The hydrogen electrode when placed in a buffer solution of `CH_(3)COONa and CH_(3)COOH` in the ratio x:y and y:x has oxidation electrode potetnial `E_(1) and E_(2)` volts respectively at `25^(@)C (P_(H_(2))=1 " atm").pK_(a)` for `CH_(3)COOH` will be

Options

A. `E_(1) + E_(2)`
B. `E_(1)-E_(2)`
C. `(E_(1) + E_(2))/(0.0591 xx 2)`
D. `(E_(1)-E_(2))/(0.0591 xx 2)`

Answer

C

Explanation

`CH_(3)COOH rarr CH_(2)COO^(-) + H^(+) therefore K_(a)= ([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH]) therefore H^(+) = (K_(a) [CH_(3)COOH])/([CH_(3)COO^(-)])`
Half Cell Reaction is `(1)/(2) H_(2) rarr H^(+) + e^(-)`
`E_(1)= 0 - (0.059)/(1)log K_(a)^(y)/(x), E_(2)= 0- (0.059)/(1) "log" (K_(a)^(x))/(y)`
`E_(1) + E_(2)= 0.059 xx 2 "log" (1)/(K_(a)) therefore pK_(a)= (E_(1) + E_(2))/((0.059)2)`