Explain the decreasing solubility of Group 2 sulfates
Use the different size of sulfate and hydroxide ions to explain the sulfate solubility and solution-enthalpy trend.
Section 27 · Group 2
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 27. Compare Group 2 ions at the same charge and distinguish lattice formation, lattice dissociation, hydration and solution enthalpy. Trend explanations are qualitative unless data are supplied. Written practice is self-checked
Before you begin
- Use AS Group 2 reactivity trends, ionic charge and radius, Hess cycles, and balanced equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How does Group 2 sulfate solubility change down the group?
What you will learn
- State the Group 2 sulfate solubility trend
- Compare hydration and lattice changes for sulfate salts
- Relate solution enthalpy data to solubility without treating them as identical
Sulfate solubility decreases down the group
Group 2 sulfate solubility decreases from magnesium to barium. BaSO₄ is very sparingly soluble compared with MgSO₄. Use the qualitative trend, and use any supplied values when a question asks for a quantitative comparison.
This direction is opposite to the hydroxides. The anion differs: SO₄²⁻ is much larger than OH⁻, which changes how the lattice and hydration contributions vary across the series.
Compare the relative changes, not one energy in isolation
Down the group, hydration of M²⁺ becomes less exothermic and the lattice becomes easier to separate. These changes oppose one another in the solution-enthalpy cycle. For sulfates, the sulfate ion is large, so lattice-dissociation enthalpy changes less than the cation hydration contribution. The signed sum, ΔHsol, becomes less negative or more positive (less exothermic or more endothermic) down the group, consistent with decreasing solubility.
If values are provided, calculate or compare ΔHsol using the stated convention and the formula-unit coefficients. ΔHsol is not the same quantity as solubility: equilibrium also depends on entropy and temperature. Avoid claiming that a single enthalpy term alone determines the exact solubility.
Pause and explain
Why is hydration of Ba²⁺ less exothermic than hydration of Mg²⁺?
Worked example
Explain why BaSO₄ is less soluble than MgSO₄.
Show the worked solution
- Sulfate solubility decreases down Group 2.
- Hydration of the larger Ba²⁺ ion is less exothermic.
- Compare this loss of hydration stabilisation with the lattice change and state the net trend.
Answer BaSO₄ is less soluble. Down the group, Ba²⁺ is larger and its hydration is less exothermic. Although the lattice is also easier to separate, for sulfates the overall balance makes dissolution less favourable; the sulfate solubility therefore decreases from MgSO₄ to BaSO₄.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Order MgSO₄, SrSO₄ and BaSO₄ by decreasing solubility.
Give me a hint
Solubility decreases as you move down Group 2.
Compare my reasoning
- Mg is highest in this list.
- Sr lies below Mg and above Ba.
- Write the most soluble first.
MgSO₄ > SrSO₄ > BaSO₄ in solubility.
Look for these in your work
- Group positions correct
- Trend applied
- Order direction clear
Both lattice separation and cation hydration change down Group 2. Why must a sulfate trend explanation compare their relative magnitudes?
Give me a hint
In a Hess cycle, one contribution costs energy and the other releases it.
Compare my reasoning
- Lattice dissociation is endothermic.
- Hydration is exothermic and becomes less exothermic for a larger cation.
- The net solution enthalpy depends on the signed sum, not either term alone.
The lattice-separation cost becomes smaller down the group, while cation hydration releases less energy. Since these contributions change in opposite directions in the signed sum, their relative magnitudes determine how ΔHsol changes.
Look for these in your work
- Signs distinguished
- Opposing changes identified
- Net-sum reasoning used
A student claims ‘BaSO₄ is less soluble because its ΔHsol is positive, so it cannot dissolve.’ Improve the statement.
Give me a hint
Separate a less favourable enthalpy contribution from the equilibrium and solubility conclusion.
Compare my reasoning
- A positive ΔHsol is endothermic, not impossible.
- Entropy and temperature also influence dissolution.
- State the observed down-group sulfate trend without claiming zero solubility.
A positive ΔHsol would mean the dissolution enthalpy is endothermic, but it does not mean dissolution is impossible. Entropy and temperature also affect equilibrium. BaSO₄ is much less soluble than MgSO₄, not absolutely incapable of dissolving.
Look for these in your work
- Sign interpreted
- Thermodynamic limitation included
- Very low solubility distinguished from zero
Common mistakes
- Using the hydroxide trend for sulfates
- Saying a larger cation is more strongly hydrated
- Ignoring the large sulfate anion and the relative energy changes
- Equating ΔHsol directly with a numerical solubility
What can you explain now?
How does Group 2 sulfate solubility change down the group?
Compare with the explanation
It decreases
BaSO₄ is much less soluble than MgSO₄.
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Make it stick
What changes in the enthalpy cycle compete as Group 2 cations become larger, and how does the sulfate trend reflect their balance?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.