Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\begin{array}{rlrlrlrlrlr}& C=0.1M,& & & & & & & & & \end{array}$

$K_{h_1}=\frac{K_W}{K_{a_3}}=\frac{10^{-14}}{2\times 10^{-12}}=0.5\times 10^{-2}$

$\frac{K_{a_2}}{C}>10^{-2}$

$\Rightarrow K_{a_1}=\frac{c{h}^2}{(1-h)}$

$\Rightarrow 0.5\times 10^{-2}=\frac{0.1h^2}{(1-h)}$

$or,0.05-0.05h=h^2$

$or,h^2+0.05h-0.05=0$

$or,h=\frac{-0.05\pm \sqrt{0.0025+0.2}}{2}$

$or,h=\frac{-0.05+0.45}{2},h>0$

$or,h=0.2$

$[OH]=\mathrm{ch}=0.1\times 0.2=2\times 10^{-2}$

$pOH=2-\log 2=1.7$

$pH=14-1.7=12.3$