Kohlrausch's law

Statement. Kohlrausch's law of independent migration of ions states that at infinite dilution, ions are so far apart that they exert no electrostatic influence on one another. Each ion then migrates independently and contributes a constant amount to the molar conductance of any electrolyte it appears in, regardless of the other ion present. For an electrolyte that dissociates into $\nu_+$ cations and $\nu_-$ anions, the limiting molar conductance is the sum of the ions' own limiting ionic conductances:

where: $\lambda_+^\circ$ and $\lambda_-^\circ$ are the limiting (infinite-dilution) molar ionic conductances of the cation and anion, and $\nu_+, \nu_-$ are the number of cations and anions produced per formula unit. For a simple 1:1 salt such as NaCl, this is just $\Lambda_m^\circ(\text{NaCl}) = \lambda_{\text{Na}^+}^\circ + \lambda_{\text{Cl}^-}^\circ$; for a salt like $\text{CaCl}_2$, $\nu_+ = 1$ and $\nu_- = 2$, so $\Lambda_m^\circ = \lambda_{\text{Ca}^{2+}}^\circ + 2\lambda_{\text{Cl}^-}^\circ$.

Why additivity holds physically. At infinite dilution ions are spread so far apart that the electrostatic pull between oppositely charged neighbours vanishes. Each ion's contribution to conductance then becomes purely a property of that ion alone — its size, its charge, how easily it slips through water — with no dependence on which counter-ion happens to share its salt. This is exactly what makes $\Lambda_m^\circ$ values additive across different salts sharing a common ion, and it is the reason the combination trick below works.

Three linked applications. First, since a weak electrolyte's own $\Lambda_m^\circ$ cannot be extrapolated, combine $\Lambda_m^\circ$ values of three strong electrolytes that together supply exactly the desired ions:

The $\text{Na}^+$ introduced by CH₃COONa and the $\text{Cl}^-$ introduced by HCl are exactly cancelled by subtracting NaCl, because Kohlrausch's law guarantees their contributions are the same fixed numbers in every salt they appear in. Second, once $\Lambda_m^\circ$ is known, the degree of dissociation $\alpha$ of the weak electrolyte at some measured concentration follows from $\alpha = \Lambda_m/\Lambda_m^\circ$. Third, feeding that $\alpha$ into the Ostwald dilution law, $K_a = \dfrac{C\alpha^2}{1-\alpha}$, gives the dissociation constant of the weak acid or base.

Problem: Given $\Lambda_m^\circ(\text{CH}_3\text{COONa}) = 91.0$, $\Lambda_m^\circ(\text{HCl}) = 426.2$, and $\Lambda_m^\circ(\text{NaCl}) = 126.5$ (all in $\text{S cm}^2\text{mol}^{-1}$), find $\Lambda_m^\circ$ of acetic acid; then given a $0.01\ M$ solution of this acid has $\Lambda_m = 16.2\ \text{S cm}^2\text{mol}^{-1}$, find its degree of dissociation $\alpha$ and dissociation constant $K_a$.

Step 1. Apply the Kohlrausch combination: $\Lambda_m^\circ(\text{CH}_3\text{COOH}) = \Lambda_m^\circ(\text{CH}_3\text{COONa}) + \Lambda_m^\circ(\text{HCl}) - \Lambda_m^\circ(\text{NaCl}) = 91.0 + 426.2 - 126.5$.

Step 2. $91.0 + 426.2 = 517.2$, and $517.2 - 126.5 = 390.7\ \text{S cm}^2\text{mol}^{-1}$. This equals $\lambda_{\text{H}^+}^\circ + \lambda_{\text{CH}_3\text{COO}^-}^\circ$, confirming the sodium and chloride contributions cancelled correctly.

Step 3. Find the degree of dissociation using the measured $\Lambda_m$: $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ} = \dfrac{16.2}{390.7} = 0.0415$, meaning only about $4.15\%$ of the acid molecules are dissociated at this concentration.

Step 4. Substitute $\alpha$ and $C = 0.01\ M$ into the Ostwald dilution law: $K_a = \dfrac{C\alpha^2}{1-\alpha} = \dfrac{0.01\times(0.0415)^2}{1-0.0415} = \dfrac{0.01\times0.001722}{0.9585} = \dfrac{1.722\times10^{-5}}{0.9585}$.

Step 5. $K_a = 1.80\times10^{-5}$. This matches the well-known dissociation constant of acetic acid, so $\Lambda_m^\circ(\text{CH}_3\text{COOH}) = 390.7\ \text{S cm}^2\text{mol}^{-1}$, $\alpha = 0.0415$, and $K_a = 1.80\times10^{-5}$.

Competitive stretch

If instead you were given $\Lambda_m^\circ(\text{NH}_4\text{Cl})$, $\Lambda_m^\circ(\text{NaOH})$, and $\Lambda_m^\circ(\text{NaCl})$, the same cancellation trick gives $\Lambda_m^\circ(\text{NH}_4\text{OH}) = \Lambda_m^\circ(\text{NH}_4\text{Cl}) + \Lambda_m^\circ(\text{NaOH}) - \Lambda_m^\circ(\text{NaCl})$ — here it is $\text{Na}^+$ and $\text{Cl}^-$ again that are the common spectator ions being introduced and removed, so always check which two ions are shared by the "unwanted" strong electrolyte before assigning plus and minus signs.

• Trying to extrapolate a weak electrolyte's own $\Lambda_m$ vs $\sqrt{C}$ graph to $C=0$ — this fails because the curve rises too steeply near the axis; the Kohlrausch combination method using strong-electrolyte data must be used instead.
• Getting the sign wrong in the combination formula — you add the two strong electrolytes that between them supply the desired ions and subtract the strong electrolyte made of the two spectator ions you want removed, so their contributions cancel exactly.
• Mixing units — $\lambda^\circ$ and $\Lambda_m^\circ$ must be compared in the same units throughout a calculation ($\text{S cm}^2\text{mol}^{-1}$, sometimes tabulated per equivalent as $\text{S cm}^2\text{eq}^{-1}$ for ions); check which convention a question is using before combining values.
• Applying $\alpha = \Lambda_m/\Lambda_m^\circ$ to a strong electrolyte — this ratio is not meaningful there, since strong electrolytes are treated as fully dissociated at all concentrations; their small rise in $\Lambda_m$ on dilution reflects weakening inter-ionic attraction, not increasing dissociation.

Ask the tutor: if you would like more practice building $\Lambda_m^\circ$ for other weak acids or bases from strong-electrolyte data, or a walkthrough connecting alpha here to the Ostwald dilution law from ionic equilibrium.