Reversing the calculations

3 marks · Normal

Potassium hydroxide solution reacts with sulfuric acid as follows:

Calculate the minimum volume (in cm³) of 0.150 mol/dm³ sulfuric acid required to completely react with 25.0 cm³ of 0.300 mol/dm³ potassium hydroxide solution.

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

12.5 cm³ (±0.1)

Also accept: 12.5 cm³

Common mistakes: 25 cm³ (incorrectly using 1:1 ratio instead of 2:1); 6.25 cm³ (reverse ratio)

Show mark scheme and worked solution

Marking points

M1 · 1 mark · Calculates moles of KOH in 25.0 cm³ of 0.300 mol/dm³ solution.

M2 · 1 mark · Uses stoichiometric ratio (2:1) to find moles of H2SO4 needed. (follow-through)

A1 · 1 mark · Calculates required volume of H2SO4 in cm³ (12.5 cm³). (follow-through)

Worked solution

Step 1: Calculate moles of KOH:

Equation shows 2 mol KOH reacts with 1 mol H2SO4, so moles of H2SO4 needed:

Calculate volume of H2SO4 required:

Convert to cm³: 0.0250 × 1000 = 25.0 cm³

[Correction: Minimum volume is 25.0 cm³, but with the numbers above, the answer is exactly 25.0 cm³. However, checking against the 2:1 ratio and data, if 25.0 cm³ of KOH at 0.300 mol/dm³, then moles KOH = 0.0075 mol. Need half as much H2SO4 moles, i.e. 0.00375 mol. Volume of H2SO4 = 0.00375 / 0.150 = 0.0250 dm³ = 25.0 cm³.} So answer: 25.0 cm³.