Question

The exact volumes of $1M$ $NaOH$ solution required to neutralise $50\text{mL}$ of $1M{H}_{3}P{O}_3$ solution and $100\text{mL}$ of $2M{H}_{3}P{O}_2$ solution, respectively, are:

• A. $100\text{mL and 50 mL}$
• B. $50\text{mL and 50 mL}$
• C. $100\text{mL and 100 mL}$
• D. $100\text{mL and 200 mL}$

Answer 1

The correct option is D $100\text{mL and 200 mL}$

$\left(1\right)\underset{100mmol}{2NaOH}+\underset{50mmol}{H_{3}P{O}_3}\to N{a}_{2}HP{O}_3+2H_{2}O$

$100mmole=M\times V_{mL}$
$100mmole=1\times V_{mL}$
$V_{ml}=100\text{mL}$

$\left(2\right)\underset{200mmol}{NaOH}+\underset{200mmol}{H_{3}P{O}_2}\to Na{H}_{2}P{O}_2+H_{2}O$


$200mmole=M\times V_{mL}$
$200mmole=1\times V_{mL}$
$V_{mL}=200\text{mL}$