A hydrogen electrode placed in a solution containing sodium acetate and acetic acid in the ratio of `x:y` and `y:x` has an electrode potential value `E_(1)` and `E_(2)` volts, respectively, at `25^(@)C`. The `pK_(a)` value of acetic acid is

Explanation

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)`