Molar conductivities at infinite dilution of KCl, HCl and CH3COOK are 0.013, 0.038 and 0.009 Sm2mol–1 respectively at 291K. If conductivity of 0.001M CH3COOH is 2.72 × 10–3 Sm–1 then find % degree of ionization of CH3COOH.

∧M0=∧M0+∧M0−∧M0CH3COOKHClKCl=(0.009+0.038−0.013)=340S−cm2−mole−1∧M0=K×103/M=2.72×10−3×103×10−2/0.001=27.2Scm2mole−1α=(27.2/340)=0.08⇒8%

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Molar conductivities at infinite dilution of KCl, HCl and CH₃COOK are 0.013, 0.038 and 0.009 Sm²mol^–1 respectively at 291K. If conductivity of 0.001M CH₃COOH is 2.72 × 10^–3Sm^–1 then find % degree of ionization of CH₃COOH.

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Detailed Solution

$\begin{array}{l} \land_{M}^{0} = [\begin{array}{ccc} \land_{M}^{0} + & \land_{M}^{0} - & \land_{M}^{0} \\ {CH}_{3} COOK & HCl & KCl \end{array}] \\ = (0 .009 + 0 .038 - 0 .013) \\ = 340 S - {cm}^{2} - {mole}^{- 1} \\ \land_{M}^{0} = K \times 10^{3} / M \\ = 2 .72 \times 10^{- 3} \times 10^{3} \times 10^{- 2} / 0 .001 \\ = 27 .2 {Scm}^{2} {mole}^{- 1} \\ α = (27 .2 / 340) = 0 .08 \Rightarrow 8 % \end{array}$