Question:
The volume (in $\mathrm{mL}$ ) of $0.1 \mathrm{~N} \mathrm{NaOH}$ required to neutralise $10 \mathrm{~mL}$ of $0.1 \mathrm{~N}$ phosphinic acid is
Solution:
$\mathrm{H}_{3} \mathrm{PO}_{2}+\mathrm{NaOH} \rightarrow \mathrm{NaH}_{2} \mathrm{PO}_{2}+\mathrm{H}_{2} \mathrm{O}$
$\frac{\mathrm{n}_{\mathrm{H}_{3} \mathrm{PO}_{2}} \text { reacted }}{1}=\frac{\mathrm{n}_{\mathrm{NaOH}} \text { reacted }}{1}$
$\Rightarrow \frac{0.1 \times 10}{1}=0.1 \times \mathrm{V}_{\mathrm{NaOH}}$
$\Rightarrow \mathrm{V}_{\mathrm{NaOH}}=10 \mathrm{ml}$
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