Diluição C1·V1 = C2·V2

Dilution equation C₁V₁ = C₂V₂

When you add solvent and nothing else, the amount of solute stays put. The dilution equation is just that idea written down: C₁·V₁ = C₂·V₂, where C is the concentration (mol/L, g/L, % m/v) and V the volume. The product C·V is the moles or grams of solute, so it has the same value before and after you dilute. Example: take 50 mL of 1 mol/L HCl, bring it up to a final volume of 250 mL, and C₂ = (50·1)/250 = 0.2 mol/L. One practical note: in a volumetric flask you add solvent up to the final volume V₂. Don't compute V_water = V₂ − V₁ and pour in just that amount. Mixed volumes aren't strictly additive (excess partial molar volumes come into play), so that shortcut gets you close but not exactly there.

Applications

In analytical chemistry it comes up every time you make working solutions out of stock concentrates. The clinical laboratory leans on it for sample dilutions when an assay drifts past its linear range. Microbiology uses it to prepare culture media and to set up the serial dilutions behind colony counting. And in chemistry and pharmacy teaching, the moment stock solutions show up on the bench, this is probably the equation a student reaches for most.

FAQ

Can I use any unit for C and V? Yes, as long as the units agree on both sides. mol/L with mL is fine, % m/v with L is fine. The one rule: C₁ and C₂ must share units, and so must V₁ and V₂.

Why not compute V_water = V₂ − V₁? Because mixed volumes aren't strictly additive; water and acid poured together can actually contract. What works reliably is to put the concentrate in the volumetric flask and fill with solvent up to the V₂ mark.

Does the equation apply to dilution of strong acids? Yes, for the analytical concentration. Watch the heat, though. When the dilution is strongly exothermic, as it is with concentrated H₂SO₄, add the acid to the water slowly. Reverse that and you risk violent splashing.