Question

A hydrogen electrode is placed separately in a buffer solution of ch3coona and acetic acid in the ratio of x : y and y : x has electrode potential values e1 volt and e2volts, respectively at 25°c. the pka values of acetic acid is (e1 and e2 are ox-potential):

Answer 1

 CH3COOH → CH3COO- + H+
H2O → H+ + OH-

Ka = [H+][CH3COO−][CH3COOH]
[H+] = Ka[CH3COOH][CH3COO−]
​The reaction occuring at the hydrogen electrode
H+ + e-   → ​1/2 H2

The equation in the 1st case:
E1 =−0.0591log1[H+]     =0.059 log[H+]     =0.059log KayxSimilarly E2 =  0.059log Kaxy

Now E1+ E2 = 0.059[logKayx+ logKaxy]=0.059logKa2
                        = 0.018 logKa

       −log Ka =−E1+E20.118pKa = −E1+E20.118