Topic 12 · page 17 of 17

Group 2 solubility trends

Group 2 sulfates become less soluble down the group, and hydroxides more soluble. Both trends come from two terms, lattice energy and hydration enthalpy, which both fall as the cation gets bigger. The size of the anion decides which falls faster.

12 min read12.19

Read first: Will it dissolve?

The trends from Unit 2 12.19

Two trends from Unit 2 can now be explained.

Compound type	Down the group (Mg to Ba)	Examples
Sulfates	less soluble	MgSO₄ soluble, BaSO₄ insoluble
Hydroxides	more soluble	Mg(OH)₂ sparingly soluble, Ba(OH)₂ soluble

Down Group 2 the cation radius increases. Using page 15, both the lattice energy and the cation's hydration enthalpy become less exothermic. $\Delta_\text{sol}H = -(\text{lattice energy}) + \Sigma\Delta_\text{hyd}H$ , so these two changes pull in opposite directions, and the trend depends on which one falls faster.

Which term falls faster? 12.19

The answer depends on how big the anion is.

Sulfates: hydration falls faster

$\ce{SO4^2-}$ is a large ion. The distance between the centres of the ions in the lattice is the sum of two radii, and a large anion makes up most of that sum. A bigger cation therefore changes the distance by only a small fraction, so the lattice energy changes only slightly.

The hydration enthalpy of the cation depends on how close water molecules can get to the cation alone, so it falls by a larger amount.

Hydration falls faster than lattice energy, so $\Delta_\text{sol}H$ becomes less exothermic down the group, and solubility decreases.

Hydroxides: lattice energy falls faster

$\ce{OH-}$ is a small ion. The cation radius is now a bigger fraction of the ion-ion distance, so increasing it reduces the attraction in the lattice a lot: the lattice energy falls more quickly than the hydration enthalpy of the cation.

$\Delta_\text{sol}H$ becomes more exothermic down the group, and solubility increases.

	Large anion (sulfate)	Small anion (hydroxide)
Lattice energy, Mg to Ba	falls a little	falls a lot
Cation hydration, Mg to Ba	falls more than the lattice energy	falls less than the lattice energy
$\Delta_\text{sol}H$ down the group	less exothermic	more exothermic
Solubility down the group	decreases	increases

Entropy contributes too. A small, highly charged ion such as Mg²⁺ orders the water around it strongly, which makes $\Delta S_\text{system}$ more negative. So the simple enthalpy argument is only part of the picture, though it is the part examined at this level.

PracticalBarium chloride test for sulfate ions

The low solubility of BaSO₄ is the basis of the Unit 2 test for sulfate ions. Add dilute hydrochloric acid, then barium chloride solution. A white precipitate shows sulfate ions. The acid is added first to remove carbonate ions, which would otherwise also give a white precipitate of barium carbonate.

\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}

Check yourself

State the trend in solubility of the Group 2 sulfates and of the Group 2 hydroxides from Mg to Ba.
Why does the lattice energy of a Group 2 sulfate change only slightly from MgSO₄ to BaSO₄?
Which term changes more for the sulfates, the lattice energy or the hydration enthalpy of the cation? What is the effect on $\Delta_\text{sol}H$ ?
Repeat question 3 for the hydroxides.
A student writes: "Ba(OH)₂ is more soluble than Mg(OH)₂ because its lattice energy is less exothermic." Explain what is missing from the answer.
Write the ionic equation for the barium chloride test for sulfate ions and say why acid is added first.

Answers cover these until you have tried the questions

Sulfates: solubility decreases (MgSO₄ soluble, BaSO₄ insoluble). Hydroxides: solubility increases (Mg(OH)₂ sparingly soluble, Ba(OH)₂ soluble).
The sulfate ion is large, so it makes up most of the distance between the centres of the ions. A larger cation only increases that distance by a small fraction, so the attraction in the lattice is only slightly weaker.
The hydration enthalpy of the cation falls more. So $\Delta_\text{sol}H$ becomes less exothermic (more endothermic) down the group, and solubility falls.
The lattice energy falls more, because the hydroxide ion is small and the larger cation increases the ion-ion distance by a large fraction. So $\Delta_\text{sol}H$ becomes more exothermic down the group, and solubility rises.
The answer mentions only one term. The hydration enthalpy of the cation also becomes less exothermic. The trend follows because the lattice energy falls faster than the hydration enthalpy (the hydroxide ion is small), so $\Delta_\text{sol}H$ becomes more exothermic.
$\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}$ , a white precipitate. Dilute hydrochloric acid is added first to react with carbonate ions (and so remove them), as they would also form a white precipitate with barium ions.