A hydrogen electrode placed in a buffer solution of CH3COONa and CH3COOH in the ratios x:y and y:x has electrode potential values E1 volts and E2 volts respectively at 250C . The pKa value of acetic acid is :
[E1and E2 are oxidation potentials]
A
E1+E20.118
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B
E2−E10.118
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C
−E1+E20.118
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D
E1−E20.118
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Solution
The correct option is BE1+E20.118 The expression for the oxidation potential for hydrogen electrode is : E1=E01−0.0592nlog[H+]1=0.0592log1[H+]1 E2=E02−0.0592nlog[H+]2=0.0592log1[H+]2 But, pKa=−logKa=logyx[H+]1=logyx+log1[H+]1 Hence, log1[H+]1=pKa+logxy Similarly, log1[H+]2=pKa+logyx Hence, E1+E2=0.0592×2×pKa pKa=E1+E20.118