${10}^{-6} \mathrm{M}$ NaOH is diluted by 100 times. The pH of diluted base is

After dilution, $\left[{\mathrm{OH}}^{-}\right] = {10}^{-8} \mathrm{M}$

or   $\left[{\mathrm{H}}^{+}\right] = {10}^{-6} \mathrm{M}.$  pH cannot be 6 because total

$\left[{\mathrm{OH}}^{-}\right] = {10}^{-7} + {10}^{-8} = {10}^{-7 }\left(1 + 0.1\right)$

$= {10}^{-7} \times 1.1 \mathrm{or} \left[{\mathrm{H}}^{+}\right] = {10}^{-14}/ \left({10}^{-7 } \times 1.1\right)$

$= 9.09 \times {10}^{-8}$

$\therefore \mathrm{pH} = 8 - \log 9.09 = 8-0.9546 = 7.0454$