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

Options

A. 8
B. `-8`
C. Between 7 and 8
D. Between 6 and 7

Answer

D

Explanation

Molar conc. of HCl `= 10^(-8)`.
`:.` pH = 8. But this cannot not be possible as pH of an acidic solution can not be more than 7. So we have to consider `[H^(+)]` coming from `H_(2)O`.
Total `[H^(+)] = [H^(+)]_(HCl) + [H^(+)]_(H_(2)O)`
IOnisation of `H_(2)O : H_(2)O rarr H^(+) + OH^(-)`
`K_(w) = 10^(-14) = [H^(+)][OH^(-)]`
Let x be the conc. of `[H^(+)]` from `H_(2)O` or `[H^(+)] = x [OH^(-)]_(H_(2)O)`
`:. 10^(-14) = (x + 10^(-8))(x)` or `x = 9.5 xx 10^(-8) M`
`:. [H^(+)] = 10^(-8) + 9.5 xx 10^(-8) = 10.5 xx 10^(-8)`
or pH `= -log(10.5 xx 10^(-8)) = 6.98`