ElectrochemistryHard
Question
A hydrogen electrode placed in a buffer solution of CH3COONa and CH3COOH in the ratio’s x:y and y:x has electrode potential values E1 and E2 volts, respectively, at 25°C. The pKa value of acetic acid is
Options
A.(E1 + E2)/0.118
B.(E2 − E1)/0.118
C.−(E1 + E2)/0.118
D.(E1 − E2)/0.118
Solution
The potential of hydrogen electrode at H2(1 bar) may be expressed as E = – 0.059 PH
Now, $E_{1} = - 0.059\left\lbrack P^{Ka} + \log\frac{x}{y} \right\rbrack$
and $E_{2} = - 0.059\left\lbrack P^{Ka} + \log\frac{y}{x} \right\rbrack$
$\therefore P^{Ka} = \frac{- \left( E_{1} + E_{2} \right)}{0.118}$
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