Dissociation from Conductivity

Introduction

Conductivity can reveal more than the presence of ions. For a dilute weak electrolyte, measured molar conductivity compared with its limiting value gives an approximate ionized fraction. Combined with formal concentration, that fraction can estimate a dissociation constant. The procedure is useful, but it depends on a simple chemical equilibrium and near-limiting ionic mobility.

Core explanation

For a simple weak 1:1 electrolyte AB ⇌ A⁺ + B⁻, let the starting formal concentration be c and the degree of ionization be α. At equilibrium, an idealized concentration table gives [AB] ≈ c(1−α), [A⁺] ≈ cα, and [B⁻] ≈ cα. The concentration-form dissociation constant then is K c ≈ [A⁺][B⁻]/[AB] = cα²/(1−α). This is often called Ostwald's dilution law. It predicts that, for fixed K c, α tends to rise when c is reduced.

Experimentally, Λm is measured from κ/c. The limiting Λ°m is inferred from strong-electrolyte combinations or suitably reliable data. Under the dilute approximation that the ions' molar contributions are near their limiting values, α ≈ Λm/Λ°m. Substitute that estimate into cα²/(1−α) to obtain an approximate dissociation constant. This is an estimate, not an exact definition of α at all concentrations. Conductivity changes with ionic interactions, and equilibrium constants strictly involve activities rather than raw concentrations.

For weak acetic acid, one may derive Λ°m using Λ°m(HCl)+Λ°m(CH₃COONa)−Λ°m(NaCl). The measured κ must be corrected for a relevant solvent blank if the electrolyte signal is very small. A student should not divide a conductivity in S cm⁻¹ directly by Λ°m in S cm² mol⁻¹; first form Λm using the formal concentration with coherent units. Omitting that step makes a dimensional and numerical error.

If α is small, 1−α ≈ 1 and K c ≈ cα². However, do not apply that further simplification when α is not genuinely small; at α = 0.20, the neglected denominator changes the result by 25%. The full concentration-form expression is easy enough to use and avoids unnecessary approximation. At extreme dilution, water autoionization, impurities, and activity effects may become significant, so “more dilute is always a better measurement” is not automatically true.

The method specifically assumes one molecule produces one cation and one anion. A polyprotic acid or electrolyte with multiple ionization stages needs a more detailed equilibrium and conductivity model. Different stages may produce ions of different charge and mobility. State the chemical system before using a one-line formula.

Step-by-step reasoning

1. Obtain κ, c, and Λ°m at the same temperature and solvent. 2. Convert κ into Λm with coherent units. 3. Calculate α ≈ Λm/Λ°m and check it lies between zero and one. 4. For a simple 1:1 weak electrolyte, calculate K c ≈ cα²/(1−α). 5. Describe activity and mobility limitations on the resulting estimate.

Visual explanation

Make an initial-change-equilibrium table for AB ⇌ A⁺ + B⁻. Under it, draw an arrow from the measured Λm/Λ°m ratio to α, then an arrow into the equilibrium expression.

Real-world analogy

If a fully active machine produces a known maximum output per unit, its observed output can estimate the active fraction. The estimate assumes each active unit works nearly as effectively as in the reference state.

Real-world example

A student measures a weak-acid conductivity series and computes α for each flask. The fractions should generally grow as the solutions are diluted, though noisy blank corrections can distort the most dilute readings.

Why?

Why is Λ°m needed? It provides the approximate conductivity per mole if every weak-electrolyte unit contributed its limiting ionic products, giving a reference against which measured Λm can be compared.

Common misconception

“α equals κ divided by Λ°m.” κ has the wrong units and depends on concentration; the usable approximate ratio is Λm/Λ°m after normalizing κ by formal concentration.

Worked example

At c = 0.0200 mol L⁻¹, a weak 1:1 acid has Λm = 58.5 S cm² mol⁻¹ and Λ°m = 390 S cm² mol⁻¹. Then α ≈ 58.5/390 = 0.150. The concentration-form equilibrium estimate is K c ≈ 0.0200(0.150)²/(1−0.150) = 5.29 × 10⁻⁴ mol L⁻¹ under this simplified treatment. If a thermodynamic equilibrium constant is needed, activities and standard-state conventions must be included.

Quick check

1. What is α if Λm is one-fifth of Λ°m in the idealized model? Answer: Approximately 0.20, or 20% ionized.

Exam focus

Show the 1:1 equilibrium table and state that conductivity-derived α is approximate. Keep concentration and molar-conductivity units consistent.

Advanced insight

Conductivity is a transport measurement, while a thermodynamic dissociation constant is defined through activities. Agreement between the two methods improves when solutions are sufficiently dilute and background contributions are controlled.

Summary

For a dilute simple weak electrolyte, α is approximately Λm/Λ°m. Its formal concentration then gives K c ≈ cα²/(1−α), provided the one-stage 1:1 model and mobility assumptions hold.

Practice questions

1. Why can α not sensibly exceed one in this model? Answer: It is the fraction of formula units ionized, so its physical range is zero to one. 2. If c = 0.010 M and α = 0.10, what is [A⁺] approximately? Answer: 0.0010 M for a simple 1:1 dissociation. 3. When is replacing 1−α by one least justified? Answer: When α is appreciable rather than very small.