A 36.0 mL sample of aqueous sulfuric acid was titrated with 0.250 M NaOH(aq)until the acid-base indicator signaled that the solution was neutralized. The mixture was then carefully evaporated to dryness, the residue was dried in a drying oven. The residue was then weighed and a value of 861 mg was obtained for its mass. Calculate a value for the molarity of the sulfuric acid solution from this data. Hint: Write out the neutralization reaction before proceeding with the problem.

Question

Quizplus · Accepted Answer

The neutralization reaction is H2SO4 + 2NaOH → Na2SO4 + 2H2O. The residue is Na2SO4, with a molar mass of 142.04 g/mol. 861 mg = 0.861 g of Na2SO4 corresponds to 0.861 g / 142.04 g/mol = 0.00606 mol. Since 1 mol of H2SO4 reacts with 2 mol of NaOH to produce 1 mol of Na2SO4, the moles of H2SO4 are also 0.00606 mol. The molarity of H2SO4 is 0.00606 mol / 0.036 L = 0.168 M.

A 360 ML Sample of Aqueous Sulfuric Acid Was Titrated with with 0.250