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

Explanation

`2H_(2)O(l)hArrH_(2)O^(+)(aq)+OH^(-)(aq)`
`K_(w)=[OH^(-)][H_(3)O^(+)]`
`=10^(-14)`
Let, `x=[OH^(-)]=[H_(3)O^(+)]` from `H_(2)O`. The `H_(2)O^(+)` concentration is generated `(i)` from the ionization of `HCl` dissolved i.e., `HCl(aq)+H_(2)O(l)hArrH_(2)O^(+)(aq)+Cl^(-)(aq)`, and `(ii)` from ionization of `H_(2)O`. In these very dilute solutions, both sources of `H_(2)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=9.5xx10^(-8)`
So, `pOH=7.02` and `pH=6.98`