Core Practical 1.7C

Core Practical 1.7C: Solubility of a Solid in Water

Revision guide containing method, variables, safety, sample calculations, and model exam answers.

Edexcel IGCSE Hub Core Practicals CP 1.7C

Scientific Principles & Theory

Scientific Background: The solubility of a solute is the maximum mass of solid that can dissolve in 100 g of solvent (usually water) at a specific temperature to form a saturated solution. A saturated solution is one that contains the maximum dissolved solute at that temperature, and any further solute added will remain undissolved at the bottom.

The solubility of most solids increases with temperature. This is because higher temperatures increase the kinetic energy of water molecules, allowing them to break the electrostatic lattice forces holding the solute crystal together more effectively.

Solubility (g/100 g H₂O) = (Mass of solute dissolved / Mass of water evaporated) * 100

Experimental Variables

Independent Variable

Temperature of the water (measured in °C).

Dependent Variable

Solubility of the solid (calculated in g per 100 g of water).

Control Variables

Type of solute (potassium chlorate), volume of water (10 cm³), rate of stirring.

⚠️ Lab Risk Assessment

Hazard Associated Risk Control Measure
Bunsen burner / hot apparatus Thermal burns or scalds Use a water bath instead of heating the tube directly; handle hot tubes with a holder; wear heatproof gloves.
Potassium chlorate(V) solid Oxidiser and harmful if swallowed Wear safety goggles, avoid inhaling dust, wash hands immediately after handling.

Apparatus & Procedure

Required Apparatus

  • Boiling tube
  • Thermometer (0 to 100 °C)
  • 10 cm³ graduated pipette
  • Evaporating basin
  • Bunsen burner, tripod, gauze, and heatproof mat
  • Water bath (beaker of water on a tripod)
  • Electronic balance (0.01 g resolution)
  • Stirring rod
  • Potassium chlorate(V) solid
  • Distilled water

Step-by-Step Procedure

  1. Weigh an empty, clean, dry evaporating basin and record its mass.
  2. Use a graduated pipette to add exactly 10.0 cm³ of distilled water into a boiling tube.
  3. Add a known mass of potassium chlorate(V) solid to the boiling tube. Place the tube in a hot water bath and stir until all the solid dissolves.
  4. Remove the boiling tube from the water bath and allow it to cool slowly while stirring with a thermometer.
  5. Record the temperature at which the first crystals begin to appear. This is the temperature of saturation.
  6. Quickly pour (decant) only the clear liquid solution into the pre-weighed evaporating basin, leaving any precipitated crystals at the bottom of the tube.
  7. Weigh the evaporating basin containing the saturated solution.
  8. Heat the basin gently over a Bunsen burner. Evaporate all the water until only dry solid remains. Heat slowly near the end to prevent 'spitting' of solid.
  9. Allow the evaporating basin to cool, then weigh it.
  10. Repeat the heating, cooling, and weighing cycle until a constant mass is achieved to ensure complete dryness.
Evaporating Solution to Dryness Evaporating basin Wire gauze Bunsen burner

Fig 1. Laboratory experimental setup for Core Practical 1.7C.

Sample Data & Calculations

This representative dataset illustrates the values typically obtained when carrying out this experiment in the laboratory:

Measurement Parameter Value Obtained (g)
Mass of empty evaporating basin 45.50
Mass of evaporating basin + saturated solution 62.70
Mass of evaporating basin + dry potassium chlorate(V) 49.80
Temperature of saturation (°C) 40.0

Data Processing & Analysis

  1. Mass of dry solute = (Mass of basin + dry solid) - Mass of empty basin = 49.80 g - 45.50 g = 4.30 g
  2. Mass of water evaporated = (Mass of basin + solution) - (Mass of basin + dry solid) = 62.70 g - 49.80 g = 12.90 g
  3. Solubility = (Mass of solute / Mass of water) * 100 = (4.30 g / 12.90 g) * 100 = 33.3 g / 100 g H₂O at 40 °C

Conclusion & Evaluation

Chemical Explanation: Saturated solutions are heavily dependent on temperature. Heating shifts solubility limits, allowing more solute to form coordinate bonds or ion-dipole interactions with solvent molecules. When cooled, the reverse process happens and solute precipitates out.

Experimental Error Analysis

Error Type & Source Effect on Final Result Mitigation Strategy
Systematic Error
Incomplete evaporation of water
The mass of the dry solid appears artificially high, leading to an overestimation of the solubility. Heat the basin to a constant mass by heating, cooling, and re-weighing until the mass value no longer decreases.
Random Error
Loss of solid by spitting during heating
Some solute particles are lost, which decreases the mass of the dry solid and underestimates solubility. Heat the evaporating basin very gently, especially when the solid begins to go dry. Use a watch glass cover.

Exam Practice

Exam-Style Design Question (6 Marks)

Plan an experiment to determine the solubility of potassium chlorate(V) in water at 50 °C. Your description should include the apparatus needed, a step-by-step method, the measurements you would take, and how you would calculate the solubility.

View Model Answer & Mark Scheme

Model Answer (6/6 Marks):

  1. Apparatus and Setup: Use a boiling tube, thermometer, water bath, digital balance (0.01 g), evaporating basin, and Bunsen burner equipment.
  2. Preparation: Measure 10 cm³ of distilled water into the boiling tube. Add potassium chlorate(V) solid until no more dissolves and solid remains at the bottom, ensuring a saturated solution.
  3. Temperature Control: Heat the mixture in a water bath to slightly above 50 °C to dissolve all solid, then let it cool. When the temperature drops to exactly 50 °C and crystals just start to appear, immediately decant the liquid into a pre-weighed evaporating basin.
  4. Measurements:
    • Record the mass of the empty evaporating basin (m₁).
    • Record the mass of the basin + decanted saturated solution (m₂).
    • Record the mass of the basin + dry solid after heating to constant mass (m₃).
  5. Calculations:
    • Calculate the mass of dissolved solid: mass = m₃ - m₁.
    • Calculate the mass of evaporated water: mass = m₂ - m₃.
    • Calculate solubility: solubility = [(m₃ - m₁) / (m₂ - m₃)] * 100 in units of g per 100 g of water.
Examiner Tip:

Make sure you explicitly state the formula used for calculations and explain how you ensure the solution is saturated (by having excess solid present and noting the temperature crystals start to form).