Correct Answer - D
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.