40 cm^3 of 0.25 mol/dm^3 sulfuric acid is neutralised by 25 cm^3 of sodium hydroxide. Find the concentration of the NaOH.
Answer and explanation
Correct answer: A. 0.8 mol/dm^3
- Step 1: H2SO4 + 2NaOH -> Na2SO4 + 2H2O, so n(NaOH) = 2 x n(H2SO4).
- Step 2: n(H2SO4) = 0.25 x 40 / 1000 = 0.01 mol, so n(NaOH) = 0.02 mol.
- Step 3: c(NaOH) = 0.02 / (25/1000) = 0.8 mol/dm^3.
Answer
0.8 mol/dm^3.
Why not the others: '0.4 mol/dm^3' ignores that 1 mol of H2SO4 reacts with 2 mol of NaOH; '0.31 mol/dm^3' inverts the volume ratio; '1.6 mol/dm^3' applies the factor 2 twice.
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