Question

How many mol CuI ( ${K_{sp}} = 5 \times {10^{ - 12}}$ ) will dissolve in 1.0 L of 0.10M NaI solution?

Answer 1

Hint : A temperature-dependent solubility product, which acts as an equilibrium constant, characterises each solubility equilibrium. Solubility equilibria are significant in a variety of situations, including pharmaceuticals, the environment, and many others.

Complete Step By Step Answer:
When a chemical substance in the solid form is in chemical equilibrium with a solution of that substance, it is called solubility equilibrium. The solid may dissolve in its entirety, by dissociation, or by chemical reactivity with another solution ingredient, such as acid or alkali. When a chemical substance in the solid form is in chemical equilibrium with a solution containing the compound, it is called a solubility equilibrium. This form of equilibrium is a dynamic equilibrium in which certain individual molecules move between the solid and solution phases at the same rate, resulting in equal rates of dissolution and precipitation. The solution is considered to be saturated when it has reached equilibrium. The solubility of a substance is its concentration in a saturated solution. Solubility can be measured in molar units (mol/L). Temperature affects solubility.
It is written as
${K_{sp}} = {[{\text{A}}]^p}{[\;{\text{B}}]^q}$
here we know that
${\text{CuI}} \rightleftharpoons \mathop {{\text{C}}{{\text{u}}^{\text{ + }}}}\limits_s {\text{ + }}\mathop {{{\text{I}}^{\text{ - }}}}\limits_{s + 0.1}$
Also
${\text{NaI}} \to {\text{N}}{{\text{a}}^{\text{ + }}}{\text{ + }}\mathop {{{\text{I}}^{\text{ - }}}}\limits_{{\text{0}}{\text{.1M + s}}}$
So using solubility product we have
$\Rightarrow {{\mathbf{K}}_{{\text{sp}}}} = \left[ {{\mathbf{C}}{{\mathbf{u}}^ + }} \right]\left[ {{{\mathbf{I}}^ - }} \right]$
$\Rightarrow {\mathbf{5}} \times {\mathbf{1}}{{\mathbf{0}}^{ - {\mathbf{12}}}} = [{\mathbf{S}}][{\mathbf{0}}.{\mathbf{1}}]$
$\Rightarrow {\mathbf{S}} = {\mathbf{5}} \times {\mathbf{1}}{{\mathbf{0}}^{ - {\mathbf{7}}}}{\mathbf{M}}$
A supersaturated solution is one that has a greater concentration of solute than its solubility. The insertion of a "seed," which might be a small crystal of the solute or a tiny solid particle that triggers precipitation, can cause a supersaturated solution to come to equilibrium.

Note :
Solubility equilibria may be divided into three categories.
Dissolution is a simple process.
Dissociation response for dissolution. This is a common feature of salts. In this scenario, the equilibrium constant is referred to as a solubility product.
Ionization reaction for dissolution. This is typical of weak acids or weak bases dissolving in aqueous solutions with changing pH.