Question

The conductivity of a 0.01 M solution of a 1 : 1 weak electrolyte at 298 K is 1.5 × 10⁻⁴ S cm⁻¹.

i) molar conductivity of the solution

ii) degree of dissociation and the dissociation constant of the weak electrolyte

Given that

$$\mathrm{\lambda }{\phantom{A}}_{\mathrm{cation}}^{\circ }$$ = 248.2 S cm² mol⁻¹

$$\mathrm{\lambda }{\phantom{A}}_{\mathrm{anion}}^{\circ }$$ = 51.8 S cm² mol⁻¹

Answer 1

Given C = 0.01 M

K = 1.5 × 10⁻⁴ S cm⁻¹

$$\mathrm{\lambda }{\phantom{A}}_{\mathrm{cation}}^{\circ }$$ = 248.2 S cm² mol⁻¹

$$\mathrm{\lambda }{\phantom{A}}_{\mathrm{anion}}^{\circ }$$ = 51.8 S cm² mol⁻¹

i) Molar conductivity

K = 1.5 × 10⁻⁴ S cm⁻¹

1 cm⁻¹ = 10² m⁻¹

= 1.5 × 10²

${\displaystyle {\Lambda }_{\text{m}}^{\circ }=\frac{\text{K}\left({\text{sm}}^{-1}\right)\times {10}^{-3}}{\text{C}\left(\text{in M}\right)}}$ mol⁻¹ m³

= ${\displaystyle \frac{1.5\times {10}^2\times {10}^{-3}}{0.01}}$ S mol⁻¹ m²

= 1.5 × 10⁻³ S m² mol⁻¹

ii) Degree of dissociation α = ${\displaystyle \frac{{\Lambda }^{\circ }}{{\Lambda }_{\infty }^{\circ }}}$

${\displaystyle \left({\Lambda }_{\infty }^{\circ }\right)={\lambda }_{\text{cation}}^{\circ }+{\lambda }_{\text{anion}}^{\circ }}$

= (248.2 + 51.8) S cm² mol⁻¹

= 300 S cm² mol⁻¹

= 300 × 10⁻⁴ S m² mol⁻¹

α = ${\displaystyle \frac{1.5\times {10}^{-3} {\text{S m}}^2 {\text{mol}}^{-1}}{300\times {10}^{-4} {\text{S m}}^2 {\text{mol}}^{-1}}}$

α = 0.05

K_a = ${\displaystyle \frac{{\alpha }^2\text{C}}{1-\alpha }}$

= ${\displaystyle \frac{{\left(0.05\right)}^2\left(0.01\right)}{1-0.05}}$

= ${\displaystyle \frac{25\times {10}^{-4}\times {10}^{-2}}{95\times {10}^{-2}}}$

= 0.26 × 10⁻⁴

= 2.6 × 10⁻⁵