What volume of $0.05 M C a (O H)_{2}$ solution $(i n m L)$ is needed for complete conversion of $10 m L$ of $0.1 M H_{3} P O_{4}$ into $C a (H_{2} P O_{4})_{2} ?$

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Solution

$\begin{matrix} C a (O H)_{2} & + 2 H_{3} P O_{4} \to C a (H_{2} P O_{4})_{2} + H_{2} O n_{f} = 2 & n_{f} = 1 (2 O H^{-}) & only 1 H^{+} are replaced & is replaced \end{matrix}$
$m e q_{C a (O H)_{2}} = m e q_{H_{3} P O_{4}} 0.05 \times V \times 2 = 10 \times 0.1 \times 1 V = 10 m L$

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