Electrochemistry · Lesson 6 of 13
Molar Conductivity and Dilution
“Derive molar conductivity, handle its two unit conventions and interpret concentration plots for strong and weak electrolytes.”
• Calculate molar conductivity in SI and laboratory units. • Explain why conductivity and molar conductivity respond differently to dilution. • Interpret a square-root-concentration plot for strong electrolytes. • Estimate limiting molar conductivity from dilute-solution data.
Compare a fixed volume with a fixed amount
Conductivity answers how readily a solution of a stated composition carries current after geometry has been removed. But two solutions with different concentrations also contain different amounts of electrolyte per unit volume. To compare transport per mole of added electrolyte, divide conductivity by its amount concentration. The resulting molar conductivity, Λm, is not the same as the conductivity of one fixed-size container.
One mole occupies solution volume V = 1/c when c is expressed in mol per volume. Thus Λm can also be written κV, where V is volume per mole. In the geometric interpretation, imagine that volume between electrodes separated by unit distance, with area large enough to hold it. This is a comparison construction; it does not mean a real cell must always contain exactly one mole.
Where the factor of 1000 comes from
A litre contains 1000 cm³. If the stated molarity is c in mol L⁻¹, the concentration in mol cm⁻³ is c/1000. Dividing κ in S cm⁻¹ by that concentration gives 1000κ/c in S cm² mol⁻¹. Alternatively, a litre is 10⁻³ m³, so molarity multiplied by 1000 gives mol m⁻³. These two conversions must not be interchanged.
Problem
A 0.20 M KCl solution has κ = 0.0248 S cm⁻¹. Calculate Λm.
- 1.Use the centimetre convention: Λm = 1000(0.0248)/0.20 = 124 S cm² mol⁻¹.
- 2.Check in SI: κ = 2.48 S m⁻¹ and c = 200 mol m⁻³. Therefore Λm = 2.48/200 = 0.0124 S m² mol⁻¹.
- 3.Multiplying the SI result by 10⁴ gives 124 S cm² mol⁻¹ again. This independent route checks the unit conversion.
Problem
The calibrated cell from the preceding lesson gives κ = 0.24808 S m⁻¹ for 0.020 M KCl. Find Λm.
- 1.Convert 0.020 mol L⁻¹ to 20.0 mol m⁻³, because the conductivity is in SI units.
- 2.Λm = 0.24808/20.0 = 0.012404 S m² mol⁻¹ = 124.04 S cm² mol⁻¹.
- 3.A value written as 12.404 S m² mol⁻¹ would be 1000 times too high because it would have used molarity without converting litres.
Problem
A 0.05 M NaOH solution has κ = 0.0114706 S cm⁻¹ from its measured resistance and geometry. Find Λm.
- 1.The units match the laboratory convention, so use Λm = 1000κ/c.
- 2.Λm = 1000(0.0114706)/0.05 = 229.41 S cm² mol⁻¹ = 0.022941 S m² mol⁻¹.
- 3.Using area 0.785 cm² instead of the full π value gives about 229.6 S cm² mol⁻¹. Consistent rounding explains that small change.
Dilution changes two different comparisons
Adding water spreads the electrolyte over a larger volume. In the ordinary dilute aqueous range, the number of charge-carrying ions per unit volume falls, so conductivity falls. Molar conductivity generally rises: the transport is now being compared per mole of electrolyte, and the ions move with less interference from one another. For weak electrolytes, additional ionisation increases the number of ions produced per mole as well.
For a strong electrolyte such as dilute KCl, dissociation is already nearly complete, so the rise of Λm on dilution is mainly a mobility and interionic-interaction effect. It is misleading to explain it as creation of large numbers of previously absent ions. For a weak electrolyte such as ethanoic acid, dilution shifts the dissociation equilibrium and can cause a much larger fractional increase in Λm.
| KCl concentration / mol L⁻¹ | κ / S cm⁻¹ | Λm / S cm² mol⁻¹ |
|---|---|---|
| 1.000 | 0.1113 | 111.3 |
| 0.100 | 0.0129 | 129.0 |
| 0.010 | 0.00141 | 141.0 |
The table makes the distinction visible. Diluting KCl from 1.0 M to 0.010 M lowers κ from 0.1113 to 0.00141 S cm⁻¹, while Λm increases from 111.3 to 141.0 S cm² mol⁻¹. Concentrated solutions can have more complicated trends, and at extreme dilution the background conductivity of water and contamination matters. The simple trend should be understood in its stated solution range.
