Ionic Strength and Effective Activity

Introduction

Activity coefficients depend on how crowded a solution is with charge. A simple count of ions is not enough, because a doubly charged ion influences its neighbours far more strongly than a singly charged one. Chemists capture this with a single number, the ionic strength . Once you know the ionic strength, you can predict whether activity effects are negligible, noticeable or large, and explain some surprising observations, such as a sparingly soluble salt dissolving better in salty water.

Core explanation

Defining ionic strength. Ionic strength I is calculated from every ion in the solution:

I = ½ Σ cᵢ zᵢ²

where cᵢ is the concentration of ion i in mol dm⁻³ and zᵢ is its charge number. The charge is squared, so a 2+ or 2− ion counts four times as much as a 1+ or 1− ion at the same concentration, and a 3+ ion counts nine times.

Examples. For 0.10 mol dm⁻³ NaCl: I = ½(0.10 × 1² + 0.10 × 1²) = 0.10 mol dm⁻³. For a 1:1 salt, I equals the concentration. For 0.10 mol dm⁻³ MgSO₄: I = ½(0.10 × 4 + 0.10 × 4) = 0.40 mol dm⁻³. The same concentration gives four times the ionic strength. For 0.10 mol dm⁻³ Na₂SO₄: I = ½(0.20 × 1 + 0.10 × 4) = 0.30 mol dm⁻³.

How γ depends on I. As ionic strength rises, each ion gathers a denser ionic atmosphere and its activity coefficient falls. In very dilute solutions the Debye–Hückel limiting law describes this:

log γ = −A z² √I

where A is about 0.51 for water at 25 °C. The z² factor means highly charged ions are affected far more. At I = 0.01 mol dm⁻³, γ is about 0.89 for a singly charged ion but only about 0.63 for a doubly charged ion. The limiting law works well only below about I = 0.01; extended forms are used at higher ionic strengths, and at very high ionic strength γ can pass through a minimum and rise again.

Background electrolyte. Many experiments deliberately add an unreactive salt, such as KNO₃ or NaClO₄, in large excess. This fixes the ionic strength, so activity coefficients stay nearly constant even as the reacting species change. Constants measured this way are valid for that particular ionic medium.

The salt effect on solubility. Consider silver chloride in water. Its thermodynamic solubility product is Ksp = a(Ag⁺) × a(Cl⁻) = γ±² [Ag⁺][Cl⁻]. If an unrelated salt such as KNO₃ is added, γ± falls, so the product of concentrations [Ag⁺][Cl⁻] must rise to keep Ksp constant. More AgCl dissolves. This salt effect is small compared with the common-ion effect but real and measurable.

The salt effect on acids. Similarly, adding inert salt lowers γ(H⁺) and γ(A⁻), so a weak acid dissociates slightly more in terms of concentration. Its apparent (concentration) Ka increases.

Formulae

I = ½ Σ cᵢzᵢ². Debye–Hückel limiting law: log γ = −0.51 z² √I (water, 25 °C, dilute). a = γc/c°.

Step-by-step reasoning

To calculate ionic strength:

1. Write every ion present with its concentration, remembering stoichiometry (Na₂SO₄ gives twice as much Na⁺). 2. Square each charge number. 3. Multiply each concentration by its squared charge. 4. Add these products and halve the total.

Visual explanation

Draw two beakers of the same volume. One contains a scattering of Na⁺ and Cl⁻ ions; the other contains the same number of Mg²⁺ and SO₄²⁻ ions. Shade each ion's surrounding cloud of opposite charge: in the second beaker the clouds are drawn four times as dense, showing why ionic strength is higher.

Real-world analogy

Background noise in a room affects how well you hear a conversation. A few quiet people barely matter, but loud voices (highly charged ions) count far more than their number suggests. Ionic strength is like a noise level that weighs the loudest sources most heavily.

Real-world example

Biochemists always report the ionic strength of buffers used in enzyme studies. Protein charges interact with the surrounding ions, so enzyme activity, protein solubility and binding constants can all change when the salt content changes, even at the same pH.

Why?

Why is the charge squared in the ionic strength formula? Electrostatic interaction energy is proportional to the product of the charges of the ion and its atmosphere. A more highly charged ion both creates a stronger field and attracts a more strongly charged atmosphere, giving a dependence on z².

Common misconception

"Adding an unrelated salt cannot affect the solubility of AgCl because it shares no ions." It shares no ions, but it raises the ionic strength, lowers the activity coefficients and so slightly increases the concentrations of Ag⁺ and Cl⁻ in solution.

Worked example

Question: Calculate the ionic strength of a solution containing 0.020 mol dm⁻³ K₂SO₄ and 0.010 mol dm⁻³ KCl.

Reasoning: [K⁺] = 0.040 + 0.010 = 0.050; [SO₄²⁻] = 0.020; [Cl⁻] = 0.010. I = ½(0.050 × 1 + 0.020 × 4 + 0.010 × 1) = ½(0.140) = 0.070.

Answer: I = 0.070 mol dm⁻³.

Quick check

1. What is the ionic strength of 0.050 mol dm⁻³ sodium chloride solution, and why? Answer: 0.050 mol dm⁻³, because for a 1:1 salt of singly charged ions I equals the concentration.

Exam focus

Practise ionic strength calculations with 1:1, 1:2 and 2:2 salts, taking care with stoichiometry. Be ready to explain qualitatively why γ falls with rising I and why the effect is larger for multiply charged ions.

Advanced insight

The Debye–Hückel theory treats ions as point charges in a continuous solvent. Real ions have finite size and hydration shells, so extended equations include an ion-size term, and the Davies equation adds a term that allows γ to rise again at high ionic strength. Beyond about 0.5 mol dm⁻³, specific ion-interaction models are needed.

Summary

Ionic strength, I = ½ Σ cᵢzᵢ², measures the total effect of ionic charge in a solution and weights highly charged ions heavily. Activity coefficients fall as I rises, more steeply for ions of higher charge, as described in dilute solution by the Debye–Hückel limiting law. Inert background electrolytes fix I and cause salt effects that raise solubility and apparent dissociation.

Practice questions

1. Calculate the ionic strength of 0.10 mol dm⁻³ CaCl₂. Answer: I = ½(0.10 × 4 + 0.20 × 1) = 0.30 mol dm⁻³. 2. Using the limiting law with A = 0.51, estimate γ for a 1+ ion at I = 0.0010 mol dm⁻³. Answer: log γ = −0.51 × 1 × 0.0316 = −0.016, so γ ≈ 0.96. 3. Explain why BaSO₄ is slightly more soluble in 0.1 mol dm⁻³ NaNO₃ than in pure water. Answer: The added salt raises the ionic strength, lowering the activity coefficients of Ba²⁺ and SO₄²⁻, so their concentrations must increase to keep the activity product equal to Ksp. 4. Why do experimenters add a large excess of an inert salt when measuring equilibrium constants? Answer: It fixes the ionic strength, keeping activity coefficients nearly constant so the measured concentration constants are reproducible.