# $pK_{a1}, pK_{a2}$ and $pK_{a3}$ of $H_3PO_4$ are respectively x, y and z. $pH$ of 0.1 M $Na_2HPO_4$ solution is

(a) 1

(b) $\frac{1}{2} (x+y)$

(c) $\frac{1}{2} (y+z)$

(d) $\frac{1}{2} (x+y+z)$

Answer: $\frac{1}{2} (y + z)$

$HPO^{2-}_4$ is amphiprotic
$HPO^{2-}_4 + H_2O \rightleftharpoons PO^{3-}_4 + H_3O^+$
$HPO^{2-}_4 + H_2O \rightleftharpoons H_2PO^-_4 + OH^-$
$\therefore pH$ of 0.1 M $Na_2HPO_4= \frac{pK_{a2} + pK_{a3}}{2} = \frac{1}{2} (y + z)$