Kohlrausch's law

A pattern in the numbers

Look at four limiting molar conductivities, all in $\text{S cm}^2/\text{mol}$ at $298\ \text{K}$:

• Hydrochloric acid, HCl: $425.9$
• Potassium chloride, KCl: $149.8$
• Acetic acid, CH₃COOH: $390.5$
• Potassium acetate, CH₃COOK: $114.4$

Now take differences. For HCl and KCl, the chloride is the same, and $425.9 - 149.8 = 276.1$. For acetic acid and potassium acetate, the acetate is the same, and $390.5 - 114.4 = 276.1$. The same number appears twice.

The reason is simple. The difference between an acid and its potassium salt is the difference between a hydrogen ion and a potassium ion. It does not matter which anion is present.

Kohlrausch's law

At infinite dilution the ions are so far apart that they do not pull on each other. Each ion moves on its own. So the limiting molar conductivity of an electrolyte is the sum of the contributions of its ions, and each ion gives the same share whatever the other ion is. This is the law of independent migration of ions.

If one formula unit gives $\nu_+$ cations and $\nu_-$ anions, then

The small $\lambda^\circ$ is the limiting molar ionic conductivity of one ion. Here are the values we use, in $\text{S cm}^2/\text{mol}$ at $298\ \text{K}$:

• Cations: $\text{H}^+$ $349.6$, $\text{K}^+$ $73.5$, $\text{Na}^+$ $50.1$, $\text{Mg}^{2+}$ $106.0$, $\text{Ag}^+$ $61.9$
• Anions: $\text{Cl}^-$ $76.3$, $\text{OH}^-$ $199.1$, $\text{CH}_3\text{COO}^-$ $40.9$

Notice how large the values for $\text{H}^+$ and $\text{OH}^-$ are. These two ions pass their charge along a chain of water molecules, like a relay, so they are much quicker than other ions.

Look at one example. For $\text{MgCl}_2$ each formula unit gives one $\text{Mg}^{2+}$ ion and two $\text{Cl}^-$ ions. So $\Lambda_m^\circ = 106.0 + 2 \times 76.3 = 258.6\ \text{S cm}^2/\text{mol}$. The coefficient 2 counts the two chloride ions.

Use 1: the limiting value of a weak electrolyte

We cannot read $\Lambda_m^\circ$ of acetic acid from a graph. But the law lets us add the ions. For acetic acid, one $\text{H}^+$ and one $\text{CH}_3\text{COO}^-$ give

Sometimes the ion values are not given, but the values for strong electrolytes are. We can still do it, because the unwanted ions cancel:

Check that the ions cancel. $\text{HCl}$ gives $\text{H}^+ + \text{Cl}^-$. $\text{CH}_3\text{COOK}$ adds $\text{CH}_3\text{COO}^- + \text{K}^+$. Taking away $\text{KCl}$ removes $\text{K}^+ + \text{Cl}^-$. What is left is $\text{H}^+ + \text{CH}_3\text{COO}^-$. The numbers are $425.9 + 114.4 - 149.8 = 390.5\ \text{S cm}^2/\text{mol}$, the same as before.

Use 2: the degree of dissociation and $K_a$

Once $\Lambda_m^\circ$ is known, a conductivity measurement gives $\alpha$. Take acetic acid at $c = 0.0180\ \text{mol/L}$. The measured conductivity is $\kappa = 0.0220\ \text{S/m}$.

1. $\Lambda_m = \dfrac{10\,\kappa}{c} = \dfrac{10 \times 0.0220}{0.0180} = 12.2\ \text{S cm}^2/\text{mol}$.
2. $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ} = \dfrac{12.2}{390.5} = 0.0312$.

Now the acid. $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$. Of $c$ moles in each litre, $c\alpha$ have split. So $[\text{H}^+] = [\text{A}^-] = c\alpha$ and $[\text{HA}] = c(1 - \alpha)$. We want $K_a$, so we write

For a weak acid, $\alpha$ is small, so the bottom is close to $1$ and $K_a \approx c\alpha^2$. Here, $K_a = \dfrac{0.0180 \times (0.0312)^2}{1 - 0.0312} = 1.81 \times 10^{-5}\ \text{mol/L}$. This is close to the value in the tables. So conductivity is a good way to measure an acid constant.

Use 3: a salt that hardly dissolves

A salt such as silver chloride dissolves so little that its saturated solution is extremely dilute. Then each ion moves on its own, and $\Lambda_m \approx \Lambda_m^\circ$. We measure the conductivity of the saturated solution, and take away the conductivity of the water that was used. What remains belongs to the salt. We want the concentration $c$, which is the solubility, so $c = \dfrac{\kappa}{\Lambda_m^\circ}$. In SI units, $\kappa$ is in $\text{S/m}$ and $\Lambda_m^\circ$ in $\text{S m}^2/\text{mol}$. The answer is then in $\text{mol/m}^3$, and dividing by $1000$ gives $\text{mol/L}$. A problem below does this with numbers.

Practise with new numbers

Try it in Python

The program keeps the ion values in a table and adds them for any salt. It also finds $\Lambda_m^\circ$ of acetic acid in two ways, from its ions and from three strong electrolytes.

Common mistakes

• Forgetting the coefficient. For $\text{MgCl}_2$ the chloride counts twice.
• Using $\Lambda_m$ of a solution in place of $\Lambda_m^\circ$ in the indirect method. The law works only at infinite dilution.
• Adding when you should subtract. In the indirect method the extra salt is taken away, so that its ions cancel.
• Mixing S/cm and S/m in the solubility of a salt. Change everything to SI first.
• Dropping the $1 - \alpha$ in $K_a$ when $\alpha$ is not small.

Remember these

• Kohlrausch's law: $\Lambda_m^\circ = \nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_-$. Each ion adds its own share.
• The difference between two salts that share one ion does not depend on that ion.
• A weak electrolyte: add the ions, or add and subtract strong electrolytes so that the extra ions cancel.
• $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ}$ and $K_a = \dfrac{c\alpha^2}{1 - \alpha}$.
• Sparingly soluble salt: $c = \dfrac{\kappa_{\text{salt}}}{\Lambda_m^\circ}$, with $\kappa_{\text{salt}} = \kappa_{\text{solution}} - \kappa_{\text{water}}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the limiting molar conductivity of potassium hydroxide, KOH, in $\text{S cm}^2/\text{mol}$.

2. Why is $\Lambda_m^\circ(\text{HCl}) - \Lambda_m^\circ(\text{KCl})$ equal to $\Lambda_m^\circ(\text{CH}_3\text{COOH}) - \Lambda_m^\circ(\text{CH}_3\text{COOK})$?

3. A student finds the limiting molar conductivity of $\text{MgCl}_2$ as $106.0 + 76.3 = 182.3\ \text{S cm}^2/\text{mol}$. What is wrong?

4. A weak acid has $c = 0.200\ \text{mol/L}$ and $\alpha = 0.0500$. Find $K_a$ with the full formula.