Chemistry · Lesson

Dilution, mixing and solution stoichiometry

Conserve the amount of a solute when diluting it, then distinguish nonreacting mixing from reaction stoichiometry. Calculate with final measured volume rather than assuming liquid volumes always add.

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By the end, you can

Derive c₁V₁ = c₂V₂ from conserved solute amount.
Sum amounts for a conserved nonreacting solute.
Use balanced reactions when mixing changes the species.

Dilution spreads the same amount through more volume

Take an aliquot, meaning a measured portion, of a solution and add solvent. If the specified solute neither reacts nor is lost, its amount n stays the same. Concentration decreases because the final solution volume increases. This is a conservation argument, not disappearance of solute.

Since n = cV, the initial and final states obey c₁V₁ = c₂V₂ under these conditions. Use the same solute and amount-concentration basis on both sides. The expression is not automatically a mass-percent formula, and it does not account for chemical conversion or loss.

Dilute to a volume, not by that volume

An original example takes 25.0 mL of 0.100 mol L⁻¹ solution and dilutes it to a final 100.0 mL. The solute amount is 0.00250 mol, so final c is 0.0250 mol L⁻¹. “To 100.0 mL” describes the final mixture; it does not mean adding 100.0 mL of solvent.

The graph compares this same conserved 0.00250 mol at several stipulated final volumes. Its inverse curve is c = n/V. It is a bookkeeping prediction for no reaction or solute loss, rather than an experiment or a claim about additive volumes.

The same solute amount becomes less concentrated

Amount concentration / mol L⁻¹

Final solution volume / mL

Calculated conserved-amount states
— Conserved amount: c = n/V

Underlying data

Final solution volume / mL	Amount concentration / mol L⁻¹
25	0.1 · Calculated conserved-amount states
50	0.05 · Calculated conserved-amount states
100	0.025 · Calculated conserved-amount states
200	0.0125 · Calculated conserved-amount states

Original conserved-amount case: n = 0.00250 mol. Circular markers are selected calculated concentrations at final solution volumes; the dashed line is c = n/V. The drawing does not infer how much solvent to add from additive volumes.

Add conserved amounts before finding concentration

For the same nonreacting solute, mixing two solutions gives n_total = c₁V₁ + c₂V₂. Then divide by the independently known final solution volume. If the two concentrations differ, a simple arithmetic average works only under the extra conditions that make the weighting and final volume appropriate.

Real liquid mixtures can expand or contract on mixing, so final volume need not equal the sum of initial volumes. Use the final measured volume or state an explicit additive-volume approximation. Amount conservation and volume additivity are separate assumptions.

If the species react, use a reaction ledger

When Ag⁺ and Cl⁻ form AgCl(s), their dissolved amounts change through Ag⁺ + Cl⁻ → AgCl(s). Under an idealized quantitative-precipitation approximation, the 1:1 ratio determines the limiting ion, precipitate amount and excess. Do not first average the disappearing reactant concentrations as though nothing reacted.

For 1.00 mmol Ag⁺ and 1.50 mmol Cl⁻, this approximation gives 1.00 mmol AgCl and 0.50 mmol excess Cl⁻. Atom and charge counts balance. Finite solubility, complexes and side reactions belong to a more complete equilibrium description; the quantitative approximation is explicit here.

Worked examples

Calculate a diluted concentration

Dilute 25.0 mL of 0.100 mol L⁻¹ solution to a final 100.0 mL without solute loss or reaction. Find c₂.

Convert the aliquot volume: 25.0 mL = 0.0250 L.
Conserved amount n = 0.100 × 0.0250 = 0.00250 mol.
Use final volume 0.1000 L: c₂ = 0.00250/0.1000.

c₂ = 0.0250 mol L⁻¹.

Use a supplied final mixing volume

Mix 50.0 mL of 0.100 mol L⁻¹ and 50.0 mL of 0.200 mol L⁻¹ of the same nonreacting solute. The final measured volume is supplied as 100.0 mL. Find c.

First amount = 0.100 × 0.0500 = 0.00500 mol.
Second amount = 0.200 × 0.0500 = 0.0100 mol.
Total = 0.0150 mol; divide by supplied final 0.1000 L.

c = 0.150 mol L⁻¹. Here 100.0 mL is independently supplied; it was not inferred from a universal volume-additivity rule.

Try it yourself

Does “dilute to 100 mL” mean add 100 mL of solvent?
Show the answer and reasoning
No.
It specifies the final solution volume.
Can c₁V₁ = c₂V₂ be used unchanged if half the solute is lost?
Show the answer and reasoning
No.
Its derivation requires conservation of that solute amount.
Must two real liquid solution volumes add exactly on mixing?
Show the answer and reasoning
No.
Use the final volume or an explicitly stated approximation.
In the stated 1.00 mmol Ag⁺ / 1.50 mmol Cl⁻ quantitative model, which ion limits AgCl formation?
Show the answer and reasoning
Ag⁺.
The 1:1 ratio uses 1.00 mmol of each, leaving 0.50 mmol Cl⁻ in excess.

What to carry forward

Dilution conserves solute amount only under stated conditions.
Final solution volume differs from solvent volume added.
Mixing conservation does not imply volume additivity.
Use balanced reaction amounts when species change.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.