Concentration and weighted mixture averages

Outcome

Find the concentration of combined batches using weighted averages, and calculate how much liquid must be added to reach a target concentration.

Concept and assumptions

Concentrations are percentage by volume (% v/v): solute volume per 100 volume units of solution. All volumes are in litres. Assume homogeneous, thoroughly mixed liquids, additive volumes, and no reaction, evaporation or loss of solute. Pure water contributes no solute.

For volume V at c%:

Solute volume = V × c/100.

Mixing adds the solute amounts and the total liquid volumes separately. Therefore:

Final concentration (%) = 100 × total solute volume ÷ total mixture volume.

Equivalently, for two batches:

Final percentage = (V₁c₁ + V₂c₂)/(V₁ + V₂).

Here c₁ and c₂ are percentage numbers, not decimal fractions. Volumes are the weights because a larger batch contributes proportionately more material. Equal batch volumes allow a simple average; unequal volumes generally do not.

For mass-percent mixtures, use masses as weights instead. Do not combine mass and volume percentages without suitable conversion data. A weighted average lies between the lowest and highest component concentrations.

Adding water leaves the solute amount unchanged but increases total volume. Adding a solution usually changes both. Here, liquid is added, not removed.

Worked examples

Example 1 — Two unequal batches. Mix 3 L of 20% solution with 7 L of 40% solution.

Solute amounts = 3 × 20/100 = 0.6 L and 7 × 40/100 = 2.8 L. Total solute = 3.4 L; total mixture = 3 + 7 = 10 L. Concentration = 100 × 3.4/10 = 34%. It is nearer 40% because that batch is larger.

Example 2 — Three batches. Mix 4 L of 15%, 6 L of 25% and 10 L of 40% solution.

Solute amounts = 4 × 0.15 = 0.6 L; 6 × 0.25 = 1.5 L; 10 × 0.40 = 4 L. Total solute = 0.6 + 1.5 + 4 = 6.1 L. Total volume = 4 + 6 + 10 = 20 L. Concentration = 100 × 6.1/20 = 30.5%.

Example 3 — An unknown addition. How much 50% solution must be added to 20 L of 15% solution to obtain 30%?

Let the added volume be x L. Initial solute = 20 × 15/100 = 3 L. Added solute = 0.50x; final volume = 20 + x. Thus 3 + 0.50x = 0.30(20 + x). Expanding: 3 + 0.50x = 6 + 0.30x. Therefore 0.20x = 3, giving x = 15 L. Check: solute = 3 + 7.5 = 10.5 L; volume = 35 L; 100 × 10.5/35 = 30%.

Common mistakes

Averaging unequal batches without weights; dividing by the original rather than final volume; treating a percentage as a volume; assuming dilution destroys solute; mixing incompatible concentration units.

Practice questions

1. Mix 5 L of 12% solution and 15 L of 28% solution. Find the concentration.
2. Add 4 L of water to 16 L of 25% solution. Find the concentration.
3. Mix 6 L of 10%, 4 L of 25% and 10 L of 40% solution. Find the concentration.
4. How much water must be added to 30 L of 24% solution to obtain 18%?

Worked answers

1. Solute = 5 × 0.12 + 15 × 0.28 = 0.6 + 4.2 = 4.8 L. Total volume = 20 L. Concentration = 100 × 4.8/20 = 24%.
2. Solute stays 16 × 0.25 = 4 L. Total volume = 16 + 4 = 20 L. Concentration = 100 × 4/20 = 20%.
3. Solute = 6 × 0.10 + 4 × 0.25 + 10 × 0.40 = 5.6 L. Volume = 20 L. Concentration = 100 × 5.6/20 = 28%.
4. Solute = 30 × 0.24 = 7.2 L. Let water added be x L. Then 7.2/(30 + x) = 0.18, so 30 + x = 40. Hence x = 10 L.