When one drop of a concentrated `HCl` is added to `1L` of pure water at `25^(@)C`, the `pH` drops suddenly from `7` to `4`. When the second drop of the same acid is added, the `pH` of the solution further drops to about

Options

A. `0`
B. `1.0`
C. `2.0`
D. `3.7`

Answer

D

Explanation

When `pH = 7, [H^(o+)]` initial `=10^(-7)M`.
When `pH = 4, [H^(o+)]` Total `=10^(-4)M`.
`:. [H^(o+)]` added from `HC1 = (10^(-4) - 10^(-7))`
`=10^(-4) (1-0.001)~~ 10^(-4)M`.
When another drop of `HC1` is added:
When `pH = 4, [H^(o+)] "initial" = 10^(-4)M`
`:. [H^(o+)] added ~~ 10^(-4)`
`[H^(o+)] Total = (10^(-4)+10^(-4)) = 2 xx 10^(-4) M`
`:.pH =- log (2 xx 10^(-4)) =- 0.3 + 4 = 3.7`
Note: Change in volume on adding one drop of `HC1` is assumed to be negligible.