Question

A 36.0mL sample of aqueous sulfuric acid was titrated with 0.250 molar KOH (aq) until the chemical indicator signaled that the solution was exactly neutralized. The mixture was then carefully evaporated to dryness, then the residue was dried in a drying oven. The residue was then weighed and a value of 861 my was obtained for it's mass. Calculate the number of mL of the potassium hydroxide solution that was used in the titration from this data.
H2SO4 + 2KOH --> K2SO4 + 2H2O +heat
(hint: find the number of moles of KOH using the mole ratio, then use molarity)
a. 12.7 mL b. 19.8 mL c. 25.4 mL d. 39.5 mL e. 79.1 mL

Answer 1

mass of product obtained after evaporated to dryness = K2SO4 = 861 g

molar mass of K2SO4 = 174.25 g / mol

moles = mass / molar mass

           = 861 / 174.25

           = 4.94

2 moles KOH --------------------> 1 mole K2SO4

X mole KOH ------------------------> 4.94 mole K2SO4

moles of KOH = 2 x 4.94 / 1 = 9.88

molarity of KOH = 0.250 M

molarity = moles / volume

volume   = 9.88 / 0.250 = 39.5 mL

answer : d. 39.5 mL