Mass Balance in Ionic Equilibrium

Introduction

Equilibrium reactions redistribute atoms; they do not create them. A component mass balance counts a selected chemical component across every dissolved form that contains it. This constraint is essential when protonation, complexation, or precipitation makes one measured total differ from the free-ion concentration appearing in an equilibrium expression.

Core explanation

Suppose a monoprotic acid HA is introduced at analytical concentration CT with no other source of its A component. If HA and A⁻ are the only appreciable dissolved A-containing species and no solid or gas removes A, CT = [HA] + [A⁻]. Acid dissociation shifts the distribution but not the sum. Ka links the forms through [H₃O⁺][A⁻]/[HA], while charge balance supplies another independent relation for the aqueous ions. Using just Ka without CT does not determine both HA and A⁻ concentrations.

For a diprotic acid H₂A, the same component can occupy H₂A, HA⁻, and A²⁻. Its dissolved balance is CT = [H₂A] + [HA⁻] + [A²⁻] when those are the only forms. Each species counts once because each molecule contains one A unit. A phosphate balance likewise counts dissolved species across protonation states. If an aqueous complex contains two units of a tracked component, its concentration appears with coefficient two in that component balance.

Metal-ligand binding gives another example. If Mⁿ⁺ and ML are the only dissolved metal species, dissolved metal total is [Mⁿ⁺] + [ML]. If ML₂ exists, it still contains one metal and contributes one [ML₂] term to metal balance, but it contributes two [ML₂] to a ligand balance. Ksp for a solid uses free Mⁿ⁺; the analytical total includes all these forms. If undissolved solid remains, a global conservation balance may include the amount of solid, whereas a dissolved analytical total usually describes only the liquid phase. State the system boundary before writing a total.

Dilution matters. After mixing two solutions, an analytical concentration is total moles of the component delivered divided by final solution volume, not the original stock concentration. Adding a conjugate-base salt introduces more A component; then CT includes both acid and salt contributions. A gas exchange or sampling loss can make a nominal closed-system balance inappropriate. A good model specifies which materials are introduced, removed, or fixed by an external reservoir.

Mass balance conserves atoms or chosen chemical groups. Charge balance conserves electrical neutrality. Neither substitutes for equilibrium constants; all three kinds of equations work together to determine speciation. Carefully label free concentrations and totals to avoid putting CT directly into Ka or Ksp where a specific species is required.

Step-by-step reasoning

1. Select the component whose atoms or chemical group are tracked. 2. List all species containing it and count units per species. 3. Equate their weighted concentrations to the introduced analytical total. 4. Add equilibrium and charge equations to solve distribution.

Visual explanation

Draw a box labeled total A, split into H₂A, HA⁻, and A²⁻ compartments. Moving material between compartments changes speciation but leaves the box total unchanged.

Real-world analogy

A library can move copies among shelves, desks, and loan carts. Counting one shelf misses books, while a complete inventory sums every location and applies extra counting if a package holds multiple copies.

Real-world example

In a phosphate-containing water sample, total dissolved phosphate can be measured while its individual acid-base forms depend on pH. A mass balance links the measured total to those forms.

Why?

Why does Ka alone not determine every species concentration? It fixes a ratio or product relation, while a mass balance gives the conserved total available to distribute.

Common misconception

“The prepared acid concentration equals free A⁻ concentration.” Most of the component may remain protonated, especially when pH is below its pKa.

Worked example

A closed dilute solution contains total analytical A concentration CT = 0.0100 M. Suppose measured or calculated speciation shows [A⁻] = 0.0030 M and HA/A⁻ are the only A forms. Then [HA] = CT − [A⁻] = 0.0070 M. If an added complex H₂A were significant, that two-species calculation would be incomplete. This example illustrates why the species list must precede numerical subtraction.

Quick check

1. In an H₂A/HA⁻/A²⁻ system, what is the dissolved A balance? Answer: CT = [H₂A] + [HA⁻] + [A²⁻] when no other A forms matter.

Exam focus

Track one element or chemical group at a time. Count its multiplicity in complexes and distinguish liquid-phase totals from an inventory that also includes solid or gas.

Advanced insight

Numerical speciation codes often use component totals as input constraints and calculate all species from equilibrium constants. Selecting independent components is a bookkeeping choice; the physical predictions should agree.

Summary

Mass balance states that the introduced amount of a component is distributed among all its relevant forms. It connects analytical totals to free ions and complements charge balance and equilibrium expressions.

Practice questions

1. If total metal is 0.020 M and free metal is 0.005 M with only one complex form, how much metal is complexed? Answer: 0.015 M. 2. Does ML₂ contain one or two metal units per complex? Answer: One metal unit, so it contributes once to metal balance. 3. Why is final mixture volume used for CT? Answer: Analytical concentration is total delivered moles divided by final solution volume.