The standard calculation
3 marks · NormalIn a titration, 25.0 cm³ of sulphuric acid ( ) of unknown concentration is neutralised by 32.5 cm³ of 0.150 mol/dm³ potassium hydroxide ( ). The balanced chemical equation is:
Calculate the concentration of the sulphuric acid.
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
0.0975 mol/dm³ (±0.0005)
Must show: Working showing volume to dm³ conversion, Using balanced equation ratio 1:2
Also accept: 0.098 mol/dm³
Common mistakes: Using incorrect ratio from the equation; Not converting units
Show mark scheme and worked solution
Marking points
M1 · 1 mark · Calculate moles of KOH used: 0.150 × 0.0325 = 0.00488 mol.
M2 · 1 mark · Uses 1:2 ratio to find moles of H₂SO₄ neutralised: 0.00488 ÷ 2 = 0.00244 mol. (follow-through)
A1 · 1 mark · Finds concentration: 0.00244 ÷ 0.0250 = 0.0975 mol/dm³. (follow-through)
Worked solution
Moles of KOH = 0.150 × (32.5 ÷ 1000) = 0.00488 mol
From the equation, 2 mol KOH reacts with 1 mol H₂SO₄. Moles of H₂SO₄ = 0.00488 ÷ 2 = 0.00244 mol
Concentration of H₂SO₄ = 0.00244 ÷ 0.0250 = 0.0975 mol/dm³