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]`

Options

A. `6.90`
B. `7.20`
C. `7.5`
D. `8.2`

Answer

C

Explanation

Hint: moles of `H^(+)` produced `= 4 xx 10^(-3)`
`PO_(4)^(3-)(aq.) + H^(+)(aq.) rarr HPO_(4)^(2-)(aq.)`
`{:("Initial",,0.002,0.004,,"__"),("milli moles",,,,,),("Final","___",0.002,,0.002,),("milli moles",,,,,):}`
`HPO_(4)^(3-) + H^(+) (aq.) rarr H_(2)PO_(4)^(-) (aq.)`
`{:("Initial milli",0.006,0.002,"___"),("moles",,,),("Final milli",0.004,"__",0.002),("moles",,,):}`
finally buffr solution of `H_(2)PO_(4)^(-)` (acid) `HPO_(4)^(2-)`
(conjugate base) is formed
`pH =pK_(a_(2)) + "log" ([HPO_(4)^(2-)])/([H_(2)PO_(4)])`
`= 7.2 + "log" (0.004)/(0.002) = 7.5`