The hydrogen electrode when placed in a buffer solution of `CH_3COONa and CH_3COOH` in the ratio x:y and y: x has oxidation electrode potential `E_ and E_` volts respectively at ` 25^(@) C ` `(pH_2 =1 atm)` . `pK_a` for `CH_3COOH` 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

`E =E^(0)-(0.0591)/(1) log""(1)/([H^(+)]) , ((-E_(RP))/(0.0591)) = log (H^(oplus)) = pt `
` (+E_1)/(0.0591) = P^(Ka) + log""(x)/(y)` .......... (1) , ` (+E_2)/(0.0591) =P^(Ka) + log""(y)/(x) ` ......... (2) from (1) & (2)
`(+E_1 + E_2)/(0.0591) = 2P^(Ka) + log((x)/(y)) + log((y)/(x)) + ((E_1 + E_2)/(0.0591)) = 2P^(Ka) + log""((x)/(y) xx (y)/(x)) , P^(Ka) = + ((E_1 + E_2)/(2 xx 0.0591))`