To $100 m L$ of $0.1 M$ $H_{3} P O_{4}$ $200 m L$ of $0.1 M$ $N a O H$ was added. Calculate the $p H$ of the resulting solution.
( $G i v e n K_{a 1} = 10^{- 3}, K_{a 2} = 10^{- 8}, K_{a 3} = 10^{- 13}$ )

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Solution

After the addition of $200 m L$ of $0.1 M$ $N a O H$ , second equivalence point is reached. Hence, the solution contains $N a_{2} H P O_{4}$ . It is an amphiprotic salt.

$p H$ of amphiprotic salt= $\frac{p K_{a 2} + p K_{a 3}}{2}$ = $\frac{8 + 13}{2} = 10.5$

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