Solubility product
| English | Chinese | Pinyin |
|---|---|---|
| solubility product | 溶度积 | róng dù jī |
| saturated | 饱和 | bǎo hé |
| solubility | 溶解度 | róng jiě dù |
| common ion effect | 同离子效应 | tóng lí zi xiào yìng |
How much will dissolve?
- For a salt that barely dissolves, the solubility product 溶度积 $K_{\text{sp}}$ describes its saturated 饱和 solution.
- It links solubility 溶解度 to the ion concentrations.
- A common ion makes the salt less soluble.
Solubility product lab
ionic product compared with Ksp
Increase ion concentration and see when precipitation becomes likely.
The equilibrium constant for a sparingly soluble salt dissolving is the solubility ______.
Written Ksp.
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 $K_{\text{sp}}$.
Stalactites form as dissolved minerals come back out of solution — there is a limit to how much will dissolve
The solubility product Ksp is:
For a saturated solution, Ksp multiplies the ion concentrations (each to the power of its formula coefficient).
For CaF₂, the solubility product is:
There are two F⁻ per formula unit, so the [F⁻] term is squared.
Match each solubility-product idea.
Each item links the term to its correct meaning.
The common ion effect 同离子效应
- 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.
The common ion effect means a salt becomes:
The extra (common) ion pushes the dissolving equilibrium back, so less salt dissolves.
The common ion effect is explained by:
Adding a common ion shifts the dissolving equilibrium towards the solid, reducing solubility.
The common-ion effect
- Adding an ion already in the equilibrium lowers the solubility of a sparingly soluble salt.
- Ksp must still be satisfied, so less salt dissolves.
You've got it
- $K_{\text{sp}}$ = 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)