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1.1.5States of Matter

Practical- Investigate the Solubility of a Solid in Water at a Specific Te — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions

5 min read 5 parts Lesson 16 of 30 Unit quiz

📚 Cambridge IGCSE Chemistry 0620 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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Section 1 of 5Method

The method turns a saturated solution into a dry solid so that the mass of dissolved solute can be measured.

1
Weigh a clean, dry evaporating basin on a balance and record its mass accurately.
2
Warm some water in a boiling tube to just above the target temperature, then add solid and stir until some solid remains undissolved.
3
Allow the solution to cool to exactly the target temperature, keeping the undissolved solid settled at the bottom so the liquid is saturated.
4
Pour a measured portion of the clear solution into the weighed basin, taking care that no undissolved solid is transferred with it.
5
Weigh the basin and solution, evaporate the water gently, and reweigh after each heating and cooling cycle until the mass stays constant.

Section 2 of 5Variables and Calculation

Independent variable
  • Independent variable The temperature of the solution, which is set by warming or cooling and held steady during each measurement.
Dependent variable
  • Dependent variable The mass of dry solid left after evaporation, which is used to calculate the solubility at that temperature.
Control variables
  • Control variables The volume of solution sampled, the type of solid and the balance used should all be kept the same for a fair comparison.
Solubility result
  • Solubility result Mass of solute divided by mass of water, multiplied by 100, gives the solubility in grams per 100 g of water.

Section 3 of 5Errors and Improvements

Errors and Improvements
Undissolved solid
  • Undissolved solid If any undissolved solid is transferred to the basin, the mass of solute is too high and the calculated solubility is overestimated.
Not reaching constant mass
  • Not reaching constant mass Weighing once after heating can leave water behind, which makes the solute mass appear larger than it really is.
Overheating
  • Overheating Strong heating can spit solid out or decompose it, which lowers the measured mass of solute and the calculated solubility.
Temperature
  • Temperature If the solution is sampled while still warmer than the target, more solute is dissolved than at the target temperature,
    • so the result is too high.

Section 4 of 5Worked Example

Question

  • A student records these masses: empty basin 21.40 g;
    • basin plus saturated solution 58.90 g;
    • basin plus dry solid at constant mass 30.10 g.
  • Calculate the solubility of the solid in g per 100 g of water.

  1. Mass of solution = 58.90 − 21.40 = 37.50 g.
  2. Mass of solute = 30.10 − 21.40 = 8.70 g.
  3. Mass of water = 37.50 − 8.70 = 28.80 g.
  4. Solubility = 8.70 ÷ 28.80 × 100 = 30.2 g per 100 g of water.

Answer 30.2 g of solid per 100 g of water at the stated temperature.

Section 5 of 5Exam Checklist & Tips

1.1.5 – Key Points for the Exam

  • Saturated solution A solution that cannot dissolve any more solute at that temperature.
  • Heating to constant mass Heating, cooling and reweighing repeatedly until two consecutive masses are the same, which shows all the water has gone.
  • Solubility The mass of solute that dissolves in 100 g of water at a stated temperature.
  • Dependent variable The quantity measured in an experiment that changes as the independent variable is changed.

Common Mistakes

  • Using the mass of the solution instead of the mass of water in the solubility calculation.
  • Transferring undissolved solid into the basin, which inflates the mass of solute and the calculated solubility.
  • Sampling the solution before it has cooled exactly to the target temperature.
  • Stopping the evaporation after a single weighing rather than repeating until the mass stays constant.

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