Question

A 5.00 g quantity of a diprotic acid (H₂A) was dissolved in water and made up to exactly 250 mL. Calculate the molar mass (in g/mol) of the acid if 25.0 mL of this solution reuired 11.1 mL of 1.00 M KOH for neutralization.

Answer 1

moles KOH required for neutralization = 11.1mL x 1M / 1000 = 0.0111 moles

The acid is diprotic so the the ratio between H₂A and KOH is 1 : 2

So the moles of H2A required is = 0.0111/2 = 0.00555 moles

So 0.00555 moles of H2A should be present in 25 mL

SO 0.00555 moles x 10 = 0.0555 moles should be present in 250 mL

0.0555 moles is 5 g
5.00 g / 0.055 mol = 90.1 g/mol is the molar mass of the diprotic acid.