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Why molar conductivity of acetic acid cannotbe obtained by plotting lambda M versusunder root c at infinite dilution ? Limitingmolar conductivity of NaCl, HCl andCH,COONa are are 126.4, 425.9 and 91.05 cm?mol- respectively. Calculate limiting molarconductivity for HAC.[4 marks) |
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Answer» The limiting molar conductivity for acetic ACID (HAc) is 390.55 Scm²mol⁻¹ - at infinite dilution , electrolyte dissociates completely but at such LOW concentration the conductivity of the solution is so low that it cannot be measures accurately. Hence, the molar conductivity of acetic acid cannot be obtained by plotting LAMBDA M versus - Using Kohlrausch's law, we have λ°(NaCl) = λ°(Na⁺) + λ°(Cl⁻) ----1 λ°(HCl) = λ°(H⁺) + λ°(Cl⁻) ----2 λ°(AcNa) = λ°(Na⁺) + λ°(Ac⁻) ----3 λ°(HAc) = λ°(H⁺) + λ°(Ac⁻) ----4 - on adding 2 and 3 equation and subtracting 1 from it, we get equation 4 - so we have λ°(HAc) = λ°(HCl) + λ°(AcNa) - λ°(NaCl) λ°(HAc) = 425.9 + 91.05 - 126.4 λ°(HAc) = 390.55 Scm²mol⁻¹ |
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