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The Dilution Equation (C₁V₁ = C₂V₂), Explained

After reading this you can take any stock solution, decide how much to use and how much solvent to add to hit a target concentration and volume, and check your work with real numbers.

You have a bottle of concentrated stock and you need a weaker working solution. Say you have 10 M hydrochloric acid and you want 100 mL of 1 M acid. How much of the stock do you pour, and how much water do you add? The dilution equation answers this in one line. Measure out 10 mL of the stock, add water to reach 100 mL total, and you are done. This article shows why that works and how to handle every variation.

What the equation says and why it is true

When you dilute a solution you add solvent (usually water) but you add no new solute. The number of moles of solute in the flask before and after is identical. That single fact is the whole equation.

Concentration times volume gives an amount of solute. In molarity that product is moles:

C \cdot V = \frac{\text{mol}}{\text{L}} \times \text{L} = \text{mol}

Here C is concentration and V is volume. Because the moles do not change, the product before dilution equals the product after:

C_1 V_1 = C_2 V_2

C_1 and V_1 describe the concentrated stock you start with. C_2 and V_2 describe the final diluted solution. The left side is the moles of solute you take from the bottle. The right side is the moles of solute in your finished flask. They match because you never threw any solute away or added more.

The equation is not limited to molarity. It works for any concentration measure that is proportional to amount per volume: percent weight per volume, grams per liter, parts per million. The only rule is that both concentrations use the same unit and both volumes use the same unit.

When to use it and when not to

Use C₁V₁ = C₂V₂ whenever you make a weaker solution from a stronger one by adding solvent. That covers most bench dilutions: preparing working buffers, thinning a dye, making a series of standards from one stock.

Do not use it in these cases:

  • You are mixing two solutions that both contain solute. Then you must add moles from both sources, not track a single stock.
  • The solute reacts, precipitates, or evaporates. The equation assumes the amount of solute is conserved.
  • Volumes are not additive in a way you are ignoring. For concentrated stocks, mixing 10 mL of acid with 90 mL of water does not always give exactly 100 mL. This is why you fill to a mark rather than adding a measured volume of water. More on this below.

Solving for each term

The equation has four terms. Give any three and algebra returns the fourth. The four rearrangements are:

The four forms of C₁V₁ = C₂V₂
UnknownFormulaReads as
V_1V_1 = \dfrac{C_2 V_2}{C_1}how much stock to use
C_2C_2 = \dfrac{C_1 V_1}{V_2}strength you end up with
C_1C_1 = \dfrac{C_2 V_2}{V_1}stock strength needed
V_2V_2 = \dfrac{C_1 V_1}{C_2}final volume produced

The most common bench task is the first row: you know your stock, you know the target, and you need to find how much stock to pipette. Once you have V_1, the solvent to add is simply the difference:

V_{\text{solvent}} = V_2 - V_1

This is the volume of water (or other solvent) that turns your measured stock into the final solution. It is exact only when volumes add cleanly, which holds well for dilute aqueous work.

A worked example with the demo numbers

Making 100 mL of 1 M acid from 10 M stock

Use the demo values: C_1 = 10, C_2 = 1, V_2 = 100 mL, solving for V_1.

  1. Write the equation: C_1 V_1 = C_2 V_2.
  2. Solve for the stock volume: V_1 = \dfrac{C_2 V_2}{C_1}.
  3. Insert numbers: V_1 = \dfrac{1 \times 100}{10} = 10 mL.
  4. Find the solvent: V_{\text{solvent}} = 100 - 10 = 90 mL.
  5. Check the moles. Stock: 10 \times 0.010 = 0.10 mol. Final: 1 \times 0.100 = 0.10 mol. They match.

So you measure 10 mL of the 10 M stock into a 100 mL volumetric flask, then add water until the meniscus sits on the 100 mL mark. Note that the moles check used volumes in liters (0.010 L and 0.100 L), but the ratio in step 3 works in any consistent volume unit because the units cancel.

