Question

A buffer solution of `0.080M Na_(2)HPO_(4)` and `0.020 M Na_(3)PO_(4)` is prepared. The electrolytic oxidation of `1.0 mmol RNHOH` is carried out in `100mL` buffer to give
`RNHOH + H_(2)O rarr RNO_(2) + 4H^(o+) + 4e^(-)`
Calculate approximate `pH` of the solution after oxidation is complete `pK_(a_(2)), pK_(a_(2))`, and `pK_(a_(3))` of `H_(3)PO_(4)` are `2.12,7.20`, and `12.0`, respectively.

Answer 1

`[H^(o+)]` formed due to electrolyte oxidation `= 4mmol`
`{:(,Na_(2)HPO_(4),+,Na_(3)PO_(4),),("Intial mmoles",0.08xx100,,0.02xx100,),(,=8,,=2,):}`
`[H^(o+)]` will be used by `pO_(4)^(3-)` to give `H_(2)PO_(4)^(Theta)`
`{:(PO_(4)^(3-)+,2H^(o+)rarr,H_(2)PO_(4)^(Theta),,),(2,4,0,,),(0,0,2,,):}`
Thus, a new buffer containing `Na_(2)HPO_(4) (8mmol)` and `H_(2)PO_(4)^(Theta)` will remain in solution.
`:. pH = pK_(a_(H_(2)PO_(4)^(Theta)) + "log"([Na_(2)HPO_(4)])/([H_(2)PO_(4)^(Theta)])`
`= 7.2 + "log" (8)/(2) = 7.81`