Slide presentation
Solution Dilution
General Chemistry • Chemical Reactions
Solution concentration
Solution Dilution
Dilution lowers the concentration of a solution by adding solvent. The solution becomes less concentrated, but the amount of dissolved solute stays the same.
\(M_1V_1 = M_2V_2\)Learning target: calculate unknown concentration or volume using conserved moles of solute.
Concentrated stock
A smaller volume has more solute particles per liter.
Add solvent
Water or another solvent increases the total solution volume.
Dilute solution
The same solute is spread through a larger volume.
Why it matters
Dilution makes solutions usable in the lab
Laboratories often store concentrated stock solutions, then dilute them to the exact concentration needed for an experiment.
The stock solution has a higher concentration, but dilution allows a chemist to prepare a safer or more useful final solution without changing the chemical identity of the solute.
Dilution is used to prepare
- calibration standards,
- acid-base titration solutions,
- biological buffers,
- reaction mixtures,
- safe working concentrations from concentrated stocks.
Core concept
Solute moles stay constant during dilution
During dilution, solvent is added. The amount of NaCl, sugar, acid, or other solute does not change unless more solute is added or removed.
The equation comes from \(n = M \times V\). Since \(n_1 = n_2\), the product \(M \times V\) stays constant.
Vocabulary
Variables in the dilution equation
| Symbol | Meaning | Before or after dilution? | Common unit |
|---|---|---|---|
| \(M_1\) | Initial molarity or stock concentration | Before dilution | M or mol/L |
| \(V_1\) | Initial volume of stock solution used | Before dilution | L or mL |
| \(M_2\) | Final molarity after dilution | After dilution | M or mol/L |
| \(V_2\) | Final total solution volume | After dilution | L or mL |
| \(n\) | Moles of solute | Constant during dilution | mol |
Volumes may be in mL or L in \(M_1V_1 = M_2V_2\), but \(V_1\) and \(V_2\) must use the same volume unit.
Main relationship
Dilution is based on conserved solute moles
Common rearrangements:
\[ M_2 = \frac{M_1V_1}{V_2} \] \[ V_1 = \frac{M_2V_2}{M_1} \]Interactive simulation
Prepare a dilute NaCl solution from a stock solution
Dilution setup
Calculated result
This is a dilution: the final volume is larger, so concentration decreases while moles of NaCl stay constant.
Static fallback: \(25.0\ \mathrm{mL}\) of \(1.00\ \mathrm{M}\) NaCl diluted to \(250.0\ \mathrm{mL}\) gives \(0.100\ \mathrm{M}\) NaCl.
Dynamic relationship
As final volume increases, concentration decreases
The solute amount is constant, so concentration and final volume are inversely related.
Interpretation: increasing \(V_2\) while keeping \(M_1V_1\) constant makes \(M_2\) smaller.
Worked example
Prepare 250.0 mL of 0.100 M NaCl
Problem: What volume of \(1.00\ \mathrm{M}\) NaCl stock solution is needed to prepare \(250.0\ \mathrm{mL}\) of \(0.100\ \mathrm{M}\) NaCl?
- 1. Identify the known values. \(M_1 = 1.00\ \mathrm{M}\) \(M_2 = 0.100\ \mathrm{M}\) \(V_2 = 250.0\ \mathrm{mL}\)
- 2. Start with the dilution equation. \[ M_1V_1 = M_2V_2 \]
- 3. Solve for \(V_1\). \[ V_1 = \frac{M_2V_2}{M_1} = \frac{0.100\ \mathrm{M} \times 250.0\ \mathrm{mL}}{1.00\ \mathrm{M}} = 25.0\ \mathrm{mL} \]
- 4. Interpret the answer. Measure \(25.0\ \mathrm{mL}\) of \(1.00\ \mathrm{M}\) NaCl stock, then add solvent until the final solution volume is \(250.0\ \mathrm{mL}\).
Common mistake
Do not confuse added water with final volume
Incorrect
A student says: “Use \(25.0\ \mathrm{mL}\) stock and add \(250.0\ \mathrm{mL}\) water.”
That would make the final volume about \(275.0\ \mathrm{mL}\), not \(250.0\ \mathrm{mL}\).
Correct
Use \(25.0\ \mathrm{mL}\) stock solution, then add solvent until the total volume reaches \(250.0\ \mathrm{mL}\).
The solvent added is approximately \(250.0\ \mathrm{mL} - 25.0\ \mathrm{mL} = 225.0\ \mathrm{mL}\).
Key idea: \(V_2\) means final total solution volume, not the volume of solvent added.
Practice check
Calculate the final concentration
A student dilutes \(40.0\ \mathrm{mL}\) of \(0.500\ \mathrm{M}\) CuSO4 solution to a final volume of \(200.0\ \mathrm{mL}\). What is the final molarity?
Show answer and reasoning
Use the dilution equation:
\[ M_1V_1 = M_2V_2 \]
Solve for \(M_2\):
\[ M_2 = \frac{M_1V_1}{V_2} = \frac{0.500\ \mathrm{M} \times 40.0\ \mathrm{mL}}{200.0\ \mathrm{mL}} = 0.100\ \mathrm{M} \]
Answer: the final concentration is \(0.100\ \mathrm{M}\) CuSO4.
Apply the topic
A reliable strategy for dilution problems
Identify before and after
Label the stock solution as \(M_1, V_1\) and the final solution as \(M_2, V_2\).
Check units
Use matching volume units for \(V_1\) and \(V_2\).
Use \(M_1V_1 = M_2V_2\)
Rearrange the equation for the unknown variable.
Interpret physically
The final solution has the same moles of solute but a different concentration.
In solution preparation, always distinguish final solution volume from the amount of solvent added.
Summary
What to remember
Dilution adds solvent
The solution volume increases and concentration decreases.
Solute moles stay constant
Dilution changes concentration, not the amount of dissolved solute.
Use the dilution equation
\(M_1V_1 = M_2V_2\) comes from \(n = M \times V\).
Final volume means total volume
Do not confuse final solution volume with solvent added.