Question

Percentage ionization of $H_{2}O$ at certain temperature is $3.6\times {10}^{-7}$. Calculate $K_w$ and pH of water.
($K_w=4\times {10}^{-14}$, ph=6.7)

Answer 1

Degree of ionization= $\alpha =3.6\times {10}^{-7}\times {10}^{-2}=3.6\times {10}^{-9}$

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

$1-\alpha$ $\alpha$ $\alpha$

$K_W=\left[O{H}^{-}\right]\left[H^{+}\right]={\alpha }^2=(3.6\times {10}^{-9}{)}^2=12.96\times {10}^{-18}$

Also, $pH=-\log \left[H^{+}\right]=-\log (3.6\times {10}^{-19})=8.44$