Question

A 0.342 gram sample of a monoprotic acid is dissolved in water and titrated with 0.200M KOH . What is the molar mass of the acid if 19.0 mL of the KOH solution is required to neutralize the sample?

Answer 1

Balanced chemical equation is:
KOH + HA ---> KA + H2O

lets calculate the mol of KOH
volume , V = 19 mL
= 1.9*10^-2 L


use:
number of mol,
n = Molarity * Volume
= 0.2*1.9*10^-2
= 3.8*10^-3 mol
According to balanced equation
mol of HA reacted = (1/1)* moles of KOH
= (1/1)*3.8*10^-3
= 3.8*10^-3 mol
This is number of moles of HA

mass(HA)= 0.342 g


use:
number of mol = mass / molar mass
3.8*10^-3 mol = (0.342 g)/molar mass
molar mass = 90 g/mol
Answer: 90 g/mol