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Q.

The hydrogen electrode when placed in a buffer solution of CH₃COONa and CH₃COOH in the ratio x:y and y:x has oxidation electrode potential E₁ and E₂ volts respectively at 25°C (pH₂ = 1 atm). pK_a for CH₃COOH will be :

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a

$(E_{1} - E_{2}) / 0.0591 \times 2$

b

$(E_{1} + E_{2}) / 0.0591 \times 2$

c

$E_{1} - E_{2}$

d

$E_{1} + E_{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} E = E^{0} - (0 .0591 / 1) \log 1 / [H^{+}] \\ + E_{1} / 0 .0591 = P^{Ka} + \log x / y - - - - (1) \\ + E_{2} / 0 .0591 = P^{Ka} + \log y / x - - - - (2) \\ from (1) \& (2) \\ P^{Ka} = + (E_{1} + E_{2} / 2 \times 0 .0591) \end{array}$

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