Question

Given that \( \Lambda^\circ_{\text{m}} \) for NH₄Cl, NaOH, and NaCl are 129.8, 217.4, and 108.9 S cm² mol⁻¹ respectively. The molar conductivity of a 1 × 10⁻² M solution of NH₄OH is 9.33 S cm² mol⁻¹. Calculate the degree of dissociation (\( \alpha \)) of the NH₄OH solution at this concentration.

Hint:

The degree of dissociation is calculated by comparing the observed molar conductivity to the limiting value (molar conductivity at infinite dilution).

Answer 1

We know the relation: \[ \Lambda_{\text{m}} = \Lambda^\circ_{\text{m}} \alpha \] The molar conductivity of NH₄OH at the given concentration is the sum of the conductivities of NH₄⁺ and OH⁻, minus the contribution from Cl⁻. Therefore, we can express this as: \[ \Lambda_{\text{m}} (\text{NH}_4\text{OH}) = \Lambda^\circ_{\text{m}} (\text{NH}_4^+) + \Lambda^\circ_{\text{m}} (\text{OH}^-) - \Lambda^\circ_{\text{m}} (\text{Cl}^-) \] Substituting the given values: \[ 9.33 = (129.8 + 217.4 - 108.9) \times \alpha \] Simplifying the right-hand side: \[ 9.33 = 238.3 \alpha \] Solving for \( \alpha \): \[ \alpha = \frac{9.33}{238.3} = 0.0392 \] Thus, the degree of dissociation \( \alpha \) is 0.0392.