Question

Given a .210 g sample of a diprotic acid of unknown molar mass: If the acid is dissolved in water and titrated with .241 M KOH it takes 10.5 mL of base to reach the equivalence point. What is the molar mass of the solution?

Answer 1

Amount of diprotic acid=0.210 g

we know, that the molar mass of the unknown diprotic acid is about 123.8 g/mol.

so number of moles in 0.210 g of diprotic acid =0.210/123.8= 0.00169 moles

You used 10.5 mL of 0.241  mol/L KOH. That gives 0.241/10.5 = 0.043 moles of base. Since the acid is diprotic, you had half that number (0.0217 moles) for the acid.

That's just the amount of acid in 25 mL. The amount in the original 200 mL solution is 0.0871 moles.

This means the molar mass will be 4.50 g / 0.0871 moles, or 51.66 g/mol. which is the molar mass of the solution