Dilution Formula M₁V₁ = M₂V₂ — When It Works and When It Fails
M₁V₁ = M₂V₂ is the friendliest formula in chemistry — one line, four symbols, no logarithms. That is exactly why students over-use it. Every year, answer scripts show it applied to neutralisation reactions and to mixtures of two different solutions, where it simply does not hold. This guide shows the correct use with four worked examples, and then the four situations where you must put it away and use something else.
The formula and the idea behind it
M₁ = concentration before dilution · V₁ = volume taken before dilution
M₂ = concentration after dilution · V₂ = total volume after dilution
The reason it works is simple: when you add pure solvent, the number of moles of solute does not change. Only the volume changes. Since moles n = M × V, we have n₁ = n₂, so M₁V₁ = M₂V₂.
Because the same volume unit appears on both sides, it cancels. You may work entirely in millilitres if you like — you do not have to convert to litres here, unlike in the molarity formula itself. This is one of the few places in chemistry where mL is safe.
Worked example 1 — Find the new concentration
25.0 mL of 2.00 M HCl is diluted with water to a final volume of 250 mL. Find M₂.
M₁ = 2.00 M, V₁ = 25.0 mL, V₂ = 250 mL
M₂ = M₁V₁ ÷ V₂ = (2.00 × 25.0) ÷ 250 = 50.0 ÷ 250 = 0.200 M
Sanity check: the volume went up 10 times, so the concentration should drop 10 times. 2.00 ÷ 10 = 0.200. ✔
Worked example 2 — Find the stock volume to pipette
This is the version you actually use in the lab.
How much concentrated H₂SO₄ (18.4 M) is needed to prepare 500 mL of 1.00 M acid?
M₁ = 18.4 M, M₂ = 1.00 M, V₂ = 500 mL
V₁ = M₂V₂ ÷ M₁ = (1.00 × 500) ÷ 18.4 = 500 ÷ 18.4 = 27.2 mL
Measure 27.2 mL of the concentrated acid and make up to 500 mL. Safety note that examiners do award marks for: always add acid to water, never water to acid — the dilution of sulphuric acid releases a large amount of heat.
Worked example 3 — How much water must I add?
Read this type carefully. The question asks for water added, not for the final volume.
What volume of water must be added to 100 mL of 0.500 M KCl to make it 0.200 M?
First find the final volume:
V₂ = M₁V₁ ÷ M₂ = (0.500 × 100) ÷ 0.200 = 50.0 ÷ 0.200 = 250 mL
Water to be added = V₂ − V₁ = 250 − 100 = 150 mL
Answering "250 mL" here is the classic half-mark loss.
Worked example 4 — Serial dilution
10.0 mL of 0.100 M solution is diluted to 100 mL. Then 10.0 mL of that is diluted to 100 mL again. Find the final concentration.
Step 1: M₂ = (0.100 × 10.0) ÷ 100 = 0.0100 M
Step 2: M₃ = (0.0100 × 10.0) ÷ 100 = 0.00100 M = 1.00 × 10⁻³ M
Each 10-fold dilution divides the concentration by 10, so two steps divide it by 100. This is how very dilute standards for spectrophotometry are made — weighing 0.001 g accurately is impossible, but two pipetting steps are easy.
When M₁V₁ = M₂V₂ FAILS
The formula assumes one thing only: moles of solute unchanged, solvent added. Break that assumption and the formula breaks with it.
1. When a chemical reaction happens (titration / neutralisation)
In a titration, solute is consumed. You must use the mole ratio from the balanced equation:
What volume of 0.100 M NaOH neutralises 25.0 mL of 0.100 M H₂SO₄?
Balanced: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O — so 1 mol acid needs 2 mol base.
Moles of H₂SO₄ = 0.100 × 0.0250 = 0.00250 mol
Moles of NaOH needed = 2 × 0.00250 = 0.00500 mol
V = 0.00500 ÷ 0.100 = 0.0500 L = 50.0 mL
Blindly using M₁V₁ = M₂V₂ would have given 25.0 mL — exactly half the correct answer.
2. When you mix two solutions of the same solute
Adding 0.6 M solution to 0.2 M solution is not dilution — nothing is pure solvent. Add moles, add volumes, then divide.
100 mL of 0.200 M NaCl is mixed with 300 mL of 0.600 M NaCl.
Moles = (0.200 × 0.100) + (0.600 × 0.300) = 0.0200 + 0.180 = 0.200 mol
Total volume = 0.100 + 0.300 = 0.400 L
M = 0.200 ÷ 0.400 = 0.500 M
3. When the concentration unit is molality, not molarity
Molality is per kilogram of solvent. Adding water changes the solvent mass, so m₁V₁ = m₂V₂ is meaningless. Recalculate moles per kg of solvent from scratch.
4. When volumes are not additive
For very concentrated solutions and for mixtures like ethanol and water, the final volume is measurably less than the sum of the parts, because of packing and hydrogen bonding. In real lab work this is why you always make up to the mark in a volumetric flask rather than adding a calculated volume of water. For school and entrance-exam numericals, volumes are taken as additive unless the question says otherwise.
Common mistakes that cost marks
- Giving the final volume when the question asked for water added. Subtract V₁ from V₂.
- Using it for a neutralisation. If two different substances react, you need the mole ratio from the balanced equation.
- Mixing units inside one equation. mL on one side and L on the other gives an answer wrong by 1000. Same unit both sides is the only rule.
- Confusing molarity with normality. N₁V₁ = N₂V₂ is a separate relation; for H₂SO₄, N = 2M, so the two forms give different numbers.
- Assuming the number of moles changed. On dilution it never does — if your working shows different moles before and after, you have made an error.
Where dilution appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11–12 | Solutions chapter, preparing standard solutions in the practical file |
| JEE / NEET | Multi-step dilution and mixing numericals, pH after dilution |
| IIT-JAM / CUET-PG | Volumetric analysis, back-titration setups |
| GATE / CSIR-NET | Analytical chemistry, preparing calibration standards for Beer–Lambert plots |
Get the pipette volume in one step. The Dilution calculator solves M₁V₁ = M₂V₂ for whichever of the four quantities you leave blank, so you can check both the concentration and the stock volume before you touch a burette.
Open the Dilution Calculator (M₁V₁ = M₂V₂) →Solutions, titration and stoichiometry taught step by step: ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India, with home tuition available in Delhi-NCR — see abcchemistry.in.