Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\left[O{H}^{-}\right]=\frac{10^{-5}\times 0.5\times 1+2.5\times 10^{-5}\times 0.5\times 2}{0.5+0.5+99}=\frac{0.5\times 10^{-5}+2.5\times 10^{-5}}{100}$

$=3\times 10^{-7}<10^{-6}\mathrm{M}$

We have to consider ${\mathrm{OH}}^{-}$coming from water.

$\begin{array}{rlrlrlrl}& H_{2}O(l)\underset{x}{\stackrel{K_W=10^{-14}}{\rightleftharpoons }}{H}^{+}(aq)+O{H}^{-}(aq)& & & & & & \end{array}$

$K_W=x\left(x+3\times 10^{-7}\right)$

$or,10^{-14}=x^2+3\times 10^{-7}x$

$or,x^2+3\times 10^{-7}x-10^{-14}=0$

$or,x=\frac{-3\times 10^{-7}+\sqrt{9\times 10^{-14}}+4\times 10^{-14}}{2}$

$or,x=\frac{[-3+\sqrt{13}]}{2}\times 10^{-7}$

$=0.3\times 10^{-7}=3\times 10^{-8}$

$\therefore pH=8-\log 3=8-0.48=7.52$