Calculate pH of a `1.0 xx 10^(-8)` M solution of HCl.

Explanation

The complete ionization occurs of strong acid HCI.
Thus, [HCl]=`[H^+]=1.0xx10^(-8)`M
Because of self ionization of water, the `[H_3O^+]` also present in solution, So, Hydrogen ion concentration of HCl is less than `[H_3O^+]` of water therefore `[H_3O^+]` of water is taken in calculation.
Self ionization of water : `2H_2O_((l)) hArr H_3O_((aq))^(+) + OH_((aq))^(-)`
`K_w=[H_3O^+][OH^-]=1.0xx10^(-14)` ....(Equation )
Suppose from water `[H_3O^+]=[OH^-]=1xx10^(-7)`M
Total `[H_3O^+]` in solution
=(`[H_3O^+]` of water ) + (`[H_3O^+]` of HCl )
`=(1xx10^(-7)+1xx10^(-8))` M
`=1.1xx10^(-7)` M
`pH=-log [H^+]=-log (1.1xx10^(-7))`
=-(0.0414-7.0)=(-6.9586)=+6.96