Question

The degree of dissociation of water is $1.8\times {10}^{-9}$ at $298K$. Calculate the ionisation constant and ionic product of water at $298K$.

Answer 1

$H_{2}O\rightleftharpoons H^{+}+{OH}^{-},\left[H_{2}O\right]=\frac{1000}{18}=55.56mol{L}^{-1}$
$\left[H^{+}\right]=\left[{OH}^{-}\right]=C\times \alpha$
$K_a=\frac{\left[H^{+}\right]\left[{OH}^{-}\right]}{\left[H_{2}O\right]}=C{\alpha }^2=55.56\times (1.8\times {10}^{-9}{)}^2=1.8\times {10}^{-16}$
$K_w=\left[H^{+}\right]\left[{OH}^{-}\right]=C\alpha \times C\alpha =(55.56{)}^2\times (1.8\times {10}^{-9}{)}^2={10}^{-}14$