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

Expert Answer:

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+  
0CH3COO+   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