91Ó°ÊÓ

The solubility product constant, denoted as \(K_{\text{sp}}\), plays a crucial role in solubility equilibria. \(K_{\text{sp}}\) is a measure of the extent to which a compound can dissolve in water. It is specifically used for sparingly soluble salts.

In simple terms, \(K_{\text{sp}}\) provides the product of the molar concentrations of the ions each raised to the power of their stoichiometric coefficients in a saturated solution. For \( \text{AgBr} \), the dissolution can be written as: \[ \text{AgBr}(s) \rightleftharpoons \text{Ag}^{+}(aq) + \text{Br}^{-}(aq) \]The \(K_{\text{sp}}\) expression for \( \text{AgBr} \) then becomes: \[K_{\text{sp}} = \text{[Ag}^{+}\text{][Br}^{-}\text{]} \]\( K_{\text{sp}} \) is given in this exercise as \( 5.2 \times 10^{-13} \). This very small value indicates that \( \text{AgBr} \) is only sparingly soluble in water, meaning only a tiny amount of \( \text{AgBr} \) dissolves to produce ions in solution.

Ionic equilibria in a solution involves the balance between the dissolved ions and the undissolved salt. When a salt dissolves, it dissociates into its respective ions, which interact with the solvent.

In this exercise, \( \text{AgBr} \) reaches an equilibrium with its ions in solution: \[ \text{AgBr}(s) \rightleftharpoons \text{Ag}^{+}(aq) + \text{Br}^{-}(aq) \]

The presence of additional ions from other sources, such as \( \text{MgBr}_{2} \) in this case, affects the equilibrium. When \( \text{MgBr}_{2} \) dissolves, it completely dissociates to give \( \text{Mg}^{2+} \) and two \( \text{Br}^{-} \) ions. This additional \( \text{Br}^{-} \) increases the concentration of \( \text{Br}^{-} \) in the solution, thereby shifting the equilibrium position of \( \text{AgBr} \) solution.

Due to the common ion effect, as the concentration of \( \text{Br}^{-} \) increases, the solubility of \( \text{AgBr} \) decreases. This is an example of how different ionic species interact and reach an equilibrium in the solution.

Concentration calculations are essential to determine how much of a solute dissolves and how it affects the overall solution. In this problem, the given concentration of \( \text{MgBr}_{2} \) is used to calculate the effect on the solubility of \( \text{AgBr} \).

Given \( \text{MgBr}_{2} \) concentration as 0.10 M, it dissociates completely to produce \( \text{Mg}^{2+} \) and \( 2\text{Br}^{-} \). Hence, \( \text{Br}^{-} \) concentration from \( \text{MgBr}_{2} \) would be:
\[ \text{[Br}^{-}\text{]} = 0.10 \text{ M} \times 2 = 0.20 \text{ M} \]

Next, we use the \( K_{\text{sp}} \) expression: \[ K_{\text{sp}} = \text{[Ag}^{+}\text{][Br}^{-}\text{]} \] Plugging in the known values: \[ 5.2 \times 10^{-13} = \text{[Ag}^{+}\text{]} (0.20 \text{ M}) \] To find \( \text{[Ag}^{+}\text{]} \), rearrange the equation: \[ \text{[Ag}^{+}\text{]} = \frac{5.2 \times 10^{-13}}{0.20} = 2.6 \times 10^{-12} \text{ M} \] This value represents the concentration of \( \text{Ag}^{+} \) ions in solution, equal to the moles of \( \text{AgBr} \) that dissolves since we start with pure \( \text{AgBr} \) in 1 L solution. Thus, 2.6 x 10^-12 moles of \( \text{AgBr} \) will dissolve.

Le Chatelier's Principle explains how a system at equilibrium responds to disturbances. If a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system adjusts itself to counteract the effect of the disturbance.

In the context of this problem, when \( \text{MgBr}_{2} \) is added to the solution, the additional \( \text{Br}^{-} \) ions shift the equilibrium of the \( \text{AgBr} \) dissolution.

According to Le Chatelier's Principle, adding \( \text{Br}^{-} \) will shift the equilibrium position to the left, meaning less \( \text{AgBr} \) dissolves to form \( \text{Ag}^{+} \) and \( \text{Br}^{-} \) ions. This is known as the 'common ion effect'.

By increasing the concentration of one of the products (in this case, \( \text{Br}^{-} \)), the system tries to reduce the disruption by favoring the formation of the reactant (\text{AgBr}). This leads to the decreased solubility of \( \text{AgBr} \) in the presence of \( \text{MgBr}_{2} \).