$50 mL$ of $0.1 M N a_{3} P O_{4}$ was titrated with $50 mL$ of $0.2 M H C l$ . Find the $p H$ of resulting mixture where $P K_{a_{1}}, p K_{a_{2}}$ and $p K_{a_{3}}$ of $H_{3} P O_{4}$ are $5, 8, 13$ :

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Solution

$\begin{matrix} P O_{4}^{3 -} + & H^{+} ⇌ & H P O_{4}^{2 -} Initially & 5 mm & 10 mm & 0 Final & 0 & 5 mm & 5 mm H P O_{4}^{2 -} + & H^{+} & ⇌ H_{2} P O_{4}^{-} Initially & 5 mm & 5 mm & 0 Final & 0 & 0 & 5 mm \end{matrix}$
As can be seen by above two equilibrium reactions, $H_{2} P O_{4}^{-}$ is an amphoteric ion. Hence, $p H$ can be calculated using the formula,
$p H = \frac{1}{2} (p K_{a_{1}} + p K_{a_{2}}) = \frac{1}{2} (5 + 8) \Rightarrow p H = \frac{5 + 8}{2} = 6.5$

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