A solubility contains one mole each of HA and HB (both are weak acids) in one litre solution. Now 1 mole of NaOH is added to this solution so that both the acids are partly neutralised. Heat of neutralisation for HA and HB are- `11.8` and- `12.4` kcal respectively and the heat produced during partial neutralisation of HA and HB is- `12.25` kcal.
pH of solution, if hydrolysis constant for `A^(-)` is `10^(-9)` :

Options

A. `4.5228`
B. `5.5228`
C. `3.4696`
D. `7.9215`

Answer

A

Explanation

Also `K_(H)` for `A^(-)=10^(-9)`
`:. K_(a)` for `HA = 10^(-5) , (K_(H) = (K_(w))/(K_(a)))`
`:. [H^(+)] = (K_(1)[HA])/([A^(-)]) = (10^(-5)xx0.75)/(0.25)= 3xx10^(-5)`
and `pH = 4.5228`