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

Options

A. `(E_(1) + E_(2))/(0.118)`
B. `(E_(2) - E_(1))/(0.118)`
C. `-(E_(1) + E_(2))/(0.118)`
D. `(E_(1) - E_(2))/(0.118)`

Answer

A

Explanation

`E_(1) = E^(@) - (0.059)/(1)log[H^(+)]_(1)`
`E_(2) = E^(@) - (0.059)/(1)log[H^(+)]_(2)`
On adding (also `E_(H)^(@) = 0`)
`E_(1) + E_(2) = -(0.059)/(1)[log[H^(+)]_(1) + log[H^(+)]_(2)]`
Now for `CH_(3)COOH hArr CH_(3)COO^(-)+H^(+)`
`[H^(+)] = (K_(a)[CH_(3)COOH])/([CH_(3)COO^(-)])`
`:. [H^(+)]_(1) = K_(a)(y)/(x)`
`[H^(+)]_(2) = K_(a)(x)/(y)`
`:. E_(1) + E_(2) = -(0.059)/(2)["log"(K_(a)y)/(x)+log'(K_(a)x)/(y)]`
`= - 0.059[2logK_(a)]`
`log K_(a) = (E_(1) + E_(2))/(2 xx (-0.059))`
`log K_(a) = (E_(1) + E_(2))/(0.118)`
or `pK_(a) = (E_(1) + E_(2))/(0.118)`