The resistance of 0.1 M weak acid HA in a conductivity cell is $2 \times 10^{3} Ohm$ . The cell constant of the cell is $0.78 {Cm}^{- 1}$ and $λ_{m}^{\circ}$ of acid HA is $390 S {cm}^{2} mol e^{- 1} .$ The pH of the solution is

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4.2

3.3

answer is D.

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Detailed Solution

$Given, C = 0.1 M$

$Λ_{m}^{\circ} = 390 {Scm}^{2} {mol}^{- 1}$

$R = 2 \times 10^{3} ohm$

$G^{⋆} = 0.78 {cm}^{- 1}$

$Now, K = \frac{R}{G^{*}} = \frac{0.78}{2 \times 10^{3}} = 3.9 \times 10^{- 4}$

$Λ_{m}^{\circ} = \frac{K \times 1000}{C} = \frac{3.9 \times 10^{- 4} \times 1000}{0.1} = 3.9$

$α = \frac{Λ_{m}}{Λ_{m}^{\circ}} = \frac{3.9}{390} = 10^{- 2}$

$[H^{+}] = C \cdot α = 0.1 \times 10^{- 2} = 10^{- 3}$

$So, pH = - \log 10^{- 3} = 3$

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