Question

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

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

Answer 1

Answer: (C)

$40mL$

$H_{3}P{O}_3+2KOH\to K_{2}HP{O}_3+2H_{2}O$

$\frac{M_1{V}_1}{n_1}=\frac{M_2{V}_2}{n_2}$

Here $M_1$ and $M_2$ are molar mass of phosphorus acid and KOH.

$V_1$ and $V_2$ are the volume of the phosphorus acid and KOH.

Here $n_1$ and $n_2$ are the number of moles of phosphorus acid and KOH.

$\frac{0.1\times 20}{1}=\frac{0.1\times V_2}{2}$

$V_2=40mL$.

Hence, the correct option is $C$