Dilution Equation

Three of the four quantities are given, and you solve for the fourth. To find how much stock solution you need, rearrange for V₁: V₁ = c₂·V₂/c₁. Example: for 250 mL of a 0.1-molar solution from a 2.0-molar stock, V₁ = 0.1·250/2.0 = 12.5 mL. You pipette these 12.5 mL into a 250 mL volumetric flask and fill up to the ring mark with water. Important: V₂ is always the final volume of the finished solution, not the amount of water added. The volume units may be mL or L as long as both sides use the same unit, because the conversion factor cancels.

Because when diluting, only solvent is added while the dissolved substance stays completely in the solution. Its amount n therefore does not change. Since molar concentration is defined as c = n/V, before dilution n = c₁·V₁ and afterwards n = c₂·V₂. Equating the two expressions for the same unchanged amount directly gives c₁·V₁ = c₂·V₂. This also makes intuitive sense: if you increase the volume tenfold, the concentration drops to one tenth. The equation breaks down as soon as a reaction occurs or substance precipitates or evaporates, because then the amount is no longer constant.

The dilution factor f states by which factor the solution is diluted: f = V₂/V₁ = c₁/c₂. A 1:10 dilution means f = 10, one part stock solution in a total of ten parts final volume. Dilution series matter in practice: if you dilute 1:10 three times in a row, the factors multiply to f = 10³ = 1000 and the concentration drops to one thousandth. This is how calibration solutions for photometry or microbial dilutions in microbiology are made from one stock. Beware of the wording: in chemistry, "dilute 1:10" usually means 1 part plus 9 parts solvent, giving a final volume of 10 parts.

Because V₂ in the equation is the exact final volume of the solution. If you simply added 250 mL of water to 12.5 mL of stock solution, the finished solution would have 262.5 mL and the concentration would be too low. Moreover, volumes are not always additive: when mixing ethanol and water, the total volume is slightly smaller than the sum of the individual volumes. The volumetric flask avoids both problems because it is calibrated to a single certified final volume: add the stock, add water to just below the mark, mix, then fill exactly to the ring mark. This way V₂ is correct regardless of mixing effects.

Only with one important addition. In a titration two substances react, and at the equivalence point the amounts have reacted in their stoichiometric ratio. For a 1:1 reaction such as HCl with NaOH the equation therefore looks formally identical: c(acid)·V(acid) = c(base)·V(base). But as soon as the ratio is not 1:1, the stoichiometric factor enters: sulfuric acid delivers two protons, so 2·c(H₂SO₄)·V(H₂SO₄) = c(NaOH)·V(NaOH). The pure dilution equation, by contrast, only describes the case without reaction, where the same portion of substance is merely spread over more volume. Mixing up the two situations makes the result wrong by the stoichiometric factor.