Laboratories buy acids and salt solutions concentrated: hydrochloric acid arrives at about 12 M, and stock solutions are often made at 1 M or 2 M. Before use they are diluted to the strength an experiment needs, often 0.1 M or less.

Dilution only adds water. The moles of solute taken from the stock are the moles in the diluted solution, and that one fact gives the dilution equation (Figure 1).

1. startStock solutionC₁, V₁
2. $n = C_1 V_1$moles taken
3. Same molesonly water is added
4. $C_2 = \frac{n}{V_2}$spread over V₂
5. answerDiluted solutionC₂, V₂

Setting the moles equal gives $C_1 V_1 = C_2 V_2$.Both volumes in the same unit.

Learning Objectives

After this lesson, you should be able to:

1. Use C1V1 = C2V2 to find a stock volume or final concentration.
2. Describe how to prepare a dilution correctly (volumetric flask, acid into water).

Key Ideas

1. Adding Water Keeps the Moles

From the previous lesson, Molarity, the amount of solute in a volume of solution is $n = C \times V$. When you dilute, the solute particles spread through more volume, but none are added or removed (Figure 2).

Key idea: moles of solute before dilution = moles of solute after dilution.

2. The Dilution Equation

Write the moles before and after as concentration times volume and set them equal:

• $C_1, V_1$: concentration and volume of the stock solution you take
• $C_2, V_2$: concentration and final volume of the diluted solution

Both volumes must be in the same unit. Milliliters are fine, because the unit cancels when you solve for $V_1$ or $C_2$ (Figure 3).

Try it: dilute a stock solution

Choose how much stock to take and the final volume. Watch the moles stay the same while the concentration falls; then find the settings of Worked Example 1.

3. Making a Dilution in the Lab

Measure the stock with a pipette, transfer it to a volumetric flask that already holds some water, then fill to the mark and mix (Figure 4). With concentrated acids, always add the acid to water: adding water to a concentrated acid can boil and spatter.

Worked Example 1: Volume of Stock Needed

Problem: How many milliliters of 2.00 M hydrochloric acid ($\mathrm{HCl}$) are needed to prepare 250.0 mL of 0.200 M HCl?

Step 1. List the known quantities.

• $C_1 = 2.00\,\text{M}$ (stock)
• $C_2 = 0.200\,\text{M}$ (diluted)
• $V_2 = 250.0\,\text{mL}$ (final volume)
• $V_1 = ?$

Step 2. Rearrange the dilution equation for $V_1$.

Step 3. Substitute and calculate.

Final Answer: 25.0 mL of 2.00 M HCl

1. Diluted solution wanted$0.200\,\text{M} \times 250.0\,\text{mL}$
2. Step 3$\times \dfrac{1}{2.00\,\text{M}}$divide by the stock concentration
3. Stock volume$25.00\,\text{mL} \approx 25.0\,\text{mL}$Final answer to 3 significant figures

Pipette 25.0 mL of the stock into a 250 mL volumetric flask that already contains water, then fill to the mark. About 225.0 mL of water ends up being added, but you fill to the mark rather than measure the water.

Your turn: volume of stock needed

Same method as Worked Example 1: rearrange for the stock volume, substitute, and keep both volumes in mL.

Worked Example 2: Concentration After Dilution

Problem: A student pipettes 10.0 mL of 1.00 M copper(II) sulfate ($\mathrm{CuSO_4}$) into a volumetric flask and dilutes it to 250.0 mL. What is the concentration of the diluted solution?

Step 1. Rearrange the dilution equation for $C_2$.

Step 2. Substitute and calculate.

Final Answer: 0.0400 M CuSO₄

1. Stock taken$1.00\,\text{M} \times 10.0\,\text{mL}$
2. Step 2$\times \dfrac{1}{250.0\,\text{mL}}$spread over the final volume
3. Diluted concentration$0.04000\,\text{M} \approx 0.0400\,\text{M}$Final answer to 3 significant figures

Exam tip: a dilution always lowers the concentration. The volume grew 25 times (10.0 mL to 250.0 mL), so the concentration must fall 25 times: $\frac{1.00\,\text{M}}{25} = 0.0400\,\text{M}$.

Your turn: concentration after dilution

Same method as Worked Example 2: the final volume goes in the denominator.

Common Mistakes

Common mistake: Swapping the concentrations. Writing $V_1 = \frac{C_1 V_2}{C_2}$ in Worked Example 1 gives 2500 mL, more stock than the whole flask holds. The stock volume is always the smaller volume.

Common mistake: Confusing the stock volume with the water added. $V_1$ is the volume of stock (25.0 mL). The water added is roughly $V_2-V_1$ = 225.0 mL, but you fill to the mark rather than measuring it.

Common mistake: Using V₁ + V₂ as the final volume. $V_2$ is already the final volume. Dividing by 10.0 mL + 250.0 mL in Worked Example 2 gives 0.0385 M instead of 0.0400 M.

Common mistake: Mixing volume units. Use mL on both sides or L on both sides. If you need the answer in mL, do not convert it to liters at the end.

Common mistake: Adding water to concentrated acid. Always add acid to water, slowly, so the heat released is absorbed by the larger volume of water.

Key Takeaway

Dilution adds water but not solute, so the moles stay the same: C₁V₁ = C₂V₂. Solve for the unknown with matching volume units, and check that the diluted solution is weaker than the stock.

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. The simulator and the practice problems are audited; see the lesson audit log.