A hydrogen electrode X was placed in a buffer solution of sodium acetate and acetic acid in the ratio `a:n` and another hydrogen electode Y was placed in a buffer solution of sodium acetate and acetic acid in the ratio `b:a`. If reduction potential values for two cells are found to be `E_(1)` and `E_(2)` respectively w.r.t standard hydrogen electrode, the `pK_(a)` value of the acid can be given as:

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

C

Explanation

`E_(H^(+)//H_(2))=-0.59xx2xxpH`
`pH_(1)=pK_(a)+"log"(a)/(b), pH_(2)=pK_(a)+"log"(b)/(a)`
`therefore pH_(1)+pH_(2)=pK_(a) " " therefore pK_(a)=(-(E_(1)+E_(2)))/(0.118)`