Question

To neutralise completely $20$ mL of $0.1M$ aqueous solution of phosphorus acid $\left(H_{3}P{O}_3\right)$, the volume of $0.1$ M aqueous $KOH$ solution required is:

• A. $10$ mL
• B. $20$ mL
• C. $40$ mL
• D. $60$ mL

Answer 1

The correct option is C $40$ mL

The neutralization reaction is $H_{3}P{O}_3+2KOH\to K_{2}HP{O}_3+2H_{2}O$.

Phosphorus acid is diprotic acid as it has two ionizable hydrogens.
Thus, $1$ mole of phosphorous acid will neutralize $2$ moles of $KOH$.

The number of moles of phosphorus acid present in $20$ mL of $0.1$ M aqueous solution is $0.1\times 20\times \frac{1}{1000}=0.002$ mol.

They will neutralize $2\times 0.002$ mol $=0.004$ moles of $KOH$.

The molarity of KOH solution is $0.1$ M.

The volume of $KOH$ solution required will be $\frac{0.004}{0.1}=0.04L=40$ ml.

Hence, option $C$ is correct.