Kohlrausch's Law, Degree of Ionisation and Ionisation Constants

1A pattern in the numbers

In the 1870s Friedrich Kohlrausch measured the limiting molar conductivities of many salts and noticed a striking regularity. The difference between a potassium salt and the matching sodium salt was always the same, whatever the anion: KCl minus NaCl, KNO₃ minus NaNO₃, both come to about 23.4 S cm² mol⁻¹. At infinite dilution, it seemed, each ion carried its share of the current without caring which partner it came with. That idea lets you calculate the limiting molar conductivity of any electrolyte from a short table, including weak electrolytes whose ${\mathrm{\Lambda }}_m^{\circ }\backslash Lambda^\{\backslash circ\}\_m$ cannot be measured, and from there, how far a weak acid is ionised.

Figure 12.3: Limiting molar conductivities of ions at 25 °C (CRC Handbook values, rounded). Per mole of ion, so a 2+ or 2− ion carries twice the charge of a 1+ or 1− ion.

2Using the law for a weak electrolyte

Because each ion contributes independently, ${\mathrm{\Lambda }}_m^{\circ }\backslash Lambda^\{\backslash circ\}\_m$ for ethanoic acid can be built from three strong electrolytes whose values are easy to measure by extrapolation:

$${\mathrm{\Lambda }}_m^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H})={\mathrm{\Lambda }}_m^{\circ }(\mathrm{H}\mathrm{C}\mathrm{l})+{\mathrm{\Lambda }}_m^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{N}\mathrm{a})-{\mathrm{\Lambda }}_m^{\circ }(\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l})\backslash Lambda^\{\backslash circ\}\_m(\backslash mathrm\{CH\_3COOH\})\; =\; \backslash Lambda^\{\backslash circ\}\_m(\backslash mathrm\{HCl\})\; +\; \backslash Lambda^\{\backslash circ\}\_m(\backslash mathrm\{CH\_3COONa\})\; -\; \backslash Lambda^\{\backslash circ\}\_m(\backslash mathrm\{NaCl\})$$

The $\mathrm{N}{\mathrm{a}}^{+}\backslash mathrm\{Na^\{+\}\}$ and $\mathrm{C}{\mathrm{l}}^{-}\backslash mathrm\{Cl^\{-\}\}$ contributions cancel, leaving ${\lambda }^{\circ }({\mathrm{H}}^{+})+{\lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}{\mathrm{O}}^{-})\backslash lambda^\{\backslash circ\}(\backslash mathrm\{H^\{+\}\})\; +\; \backslash lambda^\{\backslash circ\}(\backslash mathrm\{CH\_3COO^\{-\}\})$.

3From conductivity to the degree of ionisation and $K_{a}K\_a$

In a solution of a weak electrolyte, only the ionised fraction $\alpha \backslash alpha$ conducts. Because the ions are few and far apart, each conducts almost as it would at infinite dilution, so:

$$\alpha =\frac{{\mathrm{\Lambda }}_m}{{\mathrm{\Lambda }}_m^{\circ }}\backslash alpha\; =\; \backslash frac\{\backslash Lambda\_m\}\{\backslash Lambda^\{\backslash circ\}\_m\}$$

For a weak acid HA of concentration $cc$, at equilibrium $[{\mathrm{H}}^{+}]=[{\mathrm{A}}^{-}]=c\alpha [\backslash mathrm\{H^\{+\}\}]\; =\; [\backslash mathrm\{A^\{-\}\}]\; =\; c\backslash alpha$ and $[\mathrm{H}\mathrm{A}]=c(1-\alpha )[\backslash mathrm\{HA\}]\; =\; c(1\; -\; \backslash alpha)$, so:

$$K_a=\frac{(c\alpha {)}^2}{c(1-\alpha )}=\frac{c{\alpha }^2}{1-\alpha }K\_a\; =\; \backslash frac\{(c\backslash alpha)^2\}\{c(1\; -\; \backslash alpha)\}\; =\; \backslash frac\{c\backslash alpha^\{2\}\}\{1\; -\; \backslash alpha\}$$

This is Ostwald's dilution law. When $\alpha \backslash alpha$ is small, $1-\alpha \approx 11\; -\; \backslash alpha\; \backslash approx\; 1$ and $\alpha \approx \sqrt{\frac{K_a}{c}}\backslash alpha\; \backslash approx\; \backslash sqrt\{\backslash frac\{K\_a\}\{c\}\}$: diluting a weak acid a hundred times makes it ten times more ionised. The same relation holds for a weak base, with $K_{b}K\_b$.