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

Options

A. 8
B. between 6 and 7
C. between 7 and 8
D. None of these

Answer

C

Explanation

`pH=-log[H^(+)]=-log10^(-8)=8`
It is not possible for acid, so it is `[H^(+)]`, the `[H^(+)]` of water is also added.
Total `[H^(+)]` in solution
`=[H^(+)]"of HCl"+[H^(+)]` of water
`=(1xx10^(-8)+1xx10^(-7))M`
`=(1+10)xx10^(-8)=11xx10^(-8)M`
`therefore pH=-log[H^(+)]`
`=-log 11xx10^(-8)`
`=-log 11xx8 log 10`
`=-1.0414+8=6.9586`