Solubility product
Solubility product
- For a salt that barely dissolves, the solubility product $K_{\text{sp}}$ describes its saturated solution.
- It links solubility to the ion concentrations.
- A common ion makes the salt less soluble.
The solubility product
- $K_{\text{sp}}$ is the product of the ion concentrations, each raised to the power of its number in the formula:
$$K_{\text{sp}} = [\text{Ag}^+][\text{Cl}^-] \qquad K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^-]^2$$
- You can find $K_{\text{sp}}$ from the solubility, or the solubility from $K_{\text{sp}}$.
Practice
The solubility product Ksp is:
For a saturated solution, Ksp multiplies the ion concentrations (each to the power of its formula coefficient).
Practice
For CaF₂, the solubility product is:
There are two F⁻ per formula unit, so the [F⁻] term is squared.
The common ion effect
- A salt is less soluble in a solution that already contains one of its ions.
- The extra ion pushes the dissolving equilibrium back (Le Chatelier), so less salt dissolves.
- You can calculate the new solubility using $K_{\text{sp}}$ and the common ion's concentration.
Practice
The common ion effect means a salt becomes:
The extra (common) ion pushes the dissolving equilibrium back, so less salt dissolves.
Practice
The common ion effect is explained by:
Adding a common ion shifts the dissolving equilibrium towards the solid, reducing solubility.
You've got it
Key idea
- $K_{\text{sp}}$ = product of ion concentrations, each to the power of its formula number (e.g. $K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^-]^2$)
- find $K_{\text{sp}}$ from solubility, or solubility from $K_{\text{sp}}$
- the common ion effect: a salt is less soluble when a common ion is present (Le Chatelier)