Equivalent conductance of $0.2$ M aqueous solution of a weak monobasic acid ( $H A$ ) is $10 S c m^{2} e q u i v^{- 1}$ and that at infinite dilution is $200 S c m^{2} e q u i v^{- 1}$ . Hence $p H$ of this solution is:

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Solution

$H A ⇌ H^{+} + A^{-} a t {e q b}^{m} 0.2 (1 - α) 0.2 α 0.2 α α = \frac{Λ_{e q}}{Λ_{e q}^{\infty}} = \frac{10}{200} = 0.05 [H^{+}] = 0.2 α = 0.2 \times 0.05 = 0.01 M p H = - log [H^{+}] = - log [10^{- 2}] = 2$

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Equivalent conductance of 0.2M aqueous solution of a weak monobasic acid (HA) is 10Scm2equiv−1 and that at infinite dilution is 200Scm2equiv−1. Hence pH of this solution is:

HA⇌H++A−ateqbm0.2(1−α)0.2α0.2αα=ΛeqΛ∞eq=10200=0.05[H+]=0.2α=0.2×0.05=0.01MpH=−log[H+]=−log[10−2]=2

Equivalent conductance of $0.2$ M aqueous solution of a weak monobasic acid ( $H A$ ) is $10 S c m^{2} e q u i v^{- 1}$ and that at infinite dilution is $200 S c m^{2} e q u i v^{- 1}$ . Hence $p H$ of this solution is:

$H A ⇌ H^{+} + A^{-} a t {e q b}^{m} 0.2 (1 - α) 0.2 α 0.2 α α = \frac{Λ_{e q}}{Λ_{e q}^{\infty}} = \frac{10}{200} = 0.05 [H^{+}] = 0.2 α = 0.2 \times 0.05 = 0.01 M p H = - log [H^{+}] = - log [10^{- 2}] = 2$