The pH of `10^(-8)` M HCl solution is :

Options

A. 8
B. 6.9586
C. more than 8
D. slightly more than 7

Answer

B

Explanation

The neutral water has `[H^(+)] = 1 xx 10^(-7)M`. By adding `1.0 xx 10^(-8) M` HCl, a concentration of `1.0 xx 10^(-8)M " " H^(+)` ions has increased in solution.
Thus, total `[H^(+)] =(1 xx 10^(-7) + 1 xx 10^(-8))M`
`=(1 xx 10^(-7) + 0.1 xx 10^(-7))M`
`=1.1 xx 10^(-7)M`
`pH -log (1.1 xx 10^(-7)) = -log [1.1 + log 10^(-7)]`
`=-[0.0414 - 7.0]`
=6.9586