Question

The neutralizer 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. 10 mL
• B. 60 mL
• C. 40 mL
• D. 20 mL

Answer 1

The correct option is C 40 mL

Given

20 ml of 0.1 M Phosphoric acid

0.1 M of KOH

Solution

Neutralization reaction will be

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

Phosphoric acid is a diprotic acid therefore it will only give 2 H+ ions

Therefore i mole of phosphoric acid will neutralize 2 moles of KOH

No. of moles of phospjoric acid

0.1*20*1/1000=0.002mol

No. of moles neutralized of KOH=2*0.002 moles=0.004moles

Molarity of KOH solution is 0.1

Volume of KOH solution will be 0.004/0.1=0.04L=40ml

The correct option is C