A hydrogen electrode $(p_{H_{2}} = 1 a t m; T = 298 K)$ is placed in buffer solution of ${C H}_{3} C O O N a$ and ${C H}_{3} C O O H$ in the molar ratio $x : y$ and then in $y : x$ has reduction potential values as $E_{1}$ and $E_{2}$ volts respectively. Then $p K_{e}$ of ${C H}_{3} C O O H$ will be:

\frac{E_{1} + E_{2}}{0.118}

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\frac{E_{1} + E_{2}}{0.059}

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- \frac{E_{1} + E_{2}}{0.118}

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\frac{E_{2} - E_{1}}{0.118}

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Solution

The correct option is C $- \frac{E_{1} + E_{2}}{0.118}$
$C H_{3} C O O H ⟶ C H_{3} C O O^{-} + H^{+}$
$H_{2} O ⟶ H^{+} + O H^{-}$
$K_{a} = \frac{[H^{+}] [{C H}_{3} {C O O}^{-}]}{[{C H}_{3} C O O H]} [H^{+}] = K_{a} \frac{[{C H}_{3} C O O H]}{[{C H}_{3} {C O O}^{-}]}$
Reaction at hydrogen electrode $⟶ H^{+} + e^{-} ⟶ \frac{1}{2} H_{2}$
$E_{1} = \frac{- 0.059}{1} l o g \frac{1}{[H^{+}]} = 0.059 l o g \frac{1}{[H^{+}]} = 0.059 l o g K_{a} \frac{y}{x}$
Similarly, $E_{2} = 0.059 l o g K_{a} \frac{x}{y} E_{1} + E_{2} = 0.059 [l o g K_{a} \frac{y}{x} + l o g K_{a} \frac{x}{y}] = 0.059 l o g K_{a}^{2} = 0.018 l o g K_{a} - l o g K_{a} = \frac{- E_{1} + E_{2}}{0.118} {p K}_{a} = \frac{- {(E}_{1} + E_{2})}{0.118}$

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