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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

📚 Edexcel Linear - OL (Chemistry) Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 6Key Ideas

Solubility depends on the substance and on temperature, so the experiment must fix the temperature.

Solute and solvent
  • Solute and solvent The solute is the solid that dissolves;
    • the solvent is the liquid, here water, which does the dissolving.
Saturated solution
  • Saturated solution A saturated solution contains the maximum mass of solute that dissolves at that temperature, with some undissolved solid remaining.
Solubility
  • Solubility Solubility is the mass of solute that dissolves in 100 g of water to make a saturated solution at a stated temperature.
Temperature effect
  • Temperature effect For most solids, solubility increases as temperature rises, so the temperature must be controlled in the experiment.

Part 2 of 6Method

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.

Part 3 of 6Processing the Results

Processing the Results
Mass of solution
  • Mass of solution Subtract the mass of the empty basin from the mass of basin plus solution.
Mass of solute
  • Mass of solute Subtract the mass of the empty basin from the mass of basin plus dry solid.
Mass of water
  • Mass of water Subtract the mass of solute from the mass of solution to find the water that was evaporated.
Solubility
  • Solubility Divide mass of solute by mass of water and multiply by 100 to give g per 100 g of water.
QuantityHow it is found
m(solution)(basin + solution) − basin
m(solute)(basin + dry solid) − basin
m(water)m(solution) − m(solute)

Part 4 of 6Sources of Error and Safety

Sources of Error and Safety
Undissolved solid
  • Undissolved solid Transferring any undissolved crystals makes the mass of solute too high, so only clear liquid should be poured.
Spitting
  • Spitting If liquid spits out during heating, solute is lost and the calculated solubility is too low.
Incomplete drying
  • Incomplete drying Failing to reach constant mass leaves water in the basin, which makes the solute mass too high.
Temperature
  • Temperature If the solution is not at the stated temperature when sampled, the value belongs to a different temperature.

Part 5 of 6Worked 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.

  1. Mass of solution = 74.80 − 24.80 = 50.00 g.
  2. Mass of solute = 38.30 − 24.80 = 13.50 g.
  3. Mass of water = 50.00 − 13.50 = 36.50 g.
  4. 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

Part 6 of 6Exam 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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