Molar conductivity

Why the conductivity is not enough

Take a solution of a salt and add an equal volume of water. The conductivity falls, because the same ions are now spread through twice the volume. That tells us nothing about the ions themselves. We want a quantity that asks a fairer question: how well does one mole of the electrolyte conduct?

So we divide the conductivity by the concentration. The result is the molar conductivity, $\Lambda_m$ (the Greek capital letter lambda):

Imagine all the solution that holds one mole of the electrolyte, placed between two plates that are $1\ \text{m}$ apart and wide enough to hold it. The molar conductivity is the conductance of that column.

The units, step by step

The units need care, because there are two common ways to write them.

In SI, $\kappa$ is in $\text{S/m}$ and $c$ must be in $\text{mol/m}^3$. Since $1\ \text{L} = 10^{-3}\ \text{m}^3$, a concentration of $1\ \text{mol/L}$ is $1000\ \text{mol/m}^3$. Then

Most books give $\Lambda_m$ in $\text{S cm}^2/\text{mol}$. Because $1\ \text{m}^2 = 10^4\ \text{cm}^2$, we have $1\ \text{S m}^2/\text{mol} = 10^4\ \text{S cm}^2/\text{mol}$.

If $\kappa$ is in $\text{S/cm}$ and $c$ is in $\text{mol/L}$, there is a shortcut:

The 1000 changes $\text{mol/L}$ to $\text{mol/cm}^3$. If $\kappa$ is in $\text{S/m}$, use $\Lambda_m = \dfrac{10\,\kappa}{c}$ with $c$ in $\text{mol/L}$.

Here is an example. A solution of a 1:1 salt has $c = 0.0450\ \text{mol/L}$ and $\kappa = 0.480\ \text{S/m}$.

• The concentration is $45.0\ \text{mol/m}^3$.
• $\Lambda_m = \dfrac{0.480\ \text{S/m}}{45.0\ \text{mol/m}^3} = 1.067 \times 10^{-2}\ \text{S m}^2/\text{mol}$.
• In the other unit, $\Lambda_m = 107\ \text{S cm}^2/\text{mol}$.

Strong electrolytes: a slow rise on dilution

Now we measure $\Lambda_m$ for a strong electrolyte at many concentrations. It does not stay constant. As the solution is diluted, $\Lambda_m$ rises slowly, and it reaches a limit. If you plot $\Lambda_m$ against $\sqrt{c}$, you get a straight line. Kohlrausch found this by experiment:

• $\Lambda_m^\circ$ is the limiting molar conductivity, the value as $c$ goes to zero. We read it from the point where the line meets the vertical axis.
• $A$ is a constant. It depends on the solvent, the temperature and the type of electrolyte (1:1, 2:1), but not on which 1:1 salt it is.

Why does $\Lambda_m$ rise on dilution? In a concentrated solution each ion is surrounded by ions of the opposite charge, and they pull it back. As the ions get farther apart, this pull gets weaker. The ions move more freely, so each mole conducts better.

Here is an example. A strong 1:1 electrolyte has $\Lambda_m^\circ = 148.0\ \text{S cm}^2/\text{mol}$ and $A = 90.0\ \text{S cm}^2/\text{mol}$ for each $(\text{mol/L})^{1/2}$. At $c = 0.0225\ \text{mol/L}$, the square root is $\sqrt{c} = 0.150\ (\text{mol/L})^{1/2}$. Then $\Lambda_m = 148.0 - 90.0 \times 0.150 = 134.5\ \text{S cm}^2/\text{mol}$.

Weak electrolytes: only a fraction is in ions

A weak electrolyte behaves in another way. At ordinary concentrations only a small fraction of its molecules have split into ions, so $\Lambda_m$ is small. When the solution is diluted, more of the molecules split. The fraction that has split is the degree of dissociation, $\alpha$.

So for a weak electrolyte, $\Lambda_m$ rises steeply on dilution, because both the ions get farther apart and there are more of them. At a very low concentration $\alpha$ is close to $1$ and $\Lambda_m$ is close to $\Lambda_m^\circ$.

The curve is steep where we would need to extend it, and the measurements are poor there. There is no straight part. So the limiting value of a weak electrolyte cannot be found by extending a line. The next lesson shows how to find it from the ions.

The degree of dissociation is the ratio of the molar conductivity to the limiting value:

Two quantities that move in opposite ways

When you dilute an electrolyte, two things happen at once:

• The conductivity $\kappa$ falls. There are fewer ions in each cubic metre.
• The molar conductivity $\Lambda_m$ rises. Each mole of electrolyte has more room, and for a weak electrolyte more of it is in ions.

These do not contradict each other. The conductivity depends on how much electrolyte there is. The molar conductivity compares one mole at a time.

The weak acid in the sim is acetic acid. Its limiting value, $390.5\ \text{S cm}^2/\text{mol}$, is found from its ions in the next lesson. Use the slider for the concentration. The strong electrolyte barely changes, and the weak acid changes a lot. Then make the acid stronger by raising its dissociation constant: the red curve moves up.

Practise with new numbers

Try it in Python

The program builds a small table. It finds $\Lambda_m$ for a strong and a weak electrolyte at four concentrations. For the weak acid it solves for $\alpha$ from the dissociation constant, and then uses $\Lambda_m = \alpha\,\Lambda_m^\circ$.

Common mistakes

• Using $\Lambda_m = \kappa/c$ with the wrong units. If $\kappa$ is in S/m and $c$ in mol/L, the answer is wrong by a factor of $1000$. Write the unit of every number.
• Saying that $\kappa$ and $\Lambda_m$ both fall on dilution. The conductivity falls and the molar conductivity rises.
• Putting $c$ in the law, not $\sqrt{c}$. The straight line is against $\sqrt{c}$.
• Extending the curve of a weak electrolyte to find $\Lambda_m^\circ$. There is no straight part to extend.
• Mixing up $\Lambda_m$ and $\Lambda_m^\circ$ in $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ}$. The top is the value in the solution, the bottom is the limit.

Remember these

• Molar conductivity $\Lambda_m = \dfrac{\kappa}{c}$. Its SI unit is $\text{S m}^2/\text{mol}$, and $1\ \text{S m}^2/\text{mol} = 10^4\ \text{S cm}^2/\text{mol}$.
• With $\kappa$ in S/cm and $c$ in mol/L, $\Lambda_m = \dfrac{1000\,\kappa}{c}$. With $\kappa$ in S/m, $\Lambda_m = \dfrac{10\,\kappa}{c}$.
• A strong electrolyte gives a straight line, $\Lambda_m = \Lambda_m^\circ - A\sqrt{c}$, and $\Lambda_m^\circ$ is its intercept.
• A weak electrolyte has a steep curve and no straight part. Its $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ}$.
• On dilution $\kappa$ falls and $\Lambda_m$ rises.

Check yourself

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

1. A solution has $\kappa = 2.40\ \text{S/m}$ and $c = 0.200\ \text{mol/L}$. Find $\Lambda_m$ in $\text{S cm}^2/\text{mol}$.

2. A strong electrolyte is diluted to one quarter of its concentration. Does its conductivity rise or fall? What about its molar conductivity?

3. Why can the limiting molar conductivity of acetic acid not be found by extending a straight line?

4. A value is written as $1.45 \times 10^{-2}\ \text{S m}^2/\text{mol}$. Write it in $\text{S cm}^2/\text{mol}$.