InterviewSolution
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InterviewSolution
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A hydrogen electrode placed inbuffer solution of CH_(3)COONa and acetic acid in the ratios x:y and y:x and electrode potential values E_(1) and E_(2) respectively at 25^(@)C. Calculate the pK_(a) value of acetic acid in terms of E_(1) and E_(2) (E_(1) and E_(2) are oxidation potentials). |
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Answer» Solution :For oxidation REACTION, `(1)/(2)H_(2)TOH^(+)+e^-` `E_(1)=E^(@)-(0.0591)/(1)LOG[H^(+)]_(1)` `E_(2)=E^(@)-(0.0591)/(1)log[H^(+)]_(2)` As `E^(@)=0`, adding EQNS, (i) and (ii) `E_(1)+E_(2)=-(0.0591)/(1){log[H^(+)]_(1)+log[H^(+)]_(2)}` For `CH_(3)COOHhArrCH_(3)COO^(-)+H^(+)` `K_(1)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])` or `[H^(+)]=(K_(a)[CH_(3)COOH])/([CH_(3)COO^(-)])` `therefore[H^(+)]_(1)=K_(1)(y)/(x) and [H^(+)]_(2)=K_(a)xx(x)/(y)` `thereforeE_(1)+E_(2)=-(0.0591)/(1)["log"(K_(a)y)/(x)+"log"(K_(a)x)/(y)]=-0.0591[2logK_(a)]` `thereforelogK_(a)=(E_(1)+E_(2))/(2(-0.0591))=-(E_(1)+E_(2))/(0.118)` or `pK_(a)=-logK_(a)=(E_(1)+E_(2))/(0.118)` |
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