Question

To neutralize completely 20 mL of 0.1 M aqueous solution of phosphorous acid $\left(H_{3}P{O}_3\right).$ The volume of 0.1 M aqueous KOH solution required is :

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

Answer 1

The correct option is A 40 mL

We know
$N_1{V}_1=N_2{V}_2(H_{3}P{O}_3\text{is dibasic}\therefore nfactor=2)$

n factor for $KOH$ = 1

So,

$M_{H_{3}P{O}_3}\times {n_f}_{H_{3}P{O}_3}\times V_{H_{3}P{O}_3}=M_{KOH}\times {n_f}_{KOH}\times V_{KOH}$

putting the values,

$20\times 0.2=0.1\times V_{KOH}$
$\therefore V_{KOH}=40$ mL.