Calculate the Λ0 for acetic acid, given, Λ0 (CH3COONa), Λ0 ( NaCl), Λ0 ( HCl) is 91 S cm-2mol-1, , 126 S cm-2mol-1 , 426 S cm-2mol-1 respectively. Calculate the degree of dissociation of the acid given the molar conductivity of the acid at the given concentration is 16.5 S cm-2mol-1.
Λ0 ( CH3COOH) = ?
Asked by Topperlearning User | 16th Jun, 2014, 12:50: PM
We know from Kohlrausch law that for
Λ0 ( CH3COONa) = λ0CH3COO- + λ0 Na + = 91S cm-2mol-1 ….(i)
Λ0 ( NaCl) = λ0Cl- + λ0 Na + = 126 S cm-2mol-1 …………. (ii)
Λ0 ( HCl) = λ0H + + λ0 Cl- = 426 S cm-2mol-1 ………….(iii)
Adding (i) and (iii) and subtracting (ii) from it we get,
Λ0 ( CH3COONa) + Λ0 ( HCl) = (λ0CH3COO- + λ0 Na +) + (λ0H + + λ0 Cl-)
Λ0 ( CH3COONa) + Λ0 ( HCl) = 91S cm-2mol-1 + 426 S cm-2mol-1 ………….(iv)
subtracting (ii) from (iv) we get,
(λ0CH3COO- + λ0 Na +) + (λ0H + + λ0 Cl-) - ( λ0Cl- + λ0 Na +) = λ0CH3COO- + λ0 H+
(λ0CH3COO- + λ0 Na +) + (λ0H + + λ0 Cl-) – ( λ0Cl- + λ0 Na +) = Λ0 ( CH3COOH)
Λ0 ( CH3COONa) + Λ0 ( HCl) - Λ0 ( NaCl) = 91S cm-2mol-1 + 426 S cm-2mol-1-126 S cm-2mol-1
Λ0 ( CH3COOH) = 517 S cm-2mol-1 - 126 S cm-2mol-1 = 391 S cm-2mol-1
Answered by | 16th Jun, 2014, 02:50: PM
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