Dilution C1V1 equals C2V2

Dilution equation C₁V₁ = C₂V₂

The dilution equation is really just solute conservation in disguise. Add only solvent and the moles (or mass) of solute stay put, which gives you C₁·V₁ = C₂·V₂. Concentration C can be in mol/L, g/L or % m/v, and V in whatever volume unit you like, as long as each side sticks to the same units. Example: take 50 mL of 1 mol/L HCl, bring it up to a final volume of 250 mL, and you land at C₂ = (50·1)/250 = 0.2 mol/L. Watch out, though: volumes are not strictly additive (there are excess partial molar volumes at play). Don't compute V_water = V₂ − V₁ and pour in that much water. The right move is to transfer the concentrate into a volumetric flask and top up with solvent until you hit the V₂ mark.

Applications

You see it every day in analytical chemistry, where working solutions get made from stock concentrates, and in clinical labs, where a sample is diluted to fall inside the calibration range. It also covers the preparation of stock solutions and reagents. Regulation leans on it too: ANVISA RDC 1.156/2024 defines 70% ethanol formulations, and making those from absolute alcohol comes down to getting the dilution right.

FAQ

Can I mix units of C and V? Yes, as long as you keep them the same on both sides. mol/L with mL works. % m/v with L works. What never works is mol/L on one side and % on the other.

Why not just add V₂ − V₁ of water? Because water and concentrate can contract a little when combined. So you add solvent up to the V₂ mark in a volumetric flask instead.

Does it apply to strong acids? For the analytical concentration, yes. But with concentrated H₂SO₄ and the like, always pour the acid into the water and never the other way around, since the mixing is exothermic.