What is the effect on the amount of solid \(\mathrm{Mg}(\mathrm{OH})_{2}\) that dissolves and the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{OH}^{-}\) when each of the following are added to a mixture of solid \(\mathrm{Mg}(\mathrm{OH})_{2}\) and water at equilibrium? (a) \(\mathrm{MgCl}_{2}\) (b) KOH (c) \(\mathrm{HClO}_{4}\) (d) \(\mathrm{NaNO}_{3}\) (e) \(\mathrm{Mg}(\mathrm{OH})_{2}\)

Understanding what chemical equilibrium means is crucial to grasping how reactions in chemistry function. Imagine a crowded dance floor where for every person who exits, another one enters, maintaining a constant crowd. Similarly, chemical equilibrium occurs in a closed system when the rate of the forward reaction (reactants transforming into products) equals the rate of the reverse reaction (products transforming back into reactants). At this point, the concentrations of reactants and products remain constant over time, not because the reactions have stopped, but because they are occurring at equal rates.

Consider a simple reversible reaction represented by the equation \( A \rightleftharpoons B \). At equilibrium, the rate at which A transforms into B is exactly matched by the rate at which B changes back into A. But it's important to remember that reaching equilibrium doesn't necessarily mean the amounts of A and B are equal, rather that their concentrations are steady.

Solubility equilibrium specifically refers to the point at which a substance's dissolution in a solvent is in balance with its precipitation. Taking the example of \( \mathrm{Mg}(\mathrm{OH})_{2} \), a slightly soluble substance, we find its solubility equilibrium in water represented by the chemical equation \( \mathrm{Mg}(\mathrm{OH})_{2(s)} \rightleftharpoons \mathrm{Mg}^{2+}_{(aq)} + 2\mathrm{OH}^{-}_{(aq)} \). At this solubility equilibrium, the rate at which \( \mathrm{Mg}(\mathrm{OH})_{2} \) dissolves into its ions is equal to the rate at which these ions combine back to form the solid. This balance can be quantitatively described by the solubility product constant (Ksp), which remains constant at a given temperature unless disturbed by external factors.

Le Chatelier's Principle provides a way to predict how a system at equilibrium responds to changes in conditions. According to this principle, if a dynamic equilibrium is disturbed by altering concentration, temperature, or pressure, the system will adjust itself to counteract the effect of the disturbance and re-establish equilibrium. In the context of our \( \mathrm{Mg}(\mathrm{OH})_{2} \) example:

• Adding \( \mathrm{MgCl}_{2} \) introduces more \( \mathrm{Mg}^{2+} \) ions into the system, causing a shift to the left (reactant side) to reduce the extra \( \mathrm{Mg}^{2+} \) ions through precipitation, decreasing solubility.
• Adding KOH adds \( \mathrm{OH}^{-} \) ions, shifting the equilibrium to the left, promoting formation of the solid and reducing \( \mathrm{Mg}^{2+} \) ion concentration.
• Adding \( \mathrm{HClO}_{4} \) consumes \( \mathrm{OH}^{-} \) ions, shifting equilibrium to the right (product side), increasing solubility as the solid dissolves to replace the removed \( \mathrm{OH}^{-} \) ions.
• Adding \( \mathrm{NaNO}_{3} \) involves ions that do not interact with either \( \mathrm{Mg}^{2+} \) or \( \mathrm{OH}^{-} \) and therefore has no influence on the equilibrium.
• Introducing additional \( \mathrm{Mg}(\mathrm{OH})_{2} \) does not directly alter the ion concentrations, so the equilibrium remains unaffected, but it does mean more reactant is available to enter solution if external conditions change.

Through these illustrations, Le Chatelier's Principle helps us predict how different additions will affect a solution's equilibrium and provides insight into how the system's balance can be restored through equilibrium shifts.