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Q.The conductivity of 0.001028 mol L−1\text{mol L}^{-1} acetic acid is 4.95×10−5 S cm−14.95\times10^{-5}\,\text{S cm}^{-1}. Calculate its dissociation constant if Λm∘\Lambda^\circ_m for acetic acid is 390.5 S cm2mol−1390.5\,\text{S cm}^2\text{mol}^{-1}.

Karnataka PUCKarnataka II PUC Board 2025Subjective· 3mImportance★★★★★
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From the molar conductivity, the degree of dissociation is about 0.123 and the dissociation constant of acetic acid is Ka≈1.78×10−5K_a \approx 1.78\times10^{-5}.

Step 1 — molar conductivity:

Λm=κ×1000c=(4.95×10−5)×10000.001028=4.95×10−20.001028=48.15 S cm2mol−1\Lambda_m = \frac{\kappa \times 1000}{c} = \frac{(4.95\times10^{-5}) \times 1000}{0.001028} = \frac{4.95\times10^{-2}}{0.001028} = 48.15\,\text{S cm}^2\text{mol}^{-1}

Step 2 — degree of dissociation:

α=ΛmΛm∘=48.15390.5=0.1233\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{48.15}{390.5} = 0.1233

Step 3 — dissociation constant (for the weak acid CH3COOH⇌CH3COO−+H+\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+): …

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