The pH of ${10}^{-4} \mathrm{M}$ acid, if it is 1% ionized is

The reaction will be:

 $\mathrm{HA}\left(\mathrm{aq}\right)\iff {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$

As the acid is 1% ionized:

$\left[{\mathrm{H}}^{+}\right]=\frac{1}{100}\times \left[\text{ Acid }\right]=0.01\times {10}^{-4}={10}^{-6} \mathrm{M}$

$\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=-\log (0.000001)=\log {10}^{-6}=6$

Therefore, the correct option is (A).