A buffer solution 0.04 M in `Na_(2)HPO_(4)` and 0.02 in `Na_(3)PO_(4)` is prepared. The electrolytic oxidation of 1.0 milli-mole of the organic compound RNHOH is carried out in 100 mL of the buffer. The reaction is `RNHOH+H_(2)OrarrRNO_(2)+4H^(+)+4e^(-)` The approximate pH of solution after the oxidation is complete is :
`[Given : for H_(3)PO_(4),pK_(a1)=2.2,pK_(a2)=7.20,pK_(a3)=12]`
(a)`6.90`
(b)`7.20`
(c)`7.5`
(d)None of these

Explanation

`[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`