A dissolved substance has separated into particles so small that they mix evenly with the solvent and cannot be filtered out. A suspension holds larger insoluble particles that eventually settle and can be filtered. IGCSE Chemistry questions ask you to decide which is which from evidence, and then to justify it.
This is the next skill after explaining a separation step in water treatment. It feeds straight into reading data in interpreting pollutant evidence.
How do you tell the two apart from evidence?
Use three observations, and remember what each one shows.
| Observation | Suspension | Solution |
|---|---|---|
| Appearance | Cloudy or opaque | Clear (may be coloured) |
| Left to stand | Solid settles | Nothing settles |
| Filtered | Solid stays in the filter paper | Nothing is left in the paper |
| Filtrate evaporated | Nothing left (if only suspended solid) | Solid left behind if a solute was present |
The key idea is particle size. In a solution the solute particles are individual particles spread among the solvent particles, so they pass through filter paper. In a suspension the particles are clumps far larger than that, so the paper holds them back.
Worked example
Question (invented data): Mixture X is cloudy and white. After it stands for an hour, a white solid settles. The mixture is filtered. The residue is white. The filtrate is clear. When a little of the filtrate is heated until the water has gone, a white solid remains. What does the evidence show?
Step 1, cloudy and settles. This suggests an insoluble solid is suspended. The residue in the filter paper confirms it.
Step 2, clear filtrate. Clear does not prove nothing is dissolved. So we look at the next test.
Step 3, white solid after evaporation. The water has gone, so the solid must have been dissolved in the filtrate.
Conclusion: Mixture X contains both an insoluble solid, suspended in the water, and a dissolved solid.
A solubility calculation often sits alongside this. Invented data: solid Z dissolves to a maximum of 20 g in 100 g of water at 25 °C. A student adds 50 g of Z to 200 g of water at 25 °C and stirs well.
- The water can hold 20 × 2 = 40 g of Z, because 200 g is twice 100 g.
- So 40 g dissolve and 50 − 40 = 10 g stays undissolved.
- The undissolved 10 g forms the solid that could be filtered off.
The mistake to watch for
Mistaken answer: “The liquid is clear, so there are no particles in it.”
The fault is treating “clear” as “empty”. Everything is made of particles, and dissolved solutes are particles spread through the solvent. A coloured solution such as copper sulfate is clear, yet it holds a dissolved solid.
The correction is to use the evaporation test. If a solid remains after the water has gone, it was dissolved. If nothing remains, the liquid was pure water (or contained only a volatile solute).
Keep the word “insoluble” for particles that will not dissolve and the word “suspended” for how they are spread in the liquid. They describe the same solid in two different ways.
Check yourself
1. Sample A is clear and colourless. After filtering nothing is left in the paper. Evaporating a few drops leaves a white solid. Is A a solution or a suspension?
Show answer
It is a solution. No insoluble particles were filtered out, but a solid remained after evaporation, so a solute was dissolved in the water.
2. Invented data: solid Q dissolves to a maximum of 30 g in 100 g of water at 20 °C. How much of 75 g of Q dissolves in 200 g of water at 20 °C, and how much is left?
Show answer
200 g of water can hold 30 × 2 = 60 g. So 60 g dissolves and 75 − 60 = 15 g remains undissolved.
3. Why does a suspension settle when left to stand but a solution does not?
Show answer
Suspended particles are large insoluble clumps that are denser than water, so they sink. Dissolved particles are individual particles spread evenly among the water particles, so they do not settle.
Where this leads next
Go on to interpreting pollutant evidence from a supplied dataset, where the same kind of reasoning is applied to numbers. The mole and equation-ratio tutor is useful when a solubility or concentration question also involves a reaction. Test the lot with the module practice set.
If you keep mixing up what an observation proves, it helps to have someone ask you “what does that show about the particles?” after every answer. That is how our teachers work in online one-to-one Chemistry tuition.