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

Explanation

`2H_(2)O(1)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 ionization of HCl dissolved. i.e., `HCl (aq) +H_(2)O^(+) (1) hArr H_(3)O^(+)+Cl^(-)(aq)`, and (ii) from ionization of `H_(2)O`. in these very dilute solutions both soruces of `H_(3)O^(+)` must be considered :
`[H_(3)O^(+)]=10^(-8)+x`
`K_(w)=10^(-8)+x`
`K_(w)=(10^(-8)+x)(x)=10^(-14)`
or `x^(2)=x=9.5xx10^(-8)`
So, pOH =7.02 and pH=6.98