Question

A 0.870 g sample of a monoprotic acid is dissolved in water and titrated with 0.300 M KOH.

Answer 1

Answer: The molar mass of monoprotic acid is 126.1 g/mol

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of KOH solution = 0.300 M

Volume of solution = 23.00 mL

Putting values in above equation, we get:

$0.300M=\frac{\text{Moles of KOH}\times 1000}{23.00}\\\\\text{Moles of KOH}=\frac{0.300\times 23.00}{1000}=0.0069moles$

The chemical equation for the reaction of monoprotic acid and KOH follows:

$HA+KOH\rightarrow KA+H_2O$

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of HA

So, 0.0069 moles of KOH will react with = $\frac{1}{1}\times 0.0069=0.0069$ moles of HA

To calculate the molar mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of HA = 0.870 g

Moles of monoprotic acid = 0.0069 mol

Putting values in above equation, we get:

$0.0069mol=\frac{0.870g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.870g}{0.0069mol}=126.1g/mol$

Hence, the molar mass of monoprotic acid is 126.1 g/mol