Question

A sample of a monoprotic acid (ha) weighing 0.384 g is dissolved in water and the solution is titrated with aqueous naoh. if 30.0 ml of 0.100 m naoh is required to reach the equivalence point, what is the molar mass of ha?

Answer 1

The formula for the monoprotic acid is taken as HA, reaction with base is as follows;
HA + NaOH ---> NaA + H₂O
Stoichiometry of acid to base is 1:1
At the neutralisation point, number of HA moles = number of base moles
Number of NaOH moles reacted = 0.100M / 1000 mL /L x 30.0 mL = 0.003 mol
Therefore number of HA moles reacted = 0.003 mol
the mass of acid 0.384 g
Therefore molar mass - 0.384 g/ 0.003 mol = 128 g/mol