The pH of ${10}^{-8}$ M HCl is

The concentration of HCL is very less & along with acid there will be water molecules also present.

Therefore, $[H_3{O}^{+}{]}_{total}=[H_3{O}^{+}{]}_{acid}+[H_3{O}^{+}{]}_{water}$ -----(1)

HCL is completely ionised,

$HC{L}_{\left(aq\right)}+H_2{O}_{\left(l\right)}\to H_3{O}_{\left(aq\right)}^{+}+O{H}_{\left(aq\right)}^{-}$

$[H_3{O}^{+}{]}_{acid}=1.0\times {10}^{-8}M$

The concentration of $\left[H_3{O}^{+}\right]$ from ionisation is equal to $\left[O{H}^{-}\right]$ from water i.e $[H_3{O}^{+}{]}_{water}=[O{H}^{-}{]}_{water}=x$

From (1) we get,  $[H_3{O}^{+}{]}_{total}=1.0\times {10}^{-8}+x$

Since $K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]=1\times {10}^{-14}$ ----(2)

Substituting value of $[H_3{O}^{+}{]}_{total}$ & $\left[O{H}^{-}\right]$ in (2), we get

$(1.0\times {10}^{-8}+x)\left(x\right)=1\times {10}^{-14}$

$x^2+{10}^{-8}x-{10}^{-14}=0$, solving we get $x=9.5\times {10}^{-8}$

pOH = $-log\left[O{H}^{-}\right]$ = $-log[9.5\times {10}^{-8}]$ = 7.02

from equation pH + pOH = 14

                           pH = 14 – 7.02 = 6.98