Solubility Equilibrium Revision Notes for HSC SSCE Chemistry

Solubility Equilibrium

Introduction to solubility equilibrium

When ionic compounds dissolve in water, they don't always dissolve completely. You've learned that different ionic compounds have varying levels of solubility, and we describe them as soluble, sparingly soluble, or insoluble. The term insoluble is actually a relative term - it refers to compounds with a solubility of less than $1 \text{ g L}^{-1}$ .

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The term "insoluble" doesn't mean absolutely no dissolving occurs. It's a relative term used for compounds with very low solubility - specifically less than $1 \text{ g L}^{-1}$ . Even these compounds dissolve to some extent!

Here's an important fact: every ionic compound dissolves to some extent in water, even if only very slightly. Even compounds we call "insoluble" do dissolve a tiny amount before reaching saturation. When saturation occurs, something interesting happens - a dynamic equilibrium is established between the solid salt and its dissolved ions.

What is dynamic equilibrium?

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Dynamic equilibrium occurs when ions break away from the solid ionic lattice at exactly the same rate as ions rejoin the lattice. Even though both processes continue happening, there's no net change in the amount of solid or dissolved ions.

We can represent this equilibrium with a general equation:

$\text{AB}_2(s) \rightleftharpoons \text{A}^{2+}(aq) + 2\text{B}^-(aq)$

For example, when barium sulfate reaches equilibrium in water:

$\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$

Notice how the equation is always written with the solid ionic compound on the left side of the equilibrium arrows and the dissolved ions on the right side.

Understanding the solubility product ( $K_{sp}$ )

For these dissolution equilibria, we use a special equilibrium constant known as the solubility product, represented by the symbol $K_{sp}$ .

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The symbol $K_{sp}$ can only be used when the system is at equilibrium. It specifically applies to sparingly soluble or insoluble salts only - not to highly soluble compounds.

Because this is a heterogeneous equilibrium (involving both solid and aqueous phases), we only include the dissolved ions in the equilibrium expression - the solid is not included.

For barium sulfate, the equilibrium expression is:

$K_{sp} = [\text{Ba}^{2+}] \times [\text{SO}_4^{2-}]$

For the general equation shown earlier, the equilibrium expression is:

$K_{sp} = [\text{A}^{2+}] \times [\text{B}^-]^2$

Notice how each ion concentration is raised to the power of its coefficient in the balanced chemical equation. This is a key rule when writing $K_{sp}$ expressions.

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Exam tip: Equilibrium expressions are only written for sparingly soluble or insoluble salts. You won't write $K_{sp}$ expressions for highly soluble compounds.

How to write equilibrium expressions

Follow these steps when writing solubility equilibrium expressions:

Write the balanced chemical equation showing the solid compound dissociating into its aqueous ions
Position correctly: solid on the left, ions on the right
Include only aqueous species in the $K_{sp}$ expression (never include the solid)
Raise each ion concentration to the power of its stoichiometric coefficient from the balanced equation

Worked examples

Let's look at how to write equilibrium expressions for two different insoluble compounds.

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Worked Example: Calcium hydroxide

Answer	Logic
$\text{Ca(OH)}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq)$	• Write the balanced chemical equation for the compound and its ions.
$K_{sp} = [\text{Ca}^{2+}] \times [\text{OH}^-]^2$	• Write the equilibrium expression for this heterogeneous system. Remember to: - only include aqueous species - raise each species to the power of its coefficient in the balanced chemical equation.

In this example, notice that the hydroxide ion has a coefficient of 2 in the balanced equation, so it's squared in the $K_{sp}$ expression.

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Worked Example: Iron(II) phosphate

Answer	Logic
$\text{Fe}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Fe}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)$	• Write the balanced chemical equation for the compound and its ions.
$K_{sp} = [\text{Fe}^{2+}]^3 \times [\text{PO}_4^{3-}]^2$	• Write the equilibrium expression for this heterogeneous system. Remember to: - only include aqueous species - raise each species to the power of its coefficient in the balanced chemical equation.

In this example, there are 3 iron ions and 2 phosphate ions in the balanced equation, so these numbers become the powers in the $K_{sp}$ expression.

Key concepts

Dynamic equilibrium is established when the rate at which ions break away from the ionic lattice equals the rate at which ions rejoin the lattice. This is represented by:

$\text{AB}_2(s) \rightleftharpoons \text{A}^{2+}(aq) + 2\text{B}^-(aq)$

The solubility product ( $K_{sp}$ ) is only used for sparingly soluble or insoluble salts at equilibrium.

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Remember!

Every ionic compound dissolves to some extent in water - no compound is truly "completely insoluble"
Dynamic equilibrium occurs at saturation when dissolution and precipitation rates are equal
The solubility product ( $K_{sp}$ ) is the equilibrium constant specifically for dissolution equilibria
Only write $K_{sp}$ expressions for sparingly soluble or insoluble salts
When writing equilibrium expressions:
- Solid on left, ions on right
- Only include aqueous species
- Raise each concentration to the power of its coefficient

Solubility Equilibrium (HSC SSCE Chemistry): Revision Notes