio is the key point, because sulphuric acid is dibasic.

Step 3: Use the titration formula (C_A × V_A) / (C_B × V_B) = n_A / n_B

Here the acid is A and the base is B:

• C_A = 0.01 mol/dm³, V_A = 20 cm³, n_A = 1
• C_B = ?, V_B = 20 cm³, n_B = 2

(0.01 × 20) / (C_B × 20) = 1/2

C_B × 20 = 2 × (0.01 × 20) = 0.4

C_B = 0.4 / 20 = 0.02 mol/dm³

Check by moles:

• Moles of H₂SO₄ = 0.01 × 20/1000 = 2 × 10⁻⁴ mol
• Moles of NaOH = 2 × 2 × 10⁻⁴ = 4 × 10⁻⁴ mol
• Concentration of NaOH = 4 × 10⁻⁴ / 0.020 dm³ = 0.02 mol/dm³ ✓

Answer: B (0.020)

Exam tip: Always write the balanced equation first. Many students pick 0.010 (option A) because they forget that H₂SO₄ gives two H⁺ ions per molecule, so it neutralizes twice as many moles of NaOH. When the volumes are equal, the base concentration is simply the acid concentration multiplied by 2 in this case.