The `pH` of `10^(-8)M` solution of `HCl` in water is

Explanation

`2H_(2)O (I) hArr H_(3)O^(+) (aq) + OH^(-)(aq)`
`K_(w) = [OH^(-)][H_(3)O^(+)]`
`= 10^(-14)`
Let, `x = [OH] = [H_(3)O^(+)]` from `H_(2)O`. The `H_(3)O^(+)` concentration is generated `(i)` from the inonization of HCl dissolved i.e., `HCl(aq) + H_(2)(I) hArr H_(3)O^(+)(aq)+Cl^(-)(aq)` and (ii) from ionization of `H_(2)O`. In these very dilute solutions. both sources of `H_(3)O^(+)` must be considered.
`[H_(3)O^(+)] = 10^(-8)+x`
`K_(w) = (10^(-8) + x)(x) = 10^(-14)`
or `x^(2) + 10^(-8) x - 10^(-14) = 0`
`[OH^(-)] = x = 9.5 xx 10^(-8)`
So, `pOH = 7.02` and `pH = 6.98`