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Question

The neutralizer completely 20 mL of 0.1 M aqueous solution of phosphorous acid $$ (H_3PO_3)$$, the volume of 0.1 M aqueous KOH solution required is:  


A
10 mL
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B
60 mL
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C
40 mL
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D
20 mL
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Solution

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
$$H_{3}PO_{3}+2KOH\rightarrow K_{2}HPO_{3}+2H_{2}O$$
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



Chemistry

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