The dilution factor and how to read the result

The dilution factor tells you how many times weaker the final solution is than the stock. It is the ratio of the two concentrations, which equals the ratio of the two volumes:

\text{DF} = \frac{C_1}{C_2} = \frac{V_2}{V_1}

In the worked example, \text{DF} = 10/1 = 10. This is called a 10-fold dilution, often written 1:10. It means one part stock diluted to ten parts total volume. Be careful with notation: 1:10 on a total-volume basis means 1 part stock plus 9 parts solvent, not 1 part stock plus 10 parts solvent. The colon notation is a frequent source of error, so prefer stating the dilution factor as a plain number when you can.

Dilution factor
Final volume divided by stock volume, equal to the fold reduction in concentration.
Stock
The concentrated solution you start with, described by C_1 and V_1.
Working solution
The diluted result, described by C_2 and V_2.
Holding the stock at 10 M and the final volume at 100 mL, the required stock volume rises in proportion to the target concentration: 5 mL for 0.5 M, 10 mL for 1 M, up to 50 mL for 5 M.

Explore how the stock strength changes the pour

With a fixed target of 100 mL at 1 M, the stock volume you pipette equals 100 divided by the stock molarity. A 2 M stock needs 50 mL, a 5 M stock needs 20 mL, a 10 M stock needs 10 mL, and a 50 M stock would need only 2 mL. The solvent added is 100 minus that pour. As the stock gets stronger, the pour shrinks and small measuring errors matter more.

Common mistakes

The single most common error is adding water equal to the final volume instead of filling to the final volume. If you need 100 mL of final solution and you pour 10 mL of stock, add water until the total reads 100 mL. Do not add 100 mL of water to the 10 mL of stock, which would give 110 mL and the wrong concentration.

Other traps to watch:

  • Mismatched units. If C_1 is in mol/L then C_2 must be in mol/L too. The same holds for the volumes. The equation only cares that each pair is consistent; the two pairs can differ from each other in unit.
  • Trying to concentrate. If your target C_2 is higher than C_1, the math gives V_1 \gt V_2, which is impossible to fit in the flask. You cannot make a solution stronger by adding solvent.
  • Non-additive volumes. For concentrated stocks, water and solute do not always sum volumes cleanly. Filling to a calibrated mark handles this; blindly adding V_2 - V_1 of water does not.
  • Very strong stocks. A 50 M stock diluted to 1 M in 100 mL needs only 2 mL of stock. A pipetting error of 0.1 mL is 5 percent of the pour, so the final concentration is off by 5 percent. Use a smaller intermediate dilution to keep volumes measurable.

Related tools

Once you know the concentration you want, other calculators pick up the thread. Convert between molarity, grams per liter, ppm and percent with the Concentration Converter. To turn a target molarity into grams to weigh out, find the molar mass with the Molar Mass Calculator and then use the Mole Calculator. For acid and base dilutions where you also want the resulting pH, the pH Calculator takes concentration as input. If your diluted solutions feed a reaction, the Stoichiometry Calculator and the Equilibrium (ICE Table) Solver continue from there. For freezing-point and boiling-point shifts of the solution you just made, use the Boiling & Freezing Point Calculator.

Frequently asked questions

Does C₁V₁ = C₂V₂ work for percent solutions?

Yes. It works for any concentration unit that scales with amount per volume, including percent weight per volume. To make 200 mL of a 5 percent solution from a 20 percent stock, V_1 = (5 \times 200) / 20 = 50 mL of stock, filled to 200 mL. Keep both percents on the same basis.

How much water do I actually add?

Add solvent equal to V_2 - V_1, but do it by filling to the final volume mark rather than measuring water separately. In the demo case that is 100 - 10 = 90 mL of water, added until the total reaches 100 mL.

What is a 1:10 dilution?

A 1:10 dilution has a dilution factor of 10, meaning 1 part stock brought up to 10 parts total. That is 1 part stock plus 9 parts solvent. The final concentration is one tenth of the stock.

Can I use it to make a stronger solution?

No. Adding solvent can only lower concentration. If you solve for V_1 and get a value larger than V_2, your target is stronger than your stock and the dilution is impossible.

What if I mix two solutions that both contain solute?

Then the equation does not apply directly. Add the moles from each source (C_a V_a + C_b V_b) and divide by the total volume to get the final concentration. The single-stock form assumes only one source of solute.

Try it yourself in the tool

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