The formula
- Dilution relation
C₁V₁ = C₂V₂, often written M₁V₁ = M₂V₂- Stock volume needed
V₁ = C₂V₂ / C₁- Concentration after dilution
C₂ = C₁V₁ / V₂- Why it works
n(solute) is unchanged, and n = C × V
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
C₁, M₁ | Concentration of the concentrated stock solution | M |
V₁ | Volume of stock taken | L or mL, as long as V₂ matches |
C₂, M₂ | Concentration of the diluted solution | M |
V₂ | Final total volume after dilution | same unit as V₁ |
n(solute) | Moles of solute, identical before and after | mol |
When it applies
- Solvent is being added to an existing solution and no reaction takes place. Once the solute reacts, the moles change and this relation no longer holds.
- Both volumes carry the same unit. Because the units cancel, milliliters work as well as liters, but mixing the two does not.
- V₂ is the final total volume, not the volume of solvent added. Diluting 4.17 mL of stock to 500.0 mL means topping up to the 500.0 mL mark.
- Any consistent concentration unit works, not just molarity, as long as both sides use the same one.
Worked example
Problem. How much 12.0 M concentrated hydrochloric acid is needed to prepare 500.0 mL of 0.100 M hydrochloric acid?
- Identify the four quantities: C₁ = 12.0 M, C₂ = 0.100 M, V₂ = 500.0 mL, and V₁ is what you want.
- Rearrange for the stock volume: V₁ = C₂V₂ / C₁ = (0.100 M × 500.0 mL) / 12.0 M.
- The numerator is 50.0, so V₁ = 50.0 / 12.0 = 4.1667 mL, which is 4.17 mL.
- Check by running it forward: C₂ = (12.0 × 4.1667) / 500.0 = 0.100 M, as required.
- In practice, measure 4.17 mL of acid into a flask already holding some water, then add water up to the 500.0 mL mark. Acid goes into water, never the reverse.
Answer. 4.17 mL of the 12.0 M stock, diluted with water to a final volume of 500.0 mL.
Common mistakes
- Adding V₂ of solvent rather than diluting up to a total of V₂, which overshoots the final volume and undershoots the concentration.
- Mixing units between the two volumes, such as liters on one side and milliliters on the other.
- Using the relation across a reaction, such as an acid neutralizing a base. That is a titration calculation, and it needs the balanced equation.
- Solving for the wrong unknown by pairing C₁ with V₂. Each concentration stays with the volume of its own solution.
Related formulas
- Titration:
n(titrant) = M(titrant) × V(titrant) - Molarity formula:
M = mol solute / L solution