Question

The specific conductance of a 0.01 M solution of acetic acid at +98 K is 1.65 x 10^-4 ohm cm^-1. The molar conductance at infinite dilution for H⁺ ion and CH₃COO^- ion is 349.1 ohm^-1 cm²mol^-1 and 40.9 ohm^-1 cm²mol^-1 respectively.

Calculate:

1) Molar conductance of the solution.

2) The degree of dissociation of CH₃COOH.

3) A dissociation constant for acetic acid

Answer 1

`K = 1.65 xx 10^(-4) Ω^(-1) "cm"^(-1)`

`lambda_(H^(+))^0 = 349.1 Ω^(-1) cm^(2) mol^(-1)`

`lamda_(CH_3COO^(-))^0 = 40.9 Ω^(-1) cm^2 mol^(-1)`

C = 0.01 M.

1) `∧ =(1000 K)/C = (1000 xx 1.65 xx 10^(-4))/(10^(-2)) = 16.5 Ω^(-1) "cm"^2 mol^(-1)`

`∴ lambda = 16.5 Ω^(-1) cm^2 mol^(-1)`

2) `∧^@ = lambda^@H^(+) + lambda^@CH_3COO^(-) = 349.1 + 40.9 = 390.0`

`alpha = ∧/∧^@ = 16.5/390 = 0.0423`

`alpha = 0.0423`

3) `K = (∧^2C)/(∧^@(∧^@ - ∧)) = ((16.5)^2 xx 0.01)/(390 (390 - 16.5))`

`K = 1.86 xx 10^(-5)`