The dissociation constant of a weak acid is ${10}^{-6}.$Then the pH of 0.01 N of that acid is

$\mathrm{Ka}={10}^{-6}$

N = 0.01N

$\mathrm{PH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right] \mathrm{Ka}={\mathrm{C\alpha }}^2$

$\left[{\mathrm{H}}^{+}\right]=C\alpha$                     ${10}^{-6}=+0.01\times {\mathrm{\alpha }}^2$

$={10}^{-2}\times {10}^{-2} {10}^{-4}={\mathrm{\alpha }}^2$

$={10}^{-4} {10}^{-2}=\mathrm{\alpha }$

$\mathrm{PH}=-{\log }_{10}{10}^{-4}=4$