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The pH of resultant solution of 10mL of $0.1 M H_{3} P O_{4}$ and 10mL of $0.1 M N a_{3} P O_{4}$ is equal to $p K_{a_{2}}$

In a buffer solution of $N a H_{2} P O_{4}$ and $N a_{2} H P O_{4}, N a H_{2} P O_{4}$ is acid and $N a_{2} H P O_{4}$ is salt

Buffer capacity is maximum when 1 liter of $1 M N H_{4} C l$ is added to 1 liter of $2 M N H_{4} O H$ solution

pH of acidic buffer increases and pH of basic buffer decreases if more salt is added

answer is A, B, D.

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Detailed Solution

   $\begin{array}{l} H_{3} P O_{4} + P O_{4}^{3 -} \to H_{2} P O_{4}^{\oplus} + H P O_{4}^{2 -} \\ [H_{2} P O_{4}^{\oplus}] = [H P O_{4}^{2 -}] ........... (1) \end{array}$
The resultant solution will be acidic buffer solution
$H_{2} P O_{4}^{\oplus}$ is more acidic than $H P O_{4}^{2 -}$ , hence $H_{2} P O_{4}^{\oplus}$ is considered as acid and    $H P O_{4}^{2 -}$ is considers as salt, so in Henderson equation, $K_{a_{2}}$ is considered
   $\begin{array}{l} H_{2} P O_{4}^{\oplus} \Rightarrow H P O_{4}^{2 -} + H^{\oplus} \\ p H = p K_{a_{2}} + \log [H P O_{4}^{2 -} / H_{2} P O_{4}^{\oplus}] ............... (2) \end{array}$
From    $e q (1) a n d (2), p H = p K_{a_{2}}$