A 36.0 mL sample of aqueous sulfuric acid was titrated with 0.250 M KOH(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 volume of the potassium hydroxide solution that was used in the titration from this data. Hint: Write out the neutralization reaction before proceeding with the problem.

Question

Quizplus · Accepted Answer

The neutralization reaction is H2SO4 + 2KOH → K2SO4 + 2H2O. The residue is K2SO4, with a molar mass of 174.26 g/mol. The mass of K2SO4 is 0.861 g, which corresponds to 0.861 g / 174.26 g/mol = 0.00494 mol. Since 2 moles of KOH are needed per mole of H2SO4, 0.00494 mol of K2SO4 corresponds to 0.00988 mol of KOH. Using the concentration of KOH (0.250 M), the volume of KOH used is 0.00988 mol / 0.250 mol/L = 0.0395 L, or 39.5 mL.

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