Conductivity and Molar Conductivity

Introduction

Dissolved ions carry electric current, so measuring how well a solution conducts reveals how many ions it contains and how mobile they are. Conductivity meters monitor water purity, follow titrations and estimate how far weak acids ionise. The raw reading, however, depends on the geometry of the cell as well as on the solution. This page shows how to strip out the geometry and the concentration so that the result says something about the ions themselves.

Core explanation

Resistance to conductance. A conductivity cell reports a resistance R (in Ω) or a conductance G = 1/R (in siemens, S).

Removing the geometry. For a column of solution of length l and cross-sectional area A, conductance is G = κA/l, where κ (kappa) is the conductivity . Rearranging:

κ = G × (l/A) = (l/A)/R

The ratio l/A is the cell constant . In practice it is not measured with a ruler; it is found by filling the cell with a standard potassium chloride solution of known conductivity and measuring its resistance: cell constant = κ(standard) × R(standard). The SI unit of the cell constant is m⁻¹ and of conductivity is S m⁻¹.

Removing the concentration. A more concentrated solution conducts better simply because it has more ions. Dividing by concentration gives the molar conductivity :

Λm = κ/c

For SI consistency, c must be in mol m⁻³: 1 mol dm⁻³ = 1000 mol m⁻³. The result is in S m² mol⁻¹. Many tables use S cm² mol⁻¹; 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹.

Strong electrolytes. For fully dissociated salts, Λm decreases only slightly with concentration, following Kohlrausch's empirical law Λm = Λ°m − K√c, where Λ°m is the limiting molar conductivity at infinite dilution. The decrease reflects ionic atmospheres that slow each ion.

Weak electrolytes. For a weak acid, Λm falls steeply as concentration rises because the fraction ionised falls. The degree of ionisation can be estimated as α ≈ Λm/Λ°m, where Λ°m is built from ionic values by Kohlrausch's law of independent migration, for example Λ°m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) ≈ 390.7 S cm² mol⁻¹. The acid dissociation constant then follows: Ka = cα²/(1 − α).

Reference data (25 °C). 0.0100 mol dm⁻³ KCl has κ = 0.1413 S m⁻¹, which gives Λm = 141.3 S cm² mol⁻¹, a useful benchmark.

Formulae

G = 1/R; κ = (l/A)/R; cell constant = κ(standard) × R(standard); Λm = κ/c (c in mol m⁻³); α ≈ Λm/Λ°m; Ka = cα²/(1 − α).

Step-by-step reasoning

1. Find the cell constant from the calibration solution. 2. Calculate κ of the sample = cell constant ÷ R(sample). 3. Convert concentration from mol dm⁻³ to mol m⁻³. 4. Calculate Λm = κ/c and convert units if the comparison data are in S cm² mol⁻¹. 5. For a weak electrolyte, compare with Λ°m to estimate α and Ka.

Visual explanation

Imagine a long, thin tube of solution between two electrodes and a short, fat one. The short, fat tube conducts better, though the solution is identical. Conductivity corresponds to a standard one-metre cube, and molar conductivity shares that cube's conductance among the moles of solute inside it.

Real-world analogy

Total sales in a shop depend on its floor area and on the number of staff. Sales per square metre removes the size effect, like conductivity; sales per member of staff removes the staffing level, like molar conductivity, telling you how effective each unit is.

Real-world example

Ultrapure water used in semiconductor fabrication is specified by conductivity: its theoretical value at 25 °C is about 5.5 × 10⁻⁶ S m⁻¹, arising only from water's own ionisation. A reading many times higher signals ionic contamination long before chemical analysis would detect it.

Why?

Why does Λm for acetic acid fall so much more steeply with concentration than for sodium chloride? In NaCl nearly every formula unit supplies ions at all concentrations. In acetic acid the equilibrium shifts towards the molecular form as concentration increases, so the fraction of solute present as ions drops.

Common misconception

"Divide κ by the concentration in mol dm⁻³." Mixing S m⁻¹ with mol dm⁻³ makes Λm a thousand times too large in S m² mol⁻¹. Always convert c to mol m⁻³ first, or work consistently in S cm⁻¹ and mol cm⁻³.

Worked example

Question: A cell has constant 100 m⁻¹. Filled with 0.0200 mol dm⁻³ NaCl, its resistance is 422 Ω. Calculate κ and Λm.

Reasoning: κ = 100 ÷ 422 = 0.237 S m⁻¹. c = 0.0200 × 1000 = 20.0 mol m⁻³. Λm = 0.237 ÷ 20.0 = 1.185 × 10⁻² S m² mol⁻¹. Converting: × 10⁴ gives 118.5 S cm² mol⁻¹.

Answer: κ = 0.237 S m⁻¹; Λm ≈ 1.19 × 10⁻² S m² mol⁻¹ (about 119 S cm² mol⁻¹).

Quick check

1. A calibration solution with κ = 0.1413 S m⁻¹ gives a resistance of 150 Ω. What is the cell constant? Answer: Cell constant = κ × R = 0.1413 × 150 ≈ 21.2 m⁻¹.

Exam focus

Keep units visible throughout: m⁻¹ for the cell constant, S m⁻¹ for κ, mol m⁻³ for c and S m² mol⁻¹ for Λm. Examiners set traps with mixed centimetre and metre units. For weak electrolytes, state the approximation α ≈ Λm/Λ°m before using it.

Advanced insight

Conductivity also depends strongly on temperature, typically rising by about 2% per kelvin because the viscosity of water falls. Commercial meters therefore report temperature-compensated values referenced to 25 °C. The exceptionally high molar conductivities of H⁺ and OH⁻ arise from proton hopping through hydrogen-bonded chains of water molecules (the Grotthuss mechanism) rather than from ordinary diffusion of the ions.

Summary

Resistance is converted to conductance, then to conductivity using a calibrated cell constant, κ = (l/A)/R. Dividing by concentration in mol m⁻³ gives molar conductivity, Λm = κ/c. Strong electrolytes show a small, √c-dependent fall in Λm; weak electrolytes show a steep fall, and α ≈ Λm/Λ°m leads to an estimate of Ka.

Practice questions

1. Using a cell constant of 21.2 m⁻¹, find κ for a solution whose resistance is 600 Ω. Answer: κ = 21.2 ÷ 600 ≈ 0.0353 S m⁻¹. 2. Convert Λm = 1.413 × 10⁻² S m² mol⁻¹ into S cm² mol⁻¹. Answer: Multiply by 10⁴: 141.3 S cm² mol⁻¹. 3. A 1.00 × 10⁻³ mol dm⁻³ solution of acetic acid has κ = 4.86 × 10⁻³ S m⁻¹. Find Λm in S cm² mol⁻¹ and α, given Λ°m = 390.7 S cm² mol⁻¹. Answer: Λm = 4.86 × 10⁻³ ÷ 1.00 = 4.86 × 10⁻³ S m² mol⁻¹ = 48.6 S cm² mol⁻¹; α = 48.6 ÷ 390.7 ≈ 0.124. 4. Use the result above to estimate Ka for acetic acid. Answer: Ka = cα²/(1 − α) = 1.00 × 10⁻³ × 0.0155 ÷ 0.876 ≈ 1.8 × 10⁻⁵.