C1V1 = C2V2: the dilution formula, explained

C1V1 = C2V2 says the amount of solute doesn't change when you add solvent. C1 and V1 are the concentration and volume of the stock solution you measure out. C2 and V2 are the concentration and final volume of the diluted solution. Know any three and you can solve for the fourth. (Some textbooks write it as M1V1 = M2V2, which is the same equation with molarity.)

Why it works

Molarity is moles of solute per liter of solution, so moles = molarity x volume. Diluting adds only solvent, so the moles of solute before and after are equal:

moles before = moles after, so C1 x V1 = C2 x V2

That's the whole derivation. It also tells you the formula's limits. It only holds when the moles of solute stay the same, which rules out anything where the solute reacts.

Solving for each value

V1 = C2 x V2 / C1 (how much stock to measure out, the most common question)
C2 = C1 x V1 / V2 (the concentration after diluting)
V2 = C1 x V1 / C2 (the final volume to dilute to)
C1 = C2 x V2 / V1 (the concentration of the stock)

Example 1: how much stock do you need?

You need 500. mL of 0.500 M HCl, and the stock bottle is 12.0 M.

V1 = (0.500 M x 500. mL) / 12.0 M = 20.8 mL of stock

Measure 20.8 mL of the 12.0 M acid and dilute it to a final volume of 500. mL. That's about 479 mL of water. For acids, add the acid to water rather than water to acid.

Notice the volume units: V2 went in as mL, so V1 came out in mL. You don't need to convert to liters as long as both volumes use the same unit.

Example 2: what's the new concentration?

25.0 mL of 2.00 M NaCl is diluted to 250. mL.

C2 = (2.00 M x 25.0 mL) / 250. mL = 0.200 M

The volume went up by a factor of 10, so the concentration went down by a factor of 10. That ratio, C1 / C2 (here 10), is the dilution factor, and checking it is a quick sanity test.

Example 3: a textbook-style problem in liters

0.850 L of a 5.00 M copper(II) nitrate solution is diluted to 1.80 L. What's the new concentration?

C2 = (5.00 M x 0.850 L) / 1.80 L = 2.36 M

The volume roughly doubled, and the concentration roughly halved. The answer checks out.

Example 4: mM and mL

You have a 100. mM stock and need 50.0 mL of 5.00 mM.

V1 = (5.00 mM x 50.0 mL) / 100. mM = 2.50 mL of stock, plus 47.5 mL of solvent

Millimolar works just like molar, provided C1 and C2 are in the same unit. The mistake to avoid is mixing them: putting 0.100 M on one side and 5.00 mM on the other gives an answer that is off by a factor of 1,000.

Example 5: make the stock, then dilute it

Dilution problems often follow a molarity step. Suppose you dissolve 5.85 g of NaCl (58.44 g/mol) in enough water to make 100. mL of solution.

Moles: 5.85 / 58.44 = 0.100 mol
Molarity: 0.100 mol / 0.100 L = 1.00 M

Now take 10.0 mL of that and dilute to 100. mL: C2 = 1.00 M x 10.0 / 100. = 0.100 M. The molar mass calculator handles the first step for any compound.

Serial dilutions

A serial dilution repeats the same dilution several times, which is how labs make very dilute solutions accurately. Take 1.00 mL of 1.00 M solution and dilute to 10.0 mL, a 1-in-10 dilution, to get 0.100 M. Do it again with the new solution and you have 0.0100 M. After n tenfold steps, the concentration is the original divided by 10^n. Making 0.0100 M in one step would mean measuring 0.100 mL into 10.0 mL, which is hard to do precisely.

"Solvent to add" versus "final volume"

V2 is the final volume of the solution, not the volume of water you add. The water you add is roughly V2 - V1, and intro courses treat it that way. In the lab, volumes aren't always exactly additive, so the correct procedure is to put the stock in a volumetric flask and fill to the V2 mark. If a problem says "added to 100. mL of water," the final volume is about V1 + 100. mL, not 100. mL. Read carefully.

When not to use C1V1 = C2V2

Titrations and other reactions. At the equivalence point, moles of acid and base are related by the balanced equation. For HCl and NaOH, which react 1:1, the numbers look like C1V1 = C2V2. For H2SO4 and NaOH, which react 1:2, they don't. Use moles and the mole ratio, as in stoichiometry step by step.
Mixing two solutions of the same solute. Add the moles from each solution, then divide by the total volume.
Anything where the solute reacts or precipitates. The moles are no longer constant, so the equation no longer applies.

Quick self-checks

C2 should be smaller than C1, and V2 larger than V1. If not, you've swapped something.
C1 / C2 should equal V2 / V1. Both are the dilution factor.
Units on both sides must match: M with M, mM with mM, mL with mL.

Practice

The dilution calculator solves for any one of the four values with mixed M, mM, L and mL units, and shows the solvent to add. Use it to check answers after you've worked the problem yourself, not before. For the rest of the unit, see what's in General Chemistry I. In the second semester, dilution comes back when you calculate pH for strong and weak acids. Biology labs lean on it as well, especially serial dilutions for counting bacteria, which pair naturally with the microbiology units of General Biology II.

Molarity and dilution have their own chapter in Encodr's free General Chemistry I flashcards.