Equilibria chemistry - Solubility equilibria - Core Idea

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Additional resources and activities

Solubility product

In a saturated solution of calcium carbonate, CaCO₃, there is an equilibrium between ions in the solid latttice and ions in solution:

CaCO₃(s) _∏ Ca²⁺(aq) + CO₃^2–(aq)

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We can write an equilibrium expression as:

$\inline \dpi{100} K\mathit{_{\textup{c}}= \frac{[\textup{Ca}^{\textup{2+}}(\textup{aq})][\textup{CO}_{\textup{3}}\! ^{\textup{2-}}(\textup{aq})]}{[\textup{CaCO}_{\textup{3}}(\textup{s})]}}$

But, because this is a heterogeneous equilibrium, [CaCO₃(s)] can be treated as a constant, so we can write

[Ca²⁺(aq)] [CO₃^2–(aq)] = K_c[CaCO₃(s)] = a new constant

The new constant is known as the solubility product and is given the symbol K_sp.

So, K_sp = [Ca²⁺(aq)] [CO₃^2–(aq)]

What are the units of K_sp in this case?

The units are mol² dm^–6

We can develop the idea of solubility product to apply to a sparingly soluble solid such as lead(II) chloride, PbCl₂, where the number of cations and anions is different.

The equilibrium equation for the solution of lead(II) chloride is:

PbCl₂(s) _∏ Pb²⁺(aq) + 2Cl^–(aq)

Notice that there are two chloride ions for each lead ion.

The expression for K_s_p is:

K _sp = [Pb²⁺(aq)] [Cl^–(aq)]²

You can see that the concentration of chloride ions is squared because there are two chloride ions for each lead ion in the equilibrium equation.

What is the solubility expression for iron(II) hydroxide, Fe(OH)₃?

Fe(OH)₃(s) _∏ Fe³⁺(aq) + 3OH^–(aq)

What is the K_sp expression for iron(II) hydroxide, Fe(OH)₃?

K _sp = [Fe³⁺(aq)] [OH^–(aq)]³

What are the units of K_spfor Fe(OH)₃?

The units are mol⁴ dm^–12

Points to remember about K_sp