91Ó°ÊÓ

The solubility product constant, often abbreviated as \( K_{sp} \), is a crucial concept in chemistry for understanding how compounds behave in aqueous solutions. It's specifically important for sparingly soluble salts like \( \text{Al(OH)}_3 \). The \( K_{sp} \) expression for such a compound reflects its dissociation into separate ions in the solution. For \( \text{Al(OH)}_3 \), the reaction can be written as:

• \( \text{Al(OH)}_3(s) \rightleftharpoons \text{Al}^{3+}(aq) + 3\text{OH}^-(aq) \)
• \( K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3 \)

Understanding this equilibrium is essential because it determines whether a precipitate will form under specific conditions. Here, knowing \( K_{sp} \) can help predict when metals like aluminum will come out of solution as their hydroxides, evidenced by the formation of a solid precipitate. The lower the \( K_{sp} \) value, the less soluble the compound is in water.

Hydroxide ion concentration \([\text{OH}^-]\) plays a pivotal role in determining the solubility of hydroxide compounds. In our example, we calculated \([\text{OH}^-]\) using the solubility product constant equation for \( \text{Al(OH)}_3 \). With a known concentration of \( \text{Al}^{3+} \), the \( K_{sp} \) equation can be rearranged to solve for \([\text{OH}^-]\) as follows:

• \([\text{OH}^-]^3 = \frac{K_{sp}}{[\text{Al}^{3+}]} \)
• By substituting \( K_{sp} = 1.0 \times 10^{-15} \) and \([\text{Al}^{3+}] = 1.0 \times 10^{-3}\, M \), we find \([\text{OH}^-]^3 = 1.0 \times 10^{-12} \).

The cube root of \( 1.0 \times 10^{-12} \) gives us \([\text{OH}^-] = 1.0 \times 10^{-4}\, M \). This concentration is pivotal for predicting precipitation and understanding the basicity of the solution.

Precipitation occurs when a solution can no longer keep dissolved ions in equilibrium, leading to solid formation. For \( \text{Al(OH)}_3 \), it begins to precipitate when the product of \([\text{Al}^{3+}][\text{OH}^-]^3\) exceeds its \( K_{sp} \). This is a helpful indicator for predicting when a buffer will no longer maintain solubility equilibrium due to pH changes. By administering a buffer, such as \( \text{NH}_{4} \text{Cl} \) and \( \text{NH}_{4} \text{OH} \), precise control over the pH levels can be achieved, facilitating the desired precipitation. When \([\text{OH}^-] \) reaches a certain level, specifically \( 1.0 \times 10^{-4}\, M \), the pH reaches a threshold where \( \text{Al(OH)}_3 \) begins to form a solid. This process is critical in applications that require the removal of aluminum from solutions.

Understanding the relationship between \( \text{pH} \) and \( \text{pOH} \) is fundamental in any aqueous solution chemistry. pH measures the concentration of hydrogen ions \([\text{H}^+]\), while pOH measures hydroxide ions \([\text{OH}^-]\). The two are tied together through the equation:

• \( \text{pH} + \text{pOH} = 14 \) at 25°C.
• This means if you know one, you can easily find the other.

In our exercise, after calculating \([\text{OH}^-] = 1.0 \times 10^{-4} \), the pOH was found using the formula \( \text{pOH} = -\log([\text{OH}^-]) \), resulting in a pOH of 4.Subsequently, pH is calculated as \( \text{pH} = 14 - \text{pOH} \), giving a value of 10 for a solution where aluminum precipitates as \( \text{Al(OH)}_3 \). This example illustrates the dynamic interplay between ion concentration and acidity/basicity, which is crucial in processes like titrations or buffer preparations.