Question

What is a solubility product? Calculate the solubility of ${A_2}{X_3}$ In pure water, assuming that neither kind of ion reacts with water. The solubility product of ${A_2}{X_3}$, ${K_{sp}} = 1.1 \times {10^{ - 23}}$.

Answer 1

The general form of the solubility product constant for the equation:
$aA \rightleftharpoons bB + cC$ is ${K_{sp}} = {[B]^b}{[C]^c}$. The smaller the solubility product, the lower the solubility.

Complete answer:
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. Solubility is expressed in terms of mass of solute per $100ml$ of solvent. Solubility products are useful in predicting whether a precipitate will form under special conditions.
${A_2}{X_3}$ is dissociated as:
${A_2}{X_3} \rightleftharpoons 2{A^{3 + }} + 3{X^{2 - }}$
Let the solubility be $= {S_{}}mo{l_{}}{L^ - }$
Applying the expression for the equilibrium concentrations of the ions into the solubility product expression:
${K_{sp}} = {[{A^{3 + }}]^2}{[{X^{2 - }}]^3}$
${K_{sp}} = {[2S]^2}{[3S]^3}$
$1.1 \times {10^{ - 23}} = 4{S^2} \times 27{S^3}$
$1.1 \times {10^{ - 23}} = 108{S^5}$
${S^5} = \dfrac{{1.1 \times {{10}^{ - 23}}}}{{108}} = 1 \times {10^{ - 25}}$
Therefore, the solubility will be $S = 1.0 \times {10^{ - 5}}mo{l_{}}{L^ - }$

Note:
Solubility of a product is determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. A substance’s solubility product ${K_{sp}}$, is the ratio of concentration at equilibrium.