Conductance of Electrolytic Solutions: Quick Revision

A brief note on Conductance of Electrolytic Solutions. Read it before a test. Try each question at the end before you open its answer.

Conductivity and the cell constant

• Metals carry current by electrons and conduct less when heated. Solutions carry current by ions and conduct more when heated.
• Conductance $G = \dfrac{1}{R}$, in S. Conductivity $\kappa = \dfrac{1}{\rho}$, in S/m. $1\ \text{S/cm} = 100\ \text{S/m}$.
• Cell constant $G^* = \dfrac{l}{A}$, in $\text{m}^{-1}$. Then $\kappa = \dfrac{G^*}{R}$.
• The cell constant is found by calibration with a solution of known conductivity: $G^* = \kappa R$.
• An alternating current is used, so that nothing is discharged at the plates.

Molar conductivity

• $\Lambda_m = \dfrac{\kappa}{c}$. The 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/m and $c$ in mol/L, $\Lambda_m = \dfrac{10\,\kappa}{c}$ in $\text{S cm}^2/\text{mol}$.
• With $\kappa$ in S/cm and $c$ in mol/L, $\Lambda_m = \dfrac{1000\,\kappa}{c}$.
• On dilution $\kappa$ falls and $\Lambda_m$ rises.
• Strong electrolyte: $\Lambda_m = \Lambda_m^\circ - A\sqrt{c}$, a straight line. The intercept is $\Lambda_m^\circ$.
• Weak electrolyte: a steep curve, no straight part. $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ}$ and $K_a = \dfrac{c\alpha^2}{1 - \alpha}$.

Kohlrausch's law

• $\Lambda_m^\circ = \nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_-$. Each ion adds its own share.
• A weak acid HA from strong electrolytes: $\Lambda_m^\circ(\text{HA}) = \Lambda_m^\circ(\text{HCl}) + \Lambda_m^\circ(\text{NaA}) - \Lambda_m^\circ(\text{NaCl})$.
• A salt that hardly dissolves: $c = \dfrac{\kappa_{\text{salt}}}{\Lambda_m^\circ}$, after the conductivity of the water is taken away.
• $\text{H}^+$ and $\text{OH}^-$ are the fastest ions, because they pass charge along water molecules.

A worked example

A cell with a constant of $92.0\ \text{m}^{-1}$ reads $340\ \Omega$ with a solution of $0.0200\ \text{mol/L}$. Find $\Lambda_m$.

1. Conductivity. $\kappa = \dfrac{92.0\ \text{m}^{-1}}{340\ \Omega} = 0.271\ \text{S/m}$.
2. Molar conductivity. $\Lambda_m = \dfrac{10 \times 0.271}{0.0200} = 135\ \text{S cm}^2/\text{mol}$.

Practice with new numbers

A program to try

The program carries out the two steps of the worked example.

Mistakes to avoid

• Dividing S/m by mol/L and calling the answer S cm²/mol. Multiply by 10.
• Saying that both $\kappa$ and $\Lambda_m$ fall on dilution. Only $\kappa$ falls.
• Extending the curve of a weak electrolyte to find $\Lambda_m^\circ$. Add the ions instead.
• Forgetting the coefficient of an ion that appears twice in the formula.
• Using a direct current in a conductivity cell.

Check yourself

1. A cell has a constant of $72\ \text{m}^{-1}$ and reads $160\ \Omega$. Find the conductivity.

2. A solution is diluted. Which rises, $\kappa$ or $\Lambda_m$?

3. Give the limiting molar conductivity of $\text{MgCl}_2$ in terms of its ions.

4. Why can the limiting molar conductivity of a weak acid not be read from a graph?