Approaching the infinite-dilution limit
The value approached by electrolyte molar conductivity as concentration tends toward zero at fixed solvent and temperature, conventionally written Λm°. It is an extrapolated limit, not the conductivity of an empty solution.
The coefficient A summarises interionic effects and depends on solvent, temperature and ion charges. Electrolytes are described by charge type: NaCl is 1:1, CaCl₂ is 2:1 and MgSO₄ is 2:2. Similar charge types share similar limiting-law behaviour under the model, but a straight line should not be forced across a large concentration range. The weak-electrolyte curve is not straight in √c because the fraction ionised changes strongly.
Problem
At c = 0.000198, 0.000309, 0.000521 and 0.000989 M, KCl has Λm = 148.61, 148.29, 147.81 and 147.09 S cm² mol⁻¹. Estimate the intercept.
- 1.Take square roots to obtain approximately 0.014071, 0.017578, 0.022825 and 0.031449 in (mol L⁻¹)½.
- 2.Plot Λm vertically against √c horizontally. The points are close to a falling straight line; the intercept at √c = 0 is the limiting value.
- 3.A least-squares line through these stated points gives Λm° ≈ 149.83 S cm² mol⁻¹ and A ≈ 87.44 S cm² mol⁻¹ (mol L⁻¹)⁻½. A coarse graphical extrapolation gives about 150; the fit does not justify pretending the graphical estimate is exact.
Dilution can lower conductivity and raise molar conductivity simultaneously. Always identify whether the comparison fixes solution geometry, volume or amount of electrolyte before explaining a trend.
Quick Quiz
Quiz
What is the SI unit of molar conductivity?
Λm = κ/c gives (S m⁻¹)/(mol m⁻³) = S m² mol⁻¹.
For κ = 0.0010 S cm⁻¹ and c = 0.010 M, what is Λm?
Use 1000κ/c = 1000 × 0.0010/0.010 = 100.
Why does dilute strong-electrolyte Λm rise when more water is added?
Strong electrolytes are already nearly fully dissociated; mobility and ion interactions explain the main change.
For Λm = Λm° − A√c, what does the vertical intercept represent?
At √c = 0, the concentration term vanishes and Λm approaches Λm°.
Which statement usually describes dilution of an aqueous electrolyte in its ordinary dilute range?
Ions become less numerous per volume, but their effective transport per mole improves; weak-electrolyte ionisation may also rise.
Practice Problems
- Calculate Λm for a 0.100 M solution with κ = 1.29 S m⁻¹, using both unit systems.
- At c = 0.001, 0.010, 0.020, 0.050 and 0.100 M, NaCl conductivities are 0.01237, 0.1185, 0.2315, 0.5553 and 1.0674 S m⁻¹. Calculate Λm and explain how to estimate its limit.
- A dilute strong electrolyte follows Λm = 150 − 80√c with c in M and Λm in S cm² mol⁻¹. Find Λm at 0.0025 M.
- Explain why the limiting value of a weak acid cannot be found by a simple straight-line Λm versus √c extrapolation.
- A student says that κ falls on dilution because the total number of ions in the beaker always falls. Correct the reasoning.
1. Convert concentration to 100 mol m⁻³. Λm = 1.29/100 = 0.0129 S m² mol⁻¹. 2. Alternatively κ = 0.0129 S cm⁻¹, so Λm = 1000(0.0129)/0.100 = 129 S cm² mol⁻¹. 3. The two answers differ only by the area-unit conversion 10⁴.
Key Takeaways
• Molar conductivity compares transport per mole of electrolyte. • Use Λm = κ/c with SI concentration in mol m⁻³. • Use Λm = 1000κ/c for κ in S cm⁻¹ and c in mol L⁻¹. • One S cm² mol⁻¹ equals 10⁻⁴ S m² mol⁻¹. • Dilution usually lowers κ but raises Λm in ordinary aqueous ranges. • Strong-electrolyte dilution trends mainly reflect ionic interactions and mobility. • The limiting molar conductivity is the zero-concentration intercept. • Weak electrolytes require a different method because their ionisation changes strongly.