To neutralise completely $20 m L$ of $0.1 M$ aqueous solution of phosphorous acid $H_{3} {P O}_{3}$ , the volume of $0.1 M$ $K O H$ solution required is:

60

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20

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40

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10

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Solution

The correct option is B $40$ mL
Meq. of $K O H =$ Meq. of $H_{3} {P O}_{3}$

$0.1 \times V = 20 \times 0.1 \times 2$ ( $H_{3} {P O}_{3}$ is dibasic)

$∴$ $V = 40 m L$

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20 m L

0.1 M

(H_{3} {P O}_{3}

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0.1 M

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To neutralize completely 20 ml of 0.1 M aqueous solution of $H_{3} P O_{3}$ , the volume of 0.1 M aqueous KOH solution required is:

(H_{3} P O_{3})

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