Correct Answer - C
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`