Introduction

Solubility product (Ksp) is a fundamental concept in solution chemistry that helps us understand and predict the behavior of ionic compounds in solution. This article explores how to calculate solubility product, determine solubility from Ksp values, and apply these concepts to real-world problems.

Understanding Solubility Product

When an ionic compound dissolves in water, it establishes an equilibrium between the solid compound and its dissolved ions. For example, when lead(II) fluoride dissolves:

$\text{PbF}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{F}^-(aq)$

The solubility product (Ksp) for this equilibrium is:

$K_{sp} = [\text{Pb}^{2+}][\text{F}^-]^2$

Key points about solubility product:

Common Solubility Product Expressions

Here are some important examples of solubility product expressions:

Compound	Dissolution Equation	Ksp Expression
MgCO₃	$\text{MgCO}_3(s) \rightleftharpoons \text{Mg}^{2+}(aq) + \text{CO}_3^{2-}(aq)$	$K_{sp} = [\text{Mg}^{2+}][\text{CO}_3^{2-}]$
Fe(OH)₂	$\text{Fe}(\text{OH})_2(s) \rightleftharpoons \text{Fe}^{2+}(aq) + 2\text{OH}^-(aq)$	$K_{sp} = [\text{Fe}^{2+}][\text{OH}^-]^2$
Ca₃(PO₄)₂	$\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)$	$K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2$

Calculating Solubility from Ksp

To calculate solubility from Ksp:

Example: For PbI₂ with Ksp = 9.8 × 10⁻⁹

$\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)$

If solubility = s:

Therefore: $9.8 \times 10^{-9} = [s][2s]^2 = 4s^3$ $s = 1.3 \times 10^{-3} \text{ mol/L}$

Calculating Ksp from Solubility

When given solubility, calculate Ksp by:

Example: PbF₂ solubility = 0.64 g/L