Practical: Investigate the Solubility of a Solid in Water at a Specific Temperature (Edexcel IGCSE Chemistry): Revision Note

Exam code: 4CH1

Stewart Hird

Written by: Stewart Hird

Reviewed by: Lucy Kirkham

Updated on

Practical: Investigate the solubility of a solid in water at a specific temperature

Aim

  • To find the solubility of a solid in water at a given temperature by preparing a saturated solution, evaporating the solvent, and measuring the mass of the solid obtained

Method

  1. Weigh an empty evaporating basin using a balance and record its mas

  2. Warm a boiling tube of water to just above 30°C using a water bath or beaker of hot water

  3. Add copper(II) sulfate crystals to the water in the boiling tube and stir thoroughly until no more will dissolve and some undissolved solid remains at the bottom.
    This ensures that the solution is saturated

  4. Allow the saturated solution to cool to exactly 30°C

  5. Pour a portion of the clear solution (without any solid) into the pre-weighed evaporating basin

    • It is important that no undissolved copper(II) sulfate is transferred

  6. Weigh the evaporating basin and the solution together and record the total mass

  7. Gently heat the evaporating basin to evaporate all the water, taking care to avoid spitting or loss of solid

  8. When the water appears to have evaporated completely, allow the basin to cool slightly and then weigh it again

  9. Reheat, cool, and reweigh the basin once more

  10. Repeat this process until the mass remains constant

    • This is called heating to constant mass and confirms that all water has been removed

  11. Calculate the solubility of copper(II) sulfate in water at 30°C using the masses recorded

Results

  • Mass of empty evaporating basin, mb = 25.00 g

  • Mass of basin + saturated solution, mb+ soln = 74.50 g

  • Mass of basin + dry copper(II) sulfate (after heating to constant mass), mb + solid = 37.00 g

Calculation

Step 1: Find the masses you need

  • Mass of solution

m(soln) = mb + soln − mb = 74.50 − 25.00 = 49.50 g

  • Mass of dissolved solute (CuSO4.5H2O)

m(solute) = mb + solid − mb = 37.00 − 25.00 = 12.00 g

  • Mass of water

m(water) = m(soln) − m(solute) = 49.50 − 12.00 = 37.50 g

Step 2: Calculate solubility (g solute per 100 g water)

Solubility at 30 °C = fraction numerator straight m open parentheses solute close parentheses over denominator straight m open parentheses water close parentheses end fraction x 100

fraction numerator 12.00 over denominator 37.50 end fraction x 100 = 32.0 g per 100 g 

Step 3: Express as mass percentage of the solution

% solute in solution = fraction numerator 12.00 over denominator 49.50 end fraction x 100 = 24.2% 

Examiner Tips and Tricks

“Heating to constant mass” means:

  1. Reheating

  2. Cooling

  3. Reweighing until two consecutive masses are the same

This confirms all water has been removed.

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Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Content Creator

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.

Lucy Kirkham

Reviewer: Lucy Kirkham

Expertise: Head of Content Creation

Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.