When you dilute a solution the moles of solute stay the same, so M1V1 = M2V2. Rearrange to find the missing quantity, keeping the volume units the same on both sides.
Water added equals the final volume minus the starting volume.
Dilution is a favourite in the Acids, Bases and Salts chapter of SPM Chemistry, and it rests on one clean idea: adding water changes the volume, not the amount of solute. Our online one-to-one teachers make sure you use M1V1 = M2V2 confidently and never confuse the final volume with the volume of water added.
When you use it
Use dilution whenever a stronger stock solution is watered down to a weaker one: making a dilute acid from a concentrated one, or preparing a working solution from a stock. It also appears as the “how much water must I add” question, which needs one extra subtraction.
The formula
M1V1 = M2V2
- M1 = starting (initial) molarity, in mol dm−3
- V1 = starting volume
- M2 = final molarity after dilution, in mol dm−3
- V2 = final volume
The volume units simply need to match on both sides (cm3 with cm3, or dm3 with dm3); they do not have to be dm cubed here, because the volumes appear on both sides and any common unit cancels.
Units
Molarity is in mol dm−3. The two volumes can be in cm3 or dm3 as long as they are the same unit. The key idea is that the moles of solute, M x V, are unchanged by dilution.
Why the moles stay the same
When you add water to a solution, you add no extra solute, every particle that was there before is still there afterwards, just spread through a larger volume. That is why the number of moles before dilution, M1 x V1, must equal the number of moles after, M2 x V2. Understanding the formula this way, rather than memorising the letters, means you can rebuild it under pressure and you immediately know that a bigger volume must give a smaller concentration. If your diluted answer comes out more concentrated than the stock, you know at once that something has gone wrong.
Worked example 1 (easy), diluting an acid
20 cm3 of 2 mol dm−3 hydrochloric acid is diluted to 100 cm3. Find the new molarity.
- Step 1, M2 = M1V1 / V2 = (2 x 20) / 100.
- Step 2, M2 = 40 / 100 = 0.4 mol dm−3.
Worked example 2 (medium), volume of stock needed
What volume of 1 mol dm−3 sodium hydroxide is needed to make 250 cm3 of 0.2 mol dm−3 solution?
- Step 1, V1 = M2V2 / M1 = (0.2 x 250) / 1.
- Step 2, V1 = 50 cm3. You measure 50 cm3 of the stock and make it up to 250 cm3 with distilled water.
Worked example 3 (SPM level), how much water to add
How much water must be added to 50 cm3 of 0.5 mol dm−3 sulfuric acid to make it 0.1 mol dm−3?
- Step 1, final volume: V2 = M1V1 / M2 = (0.5 x 50) / 0.1 = 250 cm3.
- Step 2, water added = final volume − starting volume = 250 − 50 = 200 cm3.
The trap is stopping at 250 cm3; the question asks for the water added, which is 200 cm3.
Common traps
- Answering with the final volume when the question asks for the water added (subtract the starting volume).
- Mixing volume units, keep both sides in the same unit.
- Thinking dilution changes the moles of solute; it does not, only the concentration and volume change.
- Rearranging M1V1 = M2V2 incorrectly; isolate the unknown carefully.
- Dropping the units on the final answer.
How we help
Our teachers drill all three shapes, new concentration, volume of stock, and water to add, so the “how much water” twist never catches you out. Lessons are one-to-one, taught in English, from RM50 per hour, with a paid one-hour trial so you can see the method before committing. Dilution connects straight into standard solutions and titration, so mastering M1V1 = M2V2 now pays off across the whole Acids, Bases and Salts topic.
Want a teacher to make this click?
We teach SPM Chemistry one to one, so your child understands it and scores it.
from RM50/hr · One-hour paid trial · Same-day reply