Kohlrausch's Law, Degree of Dissociation & Ka · Electrochemistry

CHEConductance & Kohlrausch's Law

Kohlrausch's Law, Degree of Dissociation & K_a

Kohlrausch's law of independent migration of ions lets you obtain the limiting molar conductivity of a weak electrolyte from those of strong electrolytes. From it you compute the degree of dissociation and the dissociation constant $K_a$ of a weak acid like acetic acid.

Kohlrausch's law

\Lambda^\circ_m = \nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_-

\nu_+,\nu_-

= number of cations/anions per formula unit;

\lambda^\circ

= limiting ionic conductivities.

Limiting

\Lambda^\circ_m

for a weak acid

\Lambda^\circ_m(CH_3COOH) = \Lambda^\circ_m(CH_3COONa) + \Lambda^\circ_m(HCl) - \Lambda^\circ_m(NaCl)

built from three strong electrolytes.

Degree of dissociation

\alpha = \dfrac{\Lambda_m}{\Lambda^\circ_m}

\Lambda_m

at the given concentration;

\alpha

is dimensionless and lies between

0

and

1

Dissociation constant

K_a = \dfrac{c\alpha^2}{1-\alpha} = \dfrac{c\,\Lambda_m^2}{\Lambda^\circ_m(\Lambda^\circ_m - \Lambda_m)}

for

\alpha\ll1

K_a \approx c\alpha^2

;

c

\text{mol L}^{-1}

Kohlrausch's law: at infinite dilution $\Lambda^\circ_m = \nu_+ \lambda^\circ_+ + \nu_- \lambda^\circ_-$ — each ion contributes a fixed value independent of the other ion present.
For a weak acid, build $\Lambda^\circ_m$ from strong electrolytes, e.g. $\Lambda^\circ_m(CH_3COOH) = \Lambda^\circ_m(CH_3COONa) + \Lambda^\circ_m(HCl) - \Lambda^\circ_m(NaCl)$ .
Degree of dissociation: $\alpha = \dfrac{\Lambda_m}{\Lambda^\circ_m}$ (ratio of molar conductivity at the given concentration to that at infinite dilution).
Dissociation constant of a weak monobasic acid: $K_a = \dfrac{c\alpha^2}{1-\alpha}$ ; when $\alpha \ll 1$ this simplifies to $K_a \approx c\alpha^2$ .
Combined form: $K_a = \dfrac{c\,\Lambda_m^2}{\Lambda^\circ_m(\Lambda^\circ_m - \Lambda_m)}$ , useful for going straight from conductivity data to $K_a$ .
Get $\Lambda_m$ from conductivity first: $\Lambda_m = \dfrac{\kappa \times 1000}{c}$ , then form the ratio $\alpha = \Lambda_m/\Lambda^\circ_m$ .
Keep units consistent: $\Lambda$ in $S\,\text{cm}^2\,\text{mol}^{-1}$ , $c$ in $\text{mol L}^{-1}$ ; $\alpha$ is dimensionless and must lie between $0$ and $1$ .
Kohlrausch's law has two parts: independent migration of ions (each ion's contribution is fixed) and additivity ( $\Lambda^\circ_m$ is the sum of the limiting ionic conductivities of the constituent ions).
It is especially valuable for weak electrolytes whose $\Lambda^\circ_m$ cannot be found by extrapolation — only Kohlrausch's law gives it.
Worked check (acetic acid, $c=0.001\,\text{M}$ , $\kappa=4.95\times10^{-5}$ ): $\Lambda_m = \dfrac{4.95\times10^{-5}\times1000}{0.001} = 49.5$ ; $\alpha = \dfrac{49.5}{390.5} \approx 0.127$ .
Continuing that example: $K_a = \dfrac{c\alpha^2}{1-\alpha} = \dfrac{0.001\times(0.127)^2}{1-0.127} \approx 1.85\times10^{-5}$ — close to the literature value for acetic acid.
$\Lambda_m$ can also be expressed as $\Lambda_m = \alpha\,\Lambda^\circ_m$ , the basis of Arrhenius' interpretation of conductivity for weak electrolytes.

Where the marks go

Getting the Kohlrausch combination wrong for the salt of a weak acid — it is salt $+$ strong acid $-$ common strong salt (e.g. $CH_3COONa + HCl - NaCl$ ), not an arbitrary sum.
Dropping $(1-\alpha)$ when $\alpha$ is not small — for moderately ionised acids use the full $K_a = \dfrac{c\alpha^2}{1-\alpha}$ , not just $c\alpha^2$ .
Forgetting the $\times1000$ when converting $\kappa$ to $\Lambda_m$ before computing $\alpha$ .
Mismatched units — mixing $\text{cm}^2$ and $\text{m}^2$ conductivities, or $c$ in $\text{mol m}^{-3}$ with $\Lambda_m$ in $S\,\text{cm}^2\,\text{mol}^{-1}$ .
Counting ions incorrectly in $\nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_-$ — e.g. for $CaCl_2$ , $\nu_-=2$ for chloride.

How the board asks it

Numericaldegree of dissociation and dissociation-constant formulas
For acetic acid the limiting molar conductivity is $\Lambda^\circ_m = 390.5\,S\,cm^2\,mol^{-1}$ and its molar conductivity at $0.1\,M$ is $5.2\,S\,cm^2\,mol^{-1}$ . Calculate the degree of dissociation $\alpha$ and the dissociation constant $K_a$ .
Numericalconverting $\kappa$ to $\Lambda_m$ first
The conductivity of a $0.001\,M$ acetic acid solution is $4.95\times10^{-5}\,S\,cm^{-1}$ . Given $\Lambda^\circ_m = 390.5\,S\,cm^2\,mol^{-1}$ , calculate its molar conductivity, its degree of dissociation $\alpha$ and its dissociation constant $K_a$ .
Numericalthe Kohlrausch combination for a weak acid
Given $\Lambda^\circ_m(CH_3COONa) = 91.0$ , $\Lambda^\circ_m(HCl) = 426.2$ and $\Lambda^\circ_m(NaCl) = 126.5\,S\,cm^2\,mol^{-1}$ , calculate the limiting molar conductivity $\Lambda^\circ_m$ of acetic acid using Kohlrausch's law.
Define / stateKohlrausch's law of independent migration of ions
State Kohlrausch's law of independent migration of ions and explain how it is used to determine the limiting molar conductivity of a weak electrolyte such as acetic acid.
Give reasonswhy $\Lambda^\circ_m$ of a weak electrolyte cannot be found by extrapolation
Account for the fact that the limiting molar conductivity of a weak electrolyte such as acetic acid cannot be obtained by extrapolating its $\Lambda_m$ versus $\sqrt{c}$ graph to zero concentration.
Numericaladditivity $\Lambda^\circ_m = \nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_-$ with ion counting
The limiting molar ionic conductivities of $Ca^{2+}$ and $Cl^-$ are $119.0$ and $76.3\,S\,cm^2\,mol^{-1}$ respectively. Calculate the limiting molar conductivity of $CaCl_2$ .

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.