Molarity of a Mixture of Two Solutions: Add Moles, Not Molarities
Two bottles of the same chemical, one 0.5 M and one 0.2 M, are poured together. What is the molarity now? Many students average the two numbers and write 0.35 M. That is wrong unless the two volumes are equal. This page shows the correct method, which is also the method behind most "mixing of solutions" questions in Class 11 and Class 12.
The one idea: add moles, not molarities
Molarity tells you moles per litre. When you mix solutions, the moles of solute simply add up, and so do the volumes (for the dilute solutions in school problems). The new molarity is total moles divided by total volume.
Mmix = (M1V1 + M2V2) / (V1 + V2)
This is valid when both solutions contain the same solute and do not react.
This is valid when both solutions contain the same solute and do not react.
What each symbol means
- M1, M2 are the molarities (mol/L) of the two solutions before mixing.
- V1, V2 are their volumes. Use mL for both or litre for both, but do not mix the units. The ratio then stays correct.
- M1V1 gives the amount of solute. In mL it is in millimoles (mmol), which is easier to handle. 1 mmol = 10−3 mol.
- Mmix is the final molarity of that solute in the combined solution.
Note that this is the same logic as the dilution formula M1V1 = M2V2. Dilution is just mixing with water of zero molarity.
Worked examples
Example 1: same solute, different strengths. 200 mL of 0.5 M NaCl is mixed with 300 mL of 0.2 M NaCl. Find the molarity.
Step 1: Moles from the first solution = 0.5 × 200 = 100 mmol.
Step 2: Moles from the second solution = 0.2 × 300 = 60 mmol.
Step 3: Total = 160 mmol. Total volume = 500 mL.
Step 4: Mmix = 160 / 500 = 0.32 M.
Answer: 0.32 M. The simple average (0.35 M) is wrong because there is more of the weaker solution. The answer lies between 0.2 and 0.5, closer to 0.2, as it should.
Step 1: Moles from the first solution = 0.5 × 200 = 100 mmol.
Step 2: Moles from the second solution = 0.2 × 300 = 60 mmol.
Step 3: Total = 160 mmol. Total volume = 500 mL.
Step 4: Mmix = 160 / 500 = 0.32 M.
Answer: 0.32 M. The simple average (0.35 M) is wrong because there is more of the weaker solution. The answer lies between 0.2 and 0.5, closer to 0.2, as it should.
Example 2: different solutes, no reaction. 100 mL of 0.2 M NaCl is mixed with 100 mL of 0.1 M CaCl2. Find the chloride ion concentration.
Step 1: NaCl gives one Cl− per formula unit: 0.2 × 100 = 20 mmol Cl−.
Step 2: CaCl2 gives two Cl−: 0.1 × 100 × 2 = 20 mmol Cl−.
Step 3: Total Cl− = 40 mmol in 200 mL.
Step 4: [Cl−] = 40 / 200 = 0.2 M.
Answer: 0.2 M. Always count ions, not formula units, when the question asks for an ion concentration.
Step 1: NaCl gives one Cl− per formula unit: 0.2 × 100 = 20 mmol Cl−.
Step 2: CaCl2 gives two Cl−: 0.1 × 100 × 2 = 20 mmol Cl−.
Step 3: Total Cl− = 40 mmol in 200 mL.
Step 4: [Cl−] = 40 / 200 = 0.2 M.
Answer: 0.2 M. Always count ions, not formula units, when the question asks for an ion concentration.
Example 3: mixing with a reaction. 100 mL of 0.1 M HCl is mixed with 50 mL of 0.1 M NaOH. Find the pH of the mixture at 25 °C.
Step 1: H+ available = 0.1 × 100 = 10 mmol. OH− available = 0.1 × 50 = 5 mmol.
Step 2: They react 1 : 1, so 5 mmol of H+ is left over (10 − 5).
Step 3: Total volume = 150 mL, so [H+] = 5 / 150 = 0.0333 M.
Step 4: pH = −log(0.0333) = 1.48.
Answer: pH ≈ 1.48. Here the formula Mmix above cannot be used directly, because the two solutes react. Subtract first, then divide by the total volume.
Step 1: H+ available = 0.1 × 100 = 10 mmol. OH− available = 0.1 × 50 = 5 mmol.
Step 2: They react 1 : 1, so 5 mmol of H+ is left over (10 − 5).
Step 3: Total volume = 150 mL, so [H+] = 5 / 150 = 0.0333 M.
Step 4: pH = −log(0.0333) = 1.48.
Answer: pH ≈ 1.48. Here the formula Mmix above cannot be used directly, because the two solutes react. Subtract first, then divide by the total volume.
Example 4: working backwards. In what volumes should 0.6 M and 0.1 M solutions of the same solute be mixed to make 500 mL of 0.3 M solution?
Step 1: Let the volume of the 0.6 M solution be x mL. The other is (500 − x) mL.
Step 2: Moles needed in the final solution = 0.3 × 500 = 150 mmol.
Step 3: 0.6x + 0.1(500 − x) = 150, so 0.5x + 50 = 150, so x = 200 mL.
Step 4: The 0.1 M solution is 500 − 200 = 300 mL.
Check: (0.6 × 200 + 0.1 × 300) / 500 = (120 + 30) / 500 = 0.30 M ✓
Answer: 200 mL of 0.6 M and 300 mL of 0.1 M.
Step 1: Let the volume of the 0.6 M solution be x mL. The other is (500 − x) mL.
Step 2: Moles needed in the final solution = 0.3 × 500 = 150 mmol.
Step 3: 0.6x + 0.1(500 − x) = 150, so 0.5x + 50 = 150, so x = 200 mL.
Step 4: The 0.1 M solution is 500 − 200 = 300 mL.
Check: (0.6 × 200 + 0.1 × 300) / 500 = (120 + 30) / 500 = 0.30 M ✓
Answer: 200 mL of 0.6 M and 300 mL of 0.1 M.
A quick decision guide
| What is mixed? | What to do |
|---|---|
| Same solute, no reaction | Add mmol of solute, divide by total volume |
| Different solutes, no reaction | Find mmol of each ion separately, divide each by total volume |
| Acid + base | Subtract mmol of H+ and OH−, then divide the excess by total volume |
| Solution + pure water | Use M1V1 = M2V2 (dilution) |
- Averaging the molarities. (0.5 + 0.2) / 2 = 0.35 M is wrong whenever the volumes differ. Always weight by volume.
- Forgetting that volumes add. Dividing by one of the original volumes instead of the total volume is a very common slip.
- Mixing mL and L. If one volume is in mL and the other in L, convert first.
- Ignoring the reaction. If an acid meets a base, or two solutions give a precipitate, the simple formula does not apply.
- Counting formula units instead of ions. 0.1 M CaCl2 contains 0.2 M Cl−.
- Assuming volumes always add exactly. For real, strong solutions the final volume can differ slightly. School problems tell you to take volumes as additive, and you should follow that.
Where this appears in exams
| Question type | Skill tested |
|---|---|
| "Find molarity of the mixture" | Total moles over total volume |
| "Find [Cl−] or [Na+] in the mixture" | Counting ions per formula unit |
| "pH after mixing an acid and a base" | Neutralisation first, then the excess |
| "In what ratio should two solutions be mixed?" | Setting up one equation in x |
Check your own board syllabus and your teacher's notes for the exact style of question you will face.
Do the working by hand, then use the calculator to check your molarity and amount-of-solute values.
Open the Molarity / Concentration Calculator →