1.1.5Principles of Chemistry
Practical: Investigate the Solubility of a Solid in Water at a Specific Temperature — Edexcel Linear - OL (Chemistry) Revision Notes & Topical Questions
5 min read 5 parts Lesson 19 of 25 Unit quiz
Section 1 of 5Method
Each step is designed to measure the mass of solute and the mass of water separately.
1
Weigh an empty evaporating basin and record its mass.
2
Warm water in a boiling tube slightly above 30 °C and add potassium chloride in small portions, stirring, until some solid will not dissolve.
3
Let the solution cool to exactly 30 °C so it is saturated at that temperature.
4
Pour some of the clear solution into the basin without transferring any undissolved solid, then weigh the basin plus solution.
5
Heat gently to evaporate the water, avoiding spitting, then cool and weigh the basin.
6
Reheat, cool and reweigh until two consecutive masses match; this is heating to constant mass.
Section 2 of 5Processing the Results
Processing the Results
Mass of solution
- Subtract the mass of the empty basin from the mass of basin plus solution.
Mass of solute
- Subtract the mass of the empty basin from the mass of basin plus dry solid.
Mass of water
- Subtract the mass of solute from the mass of solution to find the water that was evaporated.
Solubility
- Divide mass of solute by mass of water and multiply by 100 to give g per 100 g of water.
| Quantity | How it is found |
|---|---|
| m(solution) | (basin + solution) − basin |
| m(solute) | (basin + dry solid) − basin |
| m(water) | m(solution) − m(solute) |
Section 3 of 5Sources of Error and Safety
Sources of Error and Safety
Undissolved solid
- Transferring any undissolved crystals makes the mass of solute too high, so only clear liquid should be poured.
Spitting
- If liquid spits out during heating, solute is lost and the calculated solubility is too low.
Incomplete drying
- Failing to reach constant mass leaves water in the basin, which makes the solute mass too high.
Temperature
- If the solution is not at the stated temperature when sampled, the value belongs to a different temperature.
Section 4 of 5Worked Example
Question
- A basin has mass 24.80 g.
- With saturated potassium chloride solution it has mass 74.80 g, and after heating to constant mass it has mass 38.30 g.
- Calculate the solubility in g per 100 g of water.
- Mass of solution = 74.80 − 24.80 = 50.00 g.
- Mass of solute = 38.30 − 24.80 = 13.50 g.
- Mass of water = 50.00 − 13.50 = 36.50 g.
- Solubility = 13.50 ÷ 36.50 × 100 = 37.0 g per 100 g of water.
Answer Solubility = 37.0 g per 100 g of water at 30 °C
Section 5 of 5Exam Checklist & Tips
1.1.5 – Key Points for the Exam
- Solubility Mass of solute that dissolves in 100 g of water to give a saturated solution at a stated temperature.
- Saturated solution A solution in which no more solute can dissolve at that temperature.
- Constant mass Two consecutive weighings that are equal, showing all the water has gone.
- Evaporation Change from liquid to gas at the surface, used here to remove the water.
Common Mistakes
- Dividing the solute mass by the mass of solution instead of the mass of water for solubility.
- Stopping after one weighing instead of heating to constant mass.
- Forgetting that solubility values only make sense with a temperature quoted.
- Using a hotter solution than stated, so the sample is not saturated at the required temperature.
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