Concentration and Ionic Formula Units

Introduction

A bottle marked 0.10 M CaCl₂ specifies formula amount of calcium chloride per litre. It does not say every ion has concentration 0.10 M. In a complete-dissociation model, the formula supplies one calcium ion and two chloride ions for each formula unit, so the ion concentrations differ.

Core explanation

Begin with a balanced dissolution equation. CaCl₂(s) → Ca²⁺(aq) + 2Cl⁻(aq) shows a 1:1:2 ratio of formula units, calcium ions and chloride ions. If 0.100 mol CaCl₂ dissolves fully in 1.00 L final solution, formal CaCl₂ molarity is 0.100 M; ideal calcium-ion concentration is 0.100 M and chloride-ion concentration 0.200 M. The sum of numerical ion concentrations is 0.300 M, but it is not called “CaCl₂ molarity.”

For Al₂(SO₄)₃, one formula unit gives two Al³⁺ and three SO₄²⁻ in the simple model. A 0.0200 M formula concentration predicts 0.0400 M Al³⁺ and 0.0600 M sulfate. Charge balance checks the result: 2 × (+3) = +6 positive charge units and 3 × (−2) = −6 negative units per formula unit. If a proposed ion set violates charge conservation, a subscript or charge was likely copied incorrectly.

The calculation applies to material actually dissolved. Adding a poorly soluble salt at a nominal amount does not produce the ideal concentrations if excess solid remains. Use its dissolved concentration or a solubility equilibrium model. Even a soluble salt at high concentration can show ion pairing and nonideal behavior, so these computed concentrations are formal stoichiometric counts in an ideal complete-dissociation picture. They do not directly give ion activities.

Multiple sources of the same ion add. If a solution contains both 0.100 M NaCl and 0.0500 M CaCl₂ with no relevant reaction, formal chloride concentration is 0.100 + 2(0.0500) = 0.200 M. Na⁺ is 0.100 M and Ca²⁺ is 0.0500 M. Each formula concentration should be calculated from its own amount and the common final volume before adding shared-ion contributions.

Some dissolved compounds react with water or form complexes. The formula-based ion count is then an initial analytical balance, not a complete equilibrium speciation. A problem asking about free-ion concentration may require additional chemical data. State whether the ideal model is being used.

Step-by-step reasoning

1. Write the correct neutral ionic formula and dissociation equation. 2. Read ion coefficients from the formula, including parentheses. 3. Multiply formal formula molarity by each coefficient. 4. Add contributions from all dissolved sources of a shared ion. 5. Check charge balance and whether complete dissolution is assumed.

Visual explanation

Draw one CaCl₂ formula unit as one blue Ca²⁺ and two green Cl⁻ symbols. Repeat the unit three times, then separate the ions. A table below lists three calcium and six chloride ions, preserving the 1:2 count.

Real-world analogy

Each packaged kit has one large piece and two small pieces. Ten kits supply ten large and twenty small pieces. Kit count and piece count are different, as formula-unit and ion amounts are.

Real-world example

In a conductivity experiment, equal molarity salt solutions may supply different ideal counts and charges of mobile ions. Conductivity still also depends on ion mobility, concentration and interactions, so ion count alone does not fix the measured value.

Why?

Why is charge balance a useful check? Neutral salt formula units contribute equal positive and negative charge totals. A mistaken coefficient often breaks that equality even if atom counts appear plausible.

Common misconception

“0.10 M CaCl₂ means 0.10 M chloride.” The two chloride ions per formula unit give 0.20 M formal chloride under complete dissociation.

Worked example

Make 500 mL solution using 0.0250 mol Al₂(SO₄)₃, fully dissolved. Formula molarity is 0.0250/0.500 = 0.0500 M. Ideal Al³⁺ concentration is 2 × 0.0500 = 0.100 M and sulfate is 3 × 0.0500 = 0.150 M. Charge-equivalent concentrations balance: 3 × 0.100 = 2 × 0.150 = 0.300 mol charge per litre.

Quick check

1. What ideal sulfate concentration follows from 0.030 M Na₂SO₄? Answer: One sulfate ion occurs per formula unit, so its formal sulfate concentration is 0.030 M if fully dissociated.

Exam focus

Distinguish the salt's labeled molarity from each ion's concentration. Use subscripts and charge balance, and check whether the salt truly dissolves.

Advanced insight

Equilibrium models use activities because ions interact electrostatically in solution. Formal analytical concentration remains valuable for careful mass balance even when free-ion activity differs.

Summary

Ionic formulas specify how dissolved formula amounts translate into ion amounts. Multiply formal concentration by each ion coefficient and add shared sources. The simple result assumes complete dissolution and ideal dissociation.

Practice questions

1. Find ideal Na⁺ concentration in 0.050 M Na₂SO₄. Answer: Two sodium ions per formula unit give 2 × 0.050 = 0.100 M Na⁺. 2. Find ideal Cl⁻ concentration in 0.020 M MgCl₂. Answer: Two chloride ions per formula unit give 0.040 M Cl⁻. 3. Why cannot an added mass of insoluble salt always be turned directly into ion molarity? Answer: Only the portion actually dissolved supplies ions to the liquid; excess solid remains a separate phase.