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

Question

\(\mathrm{A}(\mathrm{OH})_{2}\) 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

The correct answer is (d) 2, 3, and 4.

Step 1: Write the dissociation equation and Ksp expression

For the substance \( \mathrm{A(OH)}_2 \), the dissociation in water can be represented as: \[ \mathrm{A(OH)}_2 \rightleftharpoons \mathrm{A}^{2+} + 2 \mathrm{OH}^- \] The solubility product (\( \mathrm{K_{sp}} \)) expression for the dissociation is given by: \[ \mathrm{K_{sp}} = [\mathrm{A}^{2+}][\mathrm{OH}^-]^2 \] Given \( \mathrm{K_{sp}} = 4 \times 10^{-12} \).

Step 2: Relate pH and solubility

Since the solubility involves the concentration of \( \mathrm{OH}^- \), and \( \mathrm{OH}^- \) concentration is linked to \( \mathrm{pH} \) via the following relation: \[ \mathrm{pOH} = -\log[\mathrm{OH}^-] \quad \text{and} \quad \mathrm{pH} + \mathrm{pOH} = 14 \] So changing \( \mathrm{pH} \) will affect the \( \mathrm{OH}^- \) concentration, thus affecting solubility.

Step 3: Analyze the solubility behavior at different pH

- At low \( \mathrm{pH} \): High concentration of \( \mathrm{H}^+ \) will shift the equilibrium left, decreasing solubility.- At high \( \mathrm{pH} \): The \( \mathrm{OH}^- \) concentration is increased, shifting equilibrium towards more dissolution, increasing solubility.

Step 4: Calculate pH of saturated solution

Given \( \mathrm{K_{sp}} = 4 \times 10^{-12} \), let the solubility be \( s \), then: \[ \mathrm{K_{sp}} = s(2s)^2 = 4s^3 = 4 \times 10^{-12} \] So, \( s = 10^{-4} \). Therefore, \([\mathrm{OH}^-] = 2 \cdot 10^{-4} \), and \( \mathrm{pOH} = 4 - \log2 \approx 3.7 \). Thus, \( \mathrm{pH} = 10.3 \).

Step 5: Evaluate statements based on analysis and calculations

- Statement 1 is incorrect as solubility is affected by \( \mathrm{pH} \).- Statement 2 is correct for low \( \mathrm{pH} \), where solubility decreases.- Statement 3 is correct, as high \( \mathrm{pH} \) increases solubility.- Statement 4 is correct, as the calculated \( \mathrm{pH} \) is \( 10.3 \).

Key Concepts

Ksp (Solubility Product Constant)

The solubility product constant, or \( K_{sp} \), is a crucial concept in understanding the solubility of ionic compounds. It refers to the product of the molar concentrations of the dissociated ions in a saturated solution, each raised to the power of their respective stoichiometric coefficients. This constant gives us an idea of how much solute can dissolve in a solution at a given temperature. For example, for the substance \( \mathrm{A(OH)}_2 \), \( K_{sp} \) is represented as:\[K_{sp} = [\mathrm{A}^{2+}][\mathrm{OH}^-]^2\]For \( \mathrm{A(OH)}_2 \) with a \( K_{sp} \) of \( 4 \times 10^{-12} \), this value lets us calculate the concentration of ions when the solution is saturated. Understanding \( K_{sp} \) helps predict whether a precipitate will form under certain conditions. If the ionic product of the dissociated ions in a solution exceeds \( K_{sp} \), the compound will precipitate. Conversely, if it is less, the compound remains dissolved.

Effect of pH on Solubility

pH can significantly affect the solubility of compounds, particularly those that dissociate into ions that can interact with \( \mathrm{H}^+ \) or \( \mathrm{OH}^- \) ions. When you adjust the pH, you will influence these interactions:

• Low pH: A high concentration of \( \mathrm{H}^+ \) ions in an acidic environment can react with \( \mathrm{OH}^- \) ions. This reaction can effectively reduce the concentration of \( \mathrm{OH}^- \) in the solution, causing the equilibrium to shift left, and decrease the solubility of \( \mathrm{A(OH)}_2 \).
• High pH: In an alkaline environment, there is an abundance of \( \mathrm{OH}^- \) ions. This increases the solubility of \( \mathrm{A(OH)}_2 \) as it shifts the equilibrium towards the right, allowing more of the solid to dissolve.

Therefore, the pH adjustment is a powerful tool to manipulate the solubility of substances like \( \mathrm{A(OH)}_2 \), either promoting or reducing its dissolution based on the pH of the medium.

Dissociation Equilibrium

When a compound like \( \mathrm{A(OH)}_2 \) dissolves in water, it undergoes dissociation to form ions. The dissociation process is reversible and can be represented by an equilibrium expression:\[\mathrm{A(OH)}_2 \rightleftharpoons \mathrm{A}^{2+} + 2 \mathrm{OH}^-\]This equation not only shows the dissociation but also the dynamic nature of equilibrium. At equilibrium, the rate of dissolution of \( \mathrm{A(OH)}_2 \) into its ions is equal to the rate of the reverse process, where ions re-form the solid compound. The equilibrium position is dependent on factors like temperature and pH.An important aspect of dissociation equilibrium is how it responds to changes applied to the system. Such changes include alterations in concentration, which is consistent with Le Chatelier's principle. For instance, increasing the concentration of \( \mathrm{OH}^- \) ions through a rise in pH will shift the equilibrium to favor more dissociation, enhancing solubility. Conversely, adding acids to decrease \( \mathrm{OH}^- \) concentration will reduce solubility by promoting the re-formation of \( \mathrm{A(OH)}_2 \). This equilibrium concept highlights the delicate balance in solubility and provides a mechanism to control it.