Problem 142 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(\mathrm{Ksp}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(1.0 \times 10^{-15} \cdot \mathrm{M}\). Find at what \(\mathrm{pH}\) does \(1.0 \times 10^{-3} \cdot \mathrm{M} \mathrm{Al}^{3+}\) precipitate on the addition of buffer of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{NH}_{4} \mathrm{OH}\) solution. (a) 10 (b) \(10.5\) (c) 11 (d) 12

Step-by-Step Solution

Answer

(a) 10

Step 1: Understanding the Solubility Product Constant

The solubility product constant (Ksp) for a compound like \(\mathrm{Al(OH)}_3\) is given, which expresses the degree to which the compound dissociates into its constituent ions in solution: \(\mathrm{Ksp} = [\mathrm{Al^{3+}}][\mathrm{OH}^-]^3\). Since we know \([\mathrm{Al^{3+}}] = 1.0 \times 10^{-3}\, \mathrm{M}\), we need to find \([\mathrm{OH}^-]\) at this concentration.

Step 2: Solve for Hydroxide Ion Concentration

Rearrange the \(\mathrm{Ksp}\) expression to solve for the hydroxide ion concentration: \([\mathrm{OH}^-]^3 = \frac{\mathrm{Ksp}}{[\mathrm{Al^{3+}}]}\). Substitute the known values into the equation: \([\mathrm{OH}^-]^3 = \frac{1.0 \times 10^{-15}}{1.0 \times 10^{-3}} = 1.0 \times 10^{-12}\). Find the cube root of \(1.0 \times 10^{-12}\) to determine \([\mathrm{OH}^-]\).

Step 3: Calculate Hydroxide Ion Concentration

To find \([\mathrm{OH}^-]\), compute the cube root of \(1.0 \times 10^{-12}\). This gives \([\mathrm{OH}^-] = (1.0 \times 10^{-12})^{1/3} = 1.0 \times 10^{-4}\, \mathrm{M}\).

Step 4: Convert Hydroxide Ion Concentration to pH

Use the relationship \(\mathrm{pOH} = -\log([\mathrm{OH}^-])\) to find \(\mathrm{pOH}\). Then calculate \(\mathrm{pH} = 14 - \mathrm{pOH}\). Substitute \([\mathrm{OH}^-] = 1.0 \times 10^{-4}\) into the equation: \(\mathrm{pOH} = 4\), so \(\mathrm{pH} = 14 - 4 = 10\).

Step 5: Conclusion

The pH at which \([\mathrm{Al}^{3+}]\) precipitates as \(\mathrm{Al(OH)}_3\) upon adding buffer is 10. This corresponds to option (a).

Key Concepts

Solubility Product Constant

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

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 of Al(OH)3

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.

pH and pOH Relationship

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.