Calcuate `pH` of `10^(-8)M HCl` solution at `25^(@)C` (Take `log1.05=0.02)`

Explanation

Here `[H^(+)]_(HCl)=10^(8)M(lt10^(-6)M)`.So `[H^(+)]` from water has to be considered .But `[H^(+)]_(f romH_(2)O)ne10^(-7)M` because of common ion effect exerted on it by `H^(+)` ions of `HCl`.So considering dissocation of `H_(2)O`:
`H_(2)OhArr H^(+)+OH^(-)H_(2)OhArr underset(10^(-8)+x)H^(+)+underset(x)OH^(-)`
`K_(W)=[H^(+)][OH^(-)]`
`10^(-14)=x(x+10^(-8))`
`rArr x^(2)+x xx10^(-8)-10^(-14)=0`
`x=(-10^(-8)+-sqrt(10^(-16)+4xx10^(-14)))/(2)=(-10^(-8)+10^(-7)sqrt((4+(1)/(100))))/(2)=((sqrt(401)-1)10^(-8))/(2)=0.95xx10^(-7)`
`[H^(+)]=10.5xx10^(-8)=1.05xx10^(-7)M`
`pH=7-log1.05~~6.98`
Note:- For `10^(-9)MHCl pH~~7`
For `10^(-12)M HCl pH~~7`