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Understanding the solubility product constant, commonly referred to as \(K_{sp}\), is crucial in solubility equilibria. It is the equilibrium constant for a solid substance dissolving in an aqueous solution. In simple terms, it's a measure of how much of the compound can dissolve in water. For a general reaction where a weakly soluble salt \(AB\) dissolves to form its ions, the expression is:

• \(AB \rightleftharpoons A^+ + B^-\)

The \(K_{sp}\) expression would be \([A^+][B^-]\). The solubility product is specific to a particular compound.

In the exercise given, the solubility product for \(Zn(OH)_{2}\) is \(3.0 \times 10^{-16}\). This very low \(K_{sp}\) value indicates that \(Zn(OH)_{2}\) is not very soluble in water. Calculating \(K_{sp}\) involves understanding the concentrations of the products of the dissolution reaction at equilibrium:

• \(K_{sp} = [Zn^{2+}][OH^{-}]^{2}\)

This equation tells us how the ions are distributed in solution. Since \(K_{sp}\) multiplied by \(OH^-\) squared yields a constant, it becomes pivotal when predicting the solubility of the compound under various conditions.

Complex ion formation involves the combination of a metal ion with ligands (typically ions or molecules) to form a complex ion. This process significantly influences the solubility of compounds by changing equilibria. When a complex forms, it can stabilize the metal ion in solution, often increasing the solubility of the original compound.

In this particular exercise, the complex ion formation is represented by the equation:

• \(Zn^{2+} + 4OH^{-} \rightleftharpoons Zn(OH)_{4}^{2-}\)

The formation constant \(K_{f}\) reflects how strongly the complex ion forms; a high \(K_{f}\)鈥攍ike the \(4.6 \times 10^{17}\) in this case鈥攊ndicates a highly stable complex.

This equilibrium shows us that as \(Zn^{2+}\) ions in the solution combine with the \(OH^-\) ions, they form the \(Zn(OH)_{4}^{2-}\) complex, thereby reducing the concentration of \(Zn^{2+}\) ions in the solution. The shift makes more \(Zn(OH)_{2}\) dissolve, altering the original solubility equilibrium.

Equilibrium calculations help predict the concentrations of substances involved in solubility equilibria at a given state. With the exercise at hand, the goal is to calculate the concentration of \(OH^-\) required to dissolve a certain amount of \(Zn(OH)_{2}\).

In this scenario, we approached the problem step-by-step:

• Identifying variables: the dissolution of \(Zn(OH)_{2}\) and the formation of \(Zn(OH)_{4}^{2-}\).
• Using expressions for \(K_{sp}\) and \(K_{f}\), connect the solubility and complex formation processes.

The system involves simultaneous equations derived from these equilibria:

• \(K_{sp} = (0.015 - x)(0.030 - 2x)^{2}\)
• \(K_{f} = \frac{x}{(0.015 - x)(0.030 - 2x)^{4}}\)

Here, \(x\) represents the amount of complex formed. Solving these equations simultaneously provides the value of \(x\),

Once \(x\) is found, substitute it back to determine the remaining \(OH^-\) concentration. This exact method ensures that the desired quantities of reactants and products are achieved, maintaining equilibrium.