titrations part 2

Question

A 0.5450-g sample of an unknown monoprotic acid is dissolved in water and titrated with standardized sodium hydroxide. The equivalence point in the titration is reached after the addition of 24.32 mL of 0.2345 M sodium hydroxide to the sample of the unknown acid. Calculate the molar mass of the acid.

Answer 1

Molar mass of acid = grams of acid / moles of acid

We know the grams = 0.5450 g, so we need to find the moles of acid.

This is accomplished by using the data from the titration of the acid with the NaOH

Let the monoprotic acid be HA

HA + NaOH ==> NaA + H2O (1:1 mol ratio of NaOH:HA)

moles NaOH used in the titration = 24.32 ml x 1 L / 1000 ml x 0.2345 mo / L = 0.005703 mols NaOH used

moles of acid = 0.005703 mols since they react in a 1:1 mol ratio

Molar mass of the acid = 0.5450 g / 0.005703 mols

Molar mass = 95.56 g / mole