4CH1

States of Matter

Principles of Chemistry

Exam Frequency Analysis

Past paper frequency (2018 to 2024)

This topic accounts for approximately 6% of your exam marks.

stable
Low
Stable6%

Appears regularly as short-answer questions on particle diagrams and state changes.

  • Solubility is how much solute a particular solvent will hold at a given temperature (and, for gases, at a given pressure)
  • Standard unit: grams of solute per 100 g of solvent (usually water)
  • e.g. KNO₃ in water at 20 °C: 32 g per 100 g of water means 32 g dissolves; anything more sinks to the bottom

Effect of temperature and pressure

SoluteEffect of raising temperatureEffect of raising pressure
SolidUsually more solubleNegligible
GasLess solubleMore soluble
  • Warm fizzy drinks lose their fizz faster than cold ones (gas leaves on warming)
  • Carbonation is achieved by forcing CO₂ into solution under pressure

Solubility curves

  • A solubility curve plots solubility (g per 100 g of water) against temperature (°C)
  • Built by measuring the saturation mass at each of several temperatures and joining the points
  • Most ionic salts: curve rises with temperature
  • Shape varies sharply between salts:
    • KNO₃: steep rise
    • Pb(NO₃)₂: moderate rise
    • NaCl: almost flat across 0–100 °C
Solubility curves plotting solubility in grams per 100 g of water against temperature for potassium nitrate (steep rise), lead(II) nitrate (moderate rise) and sodium chloride (almost flat)
Source: Solubility by Save My Exams
Worked example

How much solute crystallises when a saturated solution is cooled

A saturated solution holds 110 g of potassium nitrate per 100 g of water at 60 °C. It is cooled to 20 °C, where the solubility of potassium nitrate is 32 g per 100 g of water. Find the mass of crystals that form (per 100 g of water).

Solution:

  • At 60 °C, 100 g of water holds 110 g of dissolved KNO₃.
  • At 20 °C, 100 g of water can hold only 32 g.
  • The solute that can no longer stay dissolved comes out as crystals.
  • Mass of crystals = solubility at 60 °C − solubility at 20 °C = 110 − 32 = 78 g.