Question

A 0.782 g sample of a monoprotic acid is dissolved in water and titrated with 0.320 M NaOH. What is the molar mass of the acid if 13.5 mL of the NaOH solution is required to neutralize the sample?

Answer 1

Balanced chemical equation is:
NaOH + HA ---> NaA + H2O

lets calculate the mol of NaOH
volume , V = 13.5 mL
= 1.35*10^-2 L


use:
number of mol,
n = Molarity * Volume
= 0.32*1.35*10^-2
= 4.32*10^-3 mol
According to balanced equation
mol of HA reacted = (1/1)* moles of NaOH
= (1/1)*4.32*10^-3
= 4.32*10^-3 mol
This is number of moles of HA

mass(HA)= 0.782 g


use:
number of mol = mass / molar mass
4.32*10^-3 mol = (0.782 g)/molar mass
molar mass = 1.81*10^2 g/mol
Answer: 181 g/mol