Добуток розчинності

To determine the solubility of a material, we use the solubility product constant, $K_{sp}$.

Take the dissolution of solid barium fluoride, $\text{BaF}_2$. Barium fluoride dissolves according to the following equilibrium: $$\text{BaF}_2 (s) \leftrightharpoons \text{Ba}^{2+} (aq) + 2\text{F}^- (aq)$$

Recall the general form of the equilibrium constant, $K_{eq}$: $$K = \dfrac{[C]^m[D]^n}{[A]^k[B]^m}$$

In the case of barium fluoride, one might be led into the trap of assuming that the equilibrium constant is $\dfrac{[\text{Ba}^{2+}][\text{F}^-]^2}{[\text{BaF}_2]}$. However, since $\text{BaF}_2$ is a pure solid in the reaction, it does not have a concentration, so the equilibrium expression simply becomes: $$K_{sp} = [\text{Ba}^{2+}][\text{F}^-]^2$$ Since it is only a product of the concentrations of the products, it is called the solubility product constant. For $\text{BaF}_2$, the solubility product constant is $1.0 \times 10^{-6}$ at 298 K.

However, one must be careful, as the solubility product constant differs from the solubility of a substance. The solubility of a substance refers to the "$x$" that is found when one completes the ICE table to solve for the change in concentrations. For barium fluoride, $x \approx 6.3 \times 10^{-3} M$ (at 298 K). This is the final concentration of barium ions in solution. The final concentration of fluoride ions is twice this value, since 2 moles of $\text{F}^-$ ions and 1 mole of $\text{Ba}^{2+}$ ions are produced for every mole of $\text{BaF}_2$ that dissolves.

(Video) 17.4 Solubility and Ksp | General Chemistry

by Chad's Prep

In this video, Chad explains the solubility product constant. He shows you four example problems step-by-step. He also shows you how to relate it with molar solubility. This video will help you understand any solubility problem!