A solution is prepared by dissolving 0.1 mole of $N{a}_{3}P{O}_4$ in sufficient water to make 1  litre solution. Calculate the concentration of $O{H}^{-}$ ion in the solution. Given for $H_{3}P{O}_4$

$k_1=7.1\times {10}^{-3},k_2=6.3\times {10}^{-8},k_3=4.5\times {10}^{-13}$

Hydrolysis of $P{O}_4^{-3}$ will take place

$$\begin{gathered} \underset{(0.1-x)}{P{O}_4^{-3}}+H_{2}O\rightleftharpoons \underset{(x-y)}{HP{O}_4^{-2}}+\underset{(x+y+z)}{O{H}_{ }^{-}}[k_{h_1}=\frac{k_w}{k_3}] \\ \underset{ }{\underset{(x-y)}{HP{O}_4^{-2}}}+H_{2}O\rightleftharpoons \underset{(y-z)}{H_{2}P{O}_4^{-}}+\underset{(x+y+z)}{O{H}^{-}}[k_{h_2}=\frac{k_w}{k_2}] \\ \underset{(y-z)}{HP{O}_4^{-}}+H_{2}O\rightleftharpoons \underset{z}{H_{3}P{O}_4}+\underset{(x+y+z)}{O{H}_{ }^{-}}[k_{h_3}=\frac{k_w}{k_1}] \\ \left[O{H}^{-}\right]=x+y+z \\ \left[O{H}^{-}\right]=x \\ x>>y>>z \end{gathered}$$

$$\begin{gathered} k_{h_1}=\frac{(x+y+z)(x-y)}{(c-x)} \\ \frac{k_w}{k_3}=\frac{x.x}{(c-x)} \\ 2.2\times {10}^{-2}=\frac{x^2}{(0.1-x)} \end{gathered}$$

on solving  $x=3.73\times {10}^{-2}molar$