Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\left[{\text{OH}}^{-}\right]=0.1\times \frac{20}{100}$
$\left[{\mathrm{O}\mathrm{H}}^{–}\right]=0.1\mathrm{ }\times \mathrm{ }0.2=2\mathrm{ }\times \mathrm{ }{10}^{–2}$
pOH= -log[OH^-] = 2- log2
pOH=1.7 , pH=14 – pOH = 14 - 1.7 = 12.3