Mass Balance in Aqueous Systems

Introduction

When you dissolve 0.10 mol of ethanoic acid in a litre of water, you do not end up with 0.10 mol dm⁻³ of CH₃COOH molecules. Some of the acid becomes ethanoate ions. Yet nothing has been created or destroyed: every ethanoate group added is still there in one form or another. The mass balance turns this conservation of matter into an equation. It is one of the simplest and most powerful tools for solving ionic equilibria, and it is almost impossible to solve multi-species problems without it.

Core explanation

Analytical versus equilibrium concentration. The analytical concentration (also called formal concentration), often written c or C, is the total amount of a substance added per unit volume. The equilibrium concentration , written in square brackets, is the concentration of one particular species. For ethanoic acid at c = 0.10 mol dm⁻³:

c = [CH₃COOH] + [CH₃COO⁻] = 0.10 mol dm⁻³

Each bracket is smaller than 0.10, but together they account for everything added.

Polyprotic acids. For phosphoric acid of analytical concentration c:

c = [H₃PO₄] + [H₂PO₄⁻] + [HPO₄²⁻] + [PO₄³⁻]

All four species contain one phosphorus atom, so each counts once.

Stoichiometric factors. When a species contains more than one unit of a component, it must be multiplied by that number. In a solution of chromate and dichromate, the dichromate ion Cr₂O₇²⁻ contains two chromium atoms:

c(Cr) = [CrO₄²⁻] + [HCrO₄⁻] + 2[Cr₂O₇²⁻]

Forgetting the factor of 2 is one of the commonest errors in equilibrium algebra.

Counter-ions from salts. A salt contributes to more than one mass balance. Dissolving 0.050 mol dm⁻³ Na₂HPO₄ gives [Na⁺] = 0.100 mol dm⁻³, because sodium ions do not react, and total phosphate = 0.050 mol dm⁻³, distributed among four species.

Mass balance with a solid. When a sparingly soluble salt dissolves in pure water, the stoichiometry of the solid provides a mass balance even though the analytical concentration is unknown. For CaF₂ dissolving: every Ca²⁺ released is accompanied by two F⁻, so total fluorine = 2 × total calcium. If F⁻ is partly protonated to HF in acid, the balance becomes:

[F⁻] + [HF] = 2[Ca²⁺]

This equation, combined with Ksp and Ka, lets you calculate how acid raises the solubility of CaF₂.

Mixtures. If two solutions are mixed, each analytical concentration is first corrected for dilution, then the mass balances are written for the final volume. For example, equal volumes of 0.20 mol dm⁻³ NH₃ and 0.10 mol dm⁻³ HCl give total ammonia = 0.10 mol dm⁻³ and total chloride = 0.050 mol dm⁻³ after mixing.

What mass balances do not include. H⁺ and OH⁻ are usually not given a mass balance, because water is an effectively unlimited source of both. Their concentrations are handled through Kw and the charge balance instead.

Step-by-step reasoning

To write mass balances for any solution:

1. Identify each component that is added and conserved, such as total phosphate or total sodium. 2. Calculate its analytical concentration in the final volume, correcting for dilution. 3. List every species containing that component. 4. Multiply each species by the number of component units it contains. 5. Set the sum equal to the analytical concentration.

Visual explanation

Imagine a pie chart for "total phosphate" in a solution. The whole pie is always the same size, set by the amount added. As pH changes, the slices for H₃PO₄, H₂PO₄⁻, HPO₄²⁻ and PO₄³⁻ grow and shrink, but they always add up to the full pie.

Real-world analogy

A company has 100 employees spread across four offices. People move between offices every day, but the total headcount stays at 100. A mass balance is the headcount: it does not tell you who is in which office, but it limits the possibilities strictly.

Real-world example

Water-quality laboratories report "total ammonia nitrogen", the sum of NH₃ and NH₄⁺. The toxic form for fish is NH₃, and its share depends on pH and temperature. The mass balance lets analysts calculate the harmful NH₃ concentration from the total measured nitrogen and the pH.

Why?

Why is a mass balance always valid, even when equilibrium constants are uncertain? It rests only on conservation of atoms, which chemical reactions never violate. Equilibria redistribute a component among species, but they cannot change its total amount.

Common misconception

"The analytical concentration of an acid equals the concentration of the undissociated acid." The analytical concentration is the total of all forms; the undissociated acid is only one of them, sometimes a small one.

Worked example

Question: A solution has total ethanoate c = 0.10 mol dm⁻³ and pH 4.76, where Ka = 1.74 × 10⁻⁵ (pKa 4.76). Find [CH₃COOH] and [CH₃COO⁻].

Reasoning: At pH = pKa, [H⁺] = Ka, so [CH₃COO⁻]/[CH₃COOH] = Ka/[H⁺] = 1. The mass balance gives [CH₃COOH] + [CH₃COO⁻] = 0.10, so each is 0.050.

Answer: Both concentrations are 0.050 mol dm⁻³.

Quick check

1. Write the mass balance for sulfur in a solution of H₂S with analytical concentration 0.10 mol dm⁻³. Answer: [H₂S] + [HS⁻] + [S²⁻] = 0.10 mol dm⁻³.

Exam focus

Write mass balances before attempting any multi-species calculation, and check stoichiometric factors carefully. Remember to correct analytical concentrations for dilution when solutions are mixed; many lost marks come from using the original stock concentration.

Advanced insight

In complexation chemistry the mass balances can become long. A metal ion forming a series of complexes such as Cu(NH₃)²⁺ through Cu(NH₃)₄²⁺ appears in five species for copper, while ammonia appears in NH₃, NH₄⁺ and each complex, weighted by 1, 2, 3 and 4. Computer programs build these balances automatically from a matrix of stoichiometric coefficients.

Summary

A mass balance states that the analytical concentration of a component equals the sum of the equilibrium concentrations of all species containing it, each multiplied by its stoichiometric factor. Salts contribute to several mass balances. Solids dissolving in water give balances from their formula ratio. Mass balances rest only on conservation of matter and, with equilibrium expressions and the charge balance, complete the set of equations.

Practice questions

1. Distinguish analytical concentration from equilibrium concentration. Answer: Analytical concentration is the total amount of a substance added per unit volume; equilibrium concentration is the concentration of one specific species at equilibrium. 2. Write the mass balances for 0.020 mol dm⁻³ Na₂CO₃ solution. Answer: [Na⁺] = 0.040 mol dm⁻³; [H₂CO₃] + [HCO₃⁻] + [CO₃²⁻] = 0.020 mol dm⁻³. 3. Write the mass balance linking calcium and oxalate when CaC₂O₄ dissolves in an acidic solution. Answer: [Ca²⁺] = [H₂C₂O₄] + [HC₂O₄⁻] + [C₂O₄²⁻]. 4. Equal volumes of 0.30 mol dm⁻³ ethanoic acid and 0.10 mol dm⁻³ NaOH are mixed. State the total ethanoate and total sodium concentrations. Answer: Total ethanoate 0.15 mol dm⁻³; total sodium 0.050 mol dm⁻³.