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Henderson Hassleback Equation

The pK a valu...

Question

The $p K_{a}$ values, $H_{3} P O_{4}$ are 2, 12, 7.21 and 12.67. The pH of a 0.2 $M N a H_{2} P O_{4}$ and 0.1 $M N a_{2} H P O_{4}$ is

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Solution

Given,
$H_{3} P O_{4} ⇌ H^{+} + H_{2} P O_{4}^{-}; p K_{a} = 2.12$
$H_{2} P O_{4}^{-} ⇌ H^{+} + H P O_{4}^{2 -}; p K_{a} = 7.21 \dots \dots (i i)$
$H P O_{4}^{2 -} ⇌ H^{+} + P O_{4}^{3 -}; p K_{a} = 12.67$
Given, $\begin{matrix} N a H_{2} P O_{4} & ⇌ & N a^{+} & + & H_{2} P O_{4}^{-} 0.2 M & 0 & 0 0 & 0.2 M & 0.2 M \end{matrix}$
$∴ [H_{2} P O_{4}^{-}] = 0.2 M$
Similarly, $N a_{2} H P O_{4} ⇌ 2 N a^{+} + H P O_{4}^{2 -}$
$[H P O_{4}^{2 -}] = 0.1 M$
Now from Henderson's equation
$p H = p K_{a} + log \frac{[Acid]}{[Salt]}$
From Eq. (ii)
$p H = p K_{a} + log \frac{[H P O_{4}^{2 -}]}{[H_{2} P O_{4}^{-}]}$
$= 7.21 + log \frac{[0.1]}{[0.2]}$
$p H = 6.90$

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