Calculate pH of a `1.0xx10^(-8)` M solution of HCl.

Explanation

`2H_(2)O(l)hArrH_(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 ionization of HCl dissolved i.e.
`HCl(aq)+H_(2)O(l)hArrH_(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)-10^(-14)=0`
`[OH^(-)[=x=9.5xx10^(-8)`
so `pOH=7.02` and `pH=6.98`