Dilution is the addition of water to a solution to lower its concentration; the number of moles of solute does not change, giving M₁V₁ = M₂V₂.
Malay: Pencairan · Chinese: 稀释
Dilution is the addition of water to a solution to lower its concentration. The single idea that makes every dilution calculation work is this: adding water spreads the same amount of solute through a larger volume, so the number of moles of solute does not change. Because moles before equal moles after, and moles = molarity × volume, you get the working formula M₁V₁ = M₂V₂, where subscript 1 is before and subscript 2 is after.
A worked example shows it in action. Take 25 cm³ of 2.0 mol dm⁻³ hydrochloric acid and add water until the total volume is 100 cm³. Then M₂ = M₁V₁ ÷ V₂ = (2.0 × 25) ÷ 100 = 0.5 mol dm⁻³. The acid is now four times more dilute because the volume is four times larger, yet the amount of HCl in the flask, 0.05 mol, is exactly what it was at the start.
The confusion to avoid is thinking that dilution reduces the amount of solute. It does not, only the concentration falls, because the solute is unchanged and only the water is added. Students who try to “use up” moles during a dilution get the calculation wrong. A second slip is mismatched units: V₁ and V₂ must be in the same unit (both cm³ or both dm³), though because they appear on both sides you do not need to convert to dm³ here.
Within this chapter dilution rests on molarity and feeds into preparing solutions for titration, where a stock solution is often diluted to a working concentration. Our teachers keep the “moles stay the same” idea front and centre so the formula is something you understand rather than merely memorise.
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