A hydrogen electrode X was placed in a buffer solution of sodium acetate and acetic acid in the ratio a : b and another hydrogen electrode 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 $p K_{a}$ value of the acid can be given as

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

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

$\frac{E_{1}}{E_{2}} \times 0.118$

$- [\frac{E_{1} + E_{2}}{0.118}]$

answer is B.

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Detailed Solution

$E_{1} = - 0.059 \log \frac{1}{[H^{+}]} = - 0.059 p H_{1}, E_{2} = - 0.059 p H_{2}; p H_{1} + p H_{2} = 2 p K_{a}$

$\Rightarrow p K_{a} = \frac{1}{2} [\frac{- E_{1} - E_{2}}{0.059}] = - \frac{(E_{1} + E_{2})}{0.118}$

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