Question

The equivalent conductances of two strong electrolytes at
infinite dilution in H₂O (where ions move freely through a
solution) at 25°C are given below :
∧°CH₃COONa = 91.0 S cm² / equiv. o
∧°HCl = 426.2 S cm² / equiv. o
What additional information/ quantity one needs to calculate
∧° of an aqueous solution of acetic acid?
(a) ∧° of chloroacetic acid (ClCH₂COOH)
(b) ∧° of NaCl
(c) ∧° of CH₃COOK
(d) the limiting equivalent coductance of H⁺ (????°H⁺ ).

Answer 1

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(b) ∧° of NaCl

Answer 2

Additional information / quantity one needs to calculate ∧° of an aqueous solution of acetic acid would be :

We are given :∧°CH₃COONa = 91.0 S cm²/ equiv.o

∧°HCl = 426.2 S cm² / equiv. o

at infinite dilution in H₂O at 25°C.

• According to Kohlrausch’s law, " the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. "

• Here we can write,

∧°CH₃COONa = ∧°CH₃COO- + ∧°Na+ ... (1)

• Also,

∧°HCl = ∧°H+ + ∧°Cl- ....(2)

• For acetic acid, we need ∧°CH₃COO- and ∧°H+

=> Adding (1) & (2)

• ∧°CH₃COONa + ∧°HCl = ∧°CH₃COO- + ∧°H+ + ∧°Na+ + ∧°Cl-

• ∧°CH₃COONa + ∧°HCl = ∧°CH₃COOH + ∧°NaCl

• ∧°CH₃COOH = ∧°CH₃COONa + ∧°HCl - ∧°NaCl

• We need ∧°NaCl as additional information to calculate ∧° of an aqueous solution of acetic acid