Choosing Equations for Ionic Equilibrium

Introduction

Once you know how to write equilibrium expressions, mass balances and a charge balance, a practical question arises: which ones do you actually need? Write too few and the problem cannot be solved; write a dependent equation and you will seem to have enough equations when you do not. This page sets out the systematic treatment of equilibrium , a reliable recipe for choosing a complete, independent set of equations for any ionic system.

Core explanation

Count the unknowns. Every species whose concentration is not fixed in advance is an unknown. Spectator ions whose concentrations are known from the amount of salt added can be treated as known, but it is often clearer to include them with their own mass balance. Pure solids and water have activity 1 and are not unknowns.

Independent equilibria. For each unknown beyond those fixed by balances you need an equilibrium expression, and the equilibria must be independent. Consider carbonic acid. Three reactions might be written:

H₂CO₃ ⇌ H⁺ + HCO₃⁻ (Ka1) HCO₃⁻ ⇌ H⁺ + CO₃²⁻ (Ka2) H₂CO₃ ⇌ 2H⁺ + CO₃²⁻ (K = Ka1 × Ka2)

The third is the sum of the first two, so it adds no new information. Including it while leaving out one of the first two is fine; including all three and counting them as three equations is not. A useful rule: the number of independent equilibria equals the number of species minus the number of independent components (conserved building blocks, with H⁺ treated as a component).

Adding the balances. Mass balances come from the components added. The charge balance adds one more equation. In most acid-base systems, the equations then fall into place: equilibria plus mass balances plus charge balance equals the number of unknowns.

Example: a weak acid HA. Unknowns: [HA], [A⁻], [H⁺], [OH⁻]. Equations: Ka, Kw, mass balance for A, charge balance. Four and four.

Example: a buffer of HA and NaA. Unknowns: [HA], [A⁻], [H⁺], [OH⁻], [Na⁺]. Equations: Ka, Kw, mass balance for A (total from both substances), mass balance for Na, charge balance. Five and five.

Example: saturated AgCl in ammonia solution. Unknowns: [Ag⁺], [Ag(NH₃)⁺], [Ag(NH₃)₂⁺], [Cl⁻], [NH₃], [NH₄⁺], [H⁺], [OH⁻]: eight. Equations: Ksp, two stepwise formation constants, Kb for ammonia, Kw, mass balance for total ammonia, mass balance linking silver and chloride (all dissolved silver equals chloride), and the charge balance. Eight and eight.

Choosing wisely. When a system has a choice of equivalent equations, pick the ones with the most convenient form. A proton condition, derived by combining the charge and mass balances, often replaces the charge balance and gives simpler algebra.

Then approximate. Only after the set is complete should you decide which terms are negligible. This ordering matters: it means every approximation can be checked by substituting back into the full equation set.

Step-by-step reasoning

The systematic treatment, in order:

1. Write all relevant reactions. 2. List all unknown species. 3. Write an independent equilibrium expression for each reaction. 4. Write the mass balances. 5. Write the charge balance. 6. Check that equations equal unknowns. 7. Approximate, solve and check the approximations.

Visual explanation

Draw a two-column table. The left column lists the unknown species; the right column lists the equations. Draw a line from each equation to the species it contains. A determined system has equal column lengths, and every species is touched by at least one line.

Real-world analogy

Solving a Sudoku needs enough clues. Some clues look new but repeat information already implied by others, so they do not help. Dependent equilibria are like repeated clues: they make the puzzle look better specified than it really is.

Real-world example

Pharmaceutical chemists predict how much of a drug is ionised in the stomach and intestine by setting up the drug's acid-base equilibria, the buffer species of the body fluid and the relevant balances. Because the ionised form crosses membranes poorly, this speciation affects how well an oral drug is absorbed.

Why?

Why can a dependent equilibrium not replace a missing one? Its constant is fixed by the others, so it restates relationships already present; it cannot supply information about a species that the other equations leave undetermined.

Common misconception

"Every reaction I can write gives a new equation." Only independent reactions do. Combined or reversed reactions are mathematically redundant and lead to an apparently solvable but actually under-determined system.

Worked example

Question: Set up the equations for a solution of NH₄Cl at analytical concentration c.

Reasoning: Unknowns: [NH₄⁺], [NH₃], [Cl⁻], [H⁺], [OH⁻] (five). Equations: Ka(NH₄⁺) = [NH₃][H⁺]/[NH₄⁺]; Kw = [H⁺][OH⁻]; mass balance c = [NH₄⁺] + [NH₃]; mass balance [Cl⁻] = c; charge balance [NH₄⁺] + [H⁺] = [Cl⁻] + [OH⁻].

Answer: Five independent equations for five unknowns; the system is determined.

Quick check

1. Why is the equilibrium H₂S ⇌ 2H⁺ + S²⁻ not counted separately when Ka1 and Ka2 are already used? Answer: It is the sum of the two stepwise dissociations, so its constant Ka1 × Ka2 adds no new information.

Exam focus

Show the full set-up in extended calculations: species list, equilibria, balances and a count. Examiners award method marks even if the final algebra is simplified, and they penalise hidden assumptions.

Advanced insight

Chemical modelling software represents a system as a matrix of stoichiometric coefficients of species in terms of chosen components. Its rank tells the program how many independent components exist, and every other species is defined by one formation constant from those components. This is the mathematical form of the rule for counting independent equilibria.

Summary

To solve an ionic equilibrium, list every unknown species and match it with an independent equation. Independent equilibria, mass balances and the charge balance together should equal the number of unknowns. Dependent equilibria add no information and must not be double-counted. Approximations come last, so they can be checked against the complete set.

Practice questions

1. How many unknown species are in a solution of the weak base NH₃ in water? List them. Answer: Four: NH₃, NH₄⁺, H⁺ and OH⁻. 2. Write a complete equation set for the NH₃ solution in question 1 at analytical concentration c. Answer: Kb = [NH₄⁺][OH⁻]/[NH₃]; Kw = [H⁺][OH⁻]; c = [NH₃] + [NH₄⁺]; [NH₄⁺] + [H⁺] = [OH⁻]. 3. The reactions A ⇌ B (K₁), B ⇌ C (K₂) and A ⇌ C (K₃) are proposed. How many are independent, and how is K₃ related to the others? Answer: Two are independent; K₃ = K₁ × K₂. 4. Why should approximations be made only after the full equation set is written? Answer: So that each approximation can be tested by substituting the answer back into the complete, exact equations.