Question

The specific conductance of a 0.01 M solution of acetic acid at 298 K is 1.65 × 10⁻⁴ ohm⁻¹ cm⁻¹. The molar conductance at infinite dilution for H⁺ ion and CH₃COO⁻ ion is 349.1 ohm⁻¹ cm² mol⁻¹ and 40.9 ohm⁻¹ cm² mol⁻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

Given:

K = 1.65 × 10⁻⁴ Ω⁻¹ cm⁻¹

\[\ce{\lambda{^{\circ}_{H^{+}}}}\] = 349.1 Ω⁻¹ cm² mol⁻¹

\[\ce{\lambda^{\circ}_{CH_3COO^-}}\] = 40.9 Ω⁻¹ cm² mol⁻¹

C = 0.01 M

So, \[\ce{\Lambda{^{\circ}_{m}}CH3COOH}\] = 349.1 + 40.9 = 390 Ω⁻¹ cm² mol⁻¹

a. \[\ce{\Lambda_m = \frac{1000 \times \kappa}{C}}\]

= \[\ce{\frac{1000 \times 1.65 \times 10^{-4}}{0.01}}\]

= 16.5 Ω⁻¹ cm² mol⁻¹

b. Degree of dissociation (α) = \[\ce{\frac{\Lambda_m}{\Lambda{^{\circ}_{m}}}}\]

= \[\ce{\frac{16.5}{390}}\]

= 0.0423

c. Dissociation constant (K_α) = \[\ce{\frac{(\Lambda_{m})^2 * C}{\Lambda{^{\circ}_{m}}(\Lambda^{\circ}_{m} - \Lambda_m)}}\]

= \[\ce{\frac{(16.5)^2 \times 0.01}{390(390 - 16.5)}}\]

= \[\ce{\frac{272.25 \times 0.01}{390 \times 373.5}}\]

= \[\ce{\frac{2.7225}{145665}}\]

= 1.86 × 10⁻⁵