91Ó°ÊÓ

The solubility product constant, often denoted as \(K_{sp}\), is a key concept in understanding how salts dissolve in water. It represents the equilibrium constant for the dissolution of a sparingly soluble ionic compound. When we talk about \(K_{sp}\), we’re looking at the maximum amount of solute that can dissolve in a solution to reach equilibrium.

For example, with the substance \(\mathrm{Zn(OH)}_2\), its dissolution in water can be indicated as \(\mathrm{Zn(OH)}_2 (s) \rightleftharpoons \mathrm{Zn^{2+}} (aq) + 2\mathrm{OH^{-}} (aq)\). This equation provides us an understanding of how zinc ions and hydroxide ions form in solution. The \(K_{sp}\) expression derived from this reaction is \([\mathrm{Zn^{2+}}][\mathrm{OH^{-}}]^2\).

It's important to remember that the \(K_{sp}\) value is specific to each compound and varies with temperature. A lower \(K_{sp}\) indicates the compound is less soluble, meaning it more readily forms a precipitate. Thus, the \(K_{sp}\) value significantly influences the molar solubility calculation.

The pH and pOH relationship is crucial in determining the concentration of hydroxide ions \([\mathrm{OH^{-}}]\) in a solution, especially in buffering systems. The well-known formula \(\text{pH} + \text{pOH} = 14\) at 25°C enables this calculation. Once you know the pH, you can effortlessly find the pOH, and vice versa.

In this scenario, we were provided a solution buffered at \(\text{pH} = 11\). Using the relationship, we calculated \(\text{pOH} = 14 - \text{pH} = 3\). With the pOH known, the concentration of hydroxide ions is found by \([\mathrm{OH^{-}}] = 10^{-\text{pOH}} = 10^{-3}\, \text{M}\).

This conversion is vital to link pH and pOH to real-world measurements, allowing us to solve equilibrium expressions involving hydroxide ions.

An equilibrium expression provides a mathematical way to describe the balance between products and reactants for a reversible reaction at equilibrium. For dissolving salts, like \(\mathrm{Zn(OH)}_2\), knowing the equilibrium positions helps calculate solubility.

The equilibrium expression for \(\mathrm{Zn(OH)}_2\) given its dissolution is \(K_{sp} = [\mathrm{Zn^{2+}}][\mathrm{OH^{-}}]^2\). This formula indicates that at equilibrium, the product of the concentrations of the ions raised to the power of their stoichiometric coefficients equals \(K_{sp}\).

To find the molar solubility, or \(s\), we rearrange the expression to solve for \([\mathrm{Zn^{2+}}]\), given \([\mathrm{OH^{-}}]^2\) is known. As the hydroxide concentration is provided by pOH, we substitute it back into the equilibrium expression to find \(s\). Solving \(s \times (10^{-3})^2 = 4.1 \times 10^{-17}\), we find \(s = 4.1 \times 10^{-11} \text{M}\), which tells us the solubility of \(\mathrm{Zn(OH)}_2\) in our buffered solution.