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Henderson Hassleback Equation

25 mL of 0.1 ...

Question

$25 m L$ of $0.1 N H_{3} P O_{4}$ is titrated against $0.1 N N a O H$ . The $p H$ of the solution after addition of $62.5 m L$ of the base is :
(Given : $K_{a 1} = 1 \times 10^{- 3}, K_{a 2} = 1 \times 10^{- 8} a n d K_{a 3} = 1 \times 10^{- 13}$ )

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Solution

$H_{3} P O_{4} + N a O H \to N a H_{2} P O_{4}$

$N a H_{2} P O_{4} + N a O H \to N a_{2} H P O_{4}$
$N a_{2} H P O_{4} + N a O H \to N a_{3} P O_{4}$
First equivalence point is reached after addition of 25 mL of the base.
Second equivalence point is reached after addition of 50 ml of the base.
After addition of 62.5 mL all the $H_{3} P O_{4}$ and $N a H_{2} P O_{4}$ are consumed and the half equivalence point of third tiration is reached.

Species present in the solution at this point will be,
$N a_{2} H P O_{4}$ and $N a_{3} P O_{4}$ and its a buffer

$p H = p K a 3 + l o g \frac{[N a_{3} P O_{4}]}{[N a_{2} H P O_{4}}]$
since $[N a_{2} H P O_{4}] = [N a_{3} P O_{4}]$
$p H = p K a 3$
$K a_{3} = 1 \times 10^{- 13}$
$H e n c e, p H = 13$

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