Question

Percentage ionization of $H_{2}O$ at certain temperature is $3.6\times {10}^{-7}$. Calculate $K_w$ and $pH$ of water.

Answer 1

$\alpha =\frac{3.6\times {10}^{-7}}{100}=3.6\times {10}^{-9}$

$2H_{2}O⟶O{H}^{-}+H_3{O}^{+}100⟵Initial.(1-\alpha )\alpha \alpha ⟵Atequilibrium.$

So, $K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right]=(3.6\times {10}^{-9})(3.6\times {10}^{-18})=12.96\times {10}^{-18}$

$pH=-\log \left[H_3{O}^{+}\right]$

$\therefore pH=-\log (3.6\times {10}^{-9})$

$\therefore pH=8.4$