Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 12.19 Solubility using ΔH and ΔS-Study Notes - New Syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 12.19 Solubility using ΔH and ΔS- Study Notes- New syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 12.19 Solubility using ΔH and ΔS- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.

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Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics

12.19 Solubility of Ionic Compounds using Enthalpy and Entropy

The solubility of an ionic compound depends on the feasibility of the dissolving process, which is determined by the total entropy change. Both enthalpy change of solution and entropy change must be considered together.

Key Condition for Solubility

A compound is likely to be soluble if \( \mathrm{\Delta S_{total} > 0} \).

Since:

\( \mathrm{\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings}} \)

and:

\( \mathrm{\Delta S_{surroundings} = -\frac{\Delta H_{sol}}{T}} \)

Role of Enthalpy Change of Solution (\( \mathrm{\Delta_{sol}H} \))

Exothermic (\( \mathrm{\Delta H_{sol} < 0} \)):

    • \( \mathrm{\Delta S_{surroundings} > 0} \)
    • Favourable for solubility.

Endothermic (\( \mathrm{\Delta H_{sol} > 0} \)):

    • \( \mathrm{\Delta S_{surroundings} < 0} \)
    • Requires a sufficiently positive \( \mathrm{\Delta S_{system}} \) for dissolution.

Role of Entropy Change of the System (\( \mathrm{\Delta S_{system}} \))

  • Dissolving usually increases disorder as ions disperse in solution.
  • However, hydration may order water molecules around ions.
  • Therefore, \( \mathrm{\Delta S_{system}} \) can be positive or negative.

Overall Prediction of Solubility

  • If both \( \mathrm{\Delta H_{sol} < 0} \) and \( \mathrm{\Delta S_{system} > 0} \) → highly soluble.
  • If \( \mathrm{\Delta H_{sol} > 0} \) but \( \mathrm{\Delta S_{system}} \) is large positive → soluble at higher temperatures.
  • If both are unfavourable → low solubility.

Trends in Solubility (Unit 2 Context)

1. Effect of Ionic Charge

  • Higher charges → very high lattice energy.
  • Hydration enthalpy also increases but often not enough to compensate.
  • Therefore, compounds like \( \mathrm{MgO} \) are less soluble.

2. Effect of Ionic Radius

  • Smaller ions → stronger lattice energy and hydration enthalpy.
  • The balance determines solubility.
  • Large ions → weaker lattice → often more soluble.

3. Balance between Lattice Energy and Hydration Enthalpy

  • If hydration enthalpy ≈ lattice energy → moderate solubility.
  • If hydration enthalpy > lattice energy → exothermic dissolution → high solubility.
  • If lattice energy > hydration enthalpy → endothermic dissolution → lower solubility.

Temperature Dependence

  • Endothermic dissolution → solubility increases with temperature.
  • Exothermic dissolution → solubility may decrease with temperature.

Key Features

  • Solubility depends on total entropy change.
  • Both enthalpy and entropy must be considered together.
  • Balance between lattice energy and hydration enthalpy is crucial.
  • Temperature can change feasibility.

Example 1:

Explain why \( \mathrm{NH_4NO_3} \) dissolves endothermically but is still highly soluble in water.

▶️ Answer/Explanation

Dissolving \( \mathrm{NH_4NO_3} \) is endothermic, so \( \mathrm{\Delta H_{sol} > 0} \) and \( \mathrm{\Delta S_{surroundings}} \) is negative.

However, the ions formed are highly dispersed in solution, giving a large positive \( \mathrm{\Delta S_{system}} \).

This outweighs the negative contribution from the surroundings.

Therefore, \( \mathrm{\Delta S_{total} > 0} \), making the compound soluble.

Example 2:

Explain why \( \mathrm{MgO} \) is sparingly soluble in water.

▶️ Answer/Explanation

\( \mathrm{Mg^{2+}} \) and \( \mathrm{O^{2-}} \) have high charges, leading to very strong lattice energy.

Although hydration enthalpy is also large, it is not sufficient to overcome the lattice energy.

Therefore, \( \mathrm{\Delta H_{sol}} \) is positive and large.

The entropy increase is not enough to make \( \mathrm{\Delta S_{total}} \) positive.

Hence, \( \mathrm{MgO} \) is only sparingly soluble.

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