$2{\mathrm{H}}_2\mathrm{O} \rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+} + {\mathrm{OH}}^{-}$

${\mathrm{K}}_{\mathrm{w}} = 1 \times {10}^{-14} \mathrm{at} {25}^0\mathrm{C}.$ Hence, $\mathrm{Ka}$ is

${\mathrm{K}}_{\mathrm{w}} = \left[{\mathrm{H}}^{+}\right] \left[{\mathrm{OH}}^{-}\right]= 1 \times {10}^{-14}$

$\therefore \left[{\mathrm{H}}^{+}\right] = 1 \times {10}^{-7} \mathrm{M},$

density of water = 1 g / mL  ; or 1000 g/L    C=number of moles(n)/lit   =  1000/18 =  55.55 M

Also in weak acid $\left[{\mathrm{H}}^{+}\right] = \sqrt{{\mathrm{K}}_{\mathrm{a}}\mathrm{C}}$

$\therefore {\mathrm{K}}_{\mathrm{a}} = \frac{{\left[{\mathrm{H}}^{+}\right]}^2}{\mathrm{C}} = \frac{{10}^{-14}}{55.55}=1.8 \times {10}^{-16}$