InterviewSolution
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InterviewSolution
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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. `8.2` |
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Answer» 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` |
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