Leave one of the four boxes empty to solve for it. Concentrations can be in any unit (M, mg/mL, %, ppm) as long as C₁ and C₂ use the same one. Below, a mix-ratio tool handles "1 part to 10 parts water" instructions.
When you dilute a solution, the amount of dissolved substance stays the same; only the volume changes. That gives the dilution equation C₁V₁ = C₂V₂, where C₁ and V₁ are the stock concentration and the volume you take, and C₂ and V₂ are the concentration and volume you end up with. Solve for the missing one: V₁ = C₂V₂ ÷ C₁.
Example: to make 500 mL of a 2% solution from a 10% stock, V₁ = 2 × 500 ÷ 10 = 100 mL of stock, topped up with 400 mL of water. The dilution factor is C₁ ÷ C₂ = 5, often written 1:5 in the lab (1 part stock in 5 parts total).
Product labels usually say "1 part concentrate to N parts water". That means 1 + N parts in total, so a 1:10 mix for 5 litres is 5 ÷ 11 ≈ 0.45 L of concentrate and 4.55 L of water. Watch out: some lab protocols use 1:10 to mean 1 part in 10 total. This mix-ratio tool always uses "parts concentrate to parts water".
Stock concentration × stock volume = final concentration × final volume. Because diluting doesn’t change how much solute there is, the two products are equal, so you can find any one of the four values from the other three.
C₁ and C₂ must be in the same unit as each other (both % or both mg/mL, for example), and V₁ and V₂ must be in the same volume unit. The equation works with any consistent pair.
Diluent to add = V₂ − V₁. To make 500 mL of a 2% solution from 10% stock you need 100 mL of stock plus 400 mL of water or buffer.
On most product labels it means 1 part concentrate to 10 parts water, 11 parts in total. In some lab contexts 1:10 means 1 part in 10 total. Check which convention your instructions use; this calculator uses parts concentrate to parts water.
No. Dilution only lowers concentration, so C₂ must be less than or equal to C₁. If you get a V₁ bigger than V₂, your target is stronger than your stock.
