Problem 123 - A Complete Resource Book in Chemistry for JEE Main

Question

A \((\mathrm{OH})_{2}\) is a partially soluble substance. Its \(\mathrm{Ksp}\) value is \(4 \times 10^{-12}\), which of the following statement is correct? (1) The solubility is unaffected by \(\mathrm{pH}\) of the medium (2) Its solubility has been decreased in a buffered medium at \(\mathrm{pH}\) at \(1 \mathrm{~L}\) (3) Its solubility has been increased in a buffered medium having \(\mathrm{pH}\) at 9 (4) Its saturated solution has \(\mathrm{pH}\) is equal to \(10.3\) (a) \(1,2,3\) (b) 3 and 4 (c) 2 and 3 (d) 2,3 and 4

Step-by-Step Solution

Answer

None of the statements are correct; reevaluate each statement.

Step 1: Understanding the Ksp

The solubility product constant (\( K_{sp} \)) gives an idea about the solubility of the ionic compound. For a compound \( A(OH)_2 \), the dissolution in water can be represented as \( A(OH)_2 \rightleftharpoons A^{2+} + 2OH^- \). The expression for \( K_{sp} \) is \( K_{sp} = [A^{2+}][OH^-]^2 \). Given \( K_{sp} = 4 \times 10^{-12} \).

Step 2: Impact of pH on Solubility

The solubility of a basic compound like \( A(OH)_2 \) increases with decreasing \( pH \) because lower \( pH \) signifies more acidic conditions, leading to the consumption of \( OH^- \) ions in the equilibrium, shifting it forward to dissolve more solute. Conversely, increasing \( pH \) (more basic) can decrease solubility by providing additional \( OH^- \), shifting equilibrium backwards.

Step 3: Evaluating Each Statement

(1) Incorrect; solubility is affected by \( pH \). (2) Incorrect; solubility increases in an acidic medium or decreases if \( pH < 7 \). (3) Incorrect; solubility can be decreased in an alkaline (\( pH 9 \)) medium. (4) Incorrect; to determine if \( pH \) is 10.3, calculate \( (OH^-) \) concentration and corresponding \( pH \).

Step 4: Calculating pH for Statement (4)

To test statement (4), use \( [OH^-]^2 = \frac{K_{sp}}{[A^{2+}]} \). Assume \( x = [A^{2+}] = [OH^-]/2 \), so \([OH^-]^2 \approx 4x^3 = 4 \times 10^{-12} \). Thus, \( x^3 = 10^{-12} \), \( x = 10^{-4} \), meaning \( [OH^-] = 2 \times 10^{-4} \). Calculate \( pOH = 4.7 \), thus \( pH = 14 - 4.7 = 9.3 \). Statement (4) is incorrect.

Step 5: Conclusion and Answer

None of the given statements align with either a calculated or a conceptual understanding of how \( pH \) affects solubility based on \( K_{sp} \). Therefore, none of the options (a), (b), (c), or (d) correctly represent valid combinations of statements. Re-assess each condition.

Key Concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, represented as \( K_{sp} \), is a helpful tool in understanding the solubility of ionic compounds. When a compound like \( A(OH)_2 \) dissolves in water, it separates into its constituent ions: \( A^{2+} \) and \( 2OH^- \).
\[ A(OH)_2 \rightleftharpoons A^{2+} + 2OH^- \]
The \( K_{sp} \) is expressed as \( [A^{2+}][OH^-]^2 \). When \( K_{sp} \) is low, as in our example where \( K_{sp} = 4 \times 10^{-12} \), it means the substance is not very soluble.

In simpler terms, \( K_{sp} \) tells you how much of the compound can dissolve before the solution is saturated. Knowing this can help predict whether a precipitate will form in a given solution. It is a fundamental concept in chemistry that depicts how ionic strengths and concentrations play important roles in determining solubility limits.

pH and Solubility

pH is a measure of hydrogen ion concentration in a solution. It can have a significant impact on the solubility of compounds, especially those that can release or absorb \( OH^- \) or \( H^+ \) ions. In the case of basic compounds like \( A(OH)_2 \), solubility is influenced by the medium's acidity.

When the \( pH \) is low (acidic conditions), there are more \( H^+ \) ions available to react with \( OH^- \) ions, effectively removing them from the solution. This shift reduces the concentration of \( OH^- \) and encourages more of the \( A(OH)_2 \) compound to dissolve to restore equilibrium, thus increasing solubility. Conversely, in a high \( pH \) environment (basic conditions), the solubility decreases as an excess of \( OH^- \) ions already exist, which drives the equilibrium to favor the undissolved form of the substance.

Understanding how pH impacts solubility is crucial in fields like environmental science and pharmacology, where the behavior of compounds under different conditions must be predicted.

Equilibrium Shift due to pH Changes

The concept of equilibrium shifts in chemistry helps explain how systems respond to changes in the environment to maintain balance. For the dissolution reaction of \( A(OH)_2 \), the equilibrium can shift in response to changes such as varying \( pH \).

According to Le Chatelier's Principle, if a change occurs in the system (like a change in \( pH \)), the reaction will adjust to counteract that change. Lowering the \( pH \) increases the hydrogen ion concentration. This will consume more \( OH^- \) ions, shifting the equilibrium to the right (the product side), thus enhancing the solubility of \( A(OH)_2 \).

However, if the \( pH \) increases, the additional \( OH^- \) ions will push the equilibrium to the left (reactant side), suggesting less of the solid dissolves. The balance between reactants and products depends intricately on these external conditions, and knowing how to manipulate them can significantly impact industrial and laboratory processes.