Q25 / 75 · JEE Main 2020 (03 Sep Shift 2) · Chemistry

The volume (in $mL$ ) of $0.1 N NaOH$ required to neutralise $10 mL of 0.1 N$ phosphinic acid is ____________.

Correct answer: 10

Solution

Phosphinic acid is hypo phosphorous acid $(H_{3} {PO}_{2})$ .

$NaOH + H_{3} {PO}_{2} \to {NaH}_{2} {PO}_{2} + H_{2} O$

For neutrization

${(N_{1} V_{1})}_{acid} = {(N_{2} V_{2})}_{base}$

$0.1 \times 10 = 0.1 \times {(V_{mL})}_{NaOH}$

$V_{NaOH} = 10 mL$

Hence, the correct answer is 10.

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