Problem 47 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Read the following statement and explanation and answer as per the options given below : Assertion : A very dilute acidic solution of \(\mathrm{Cd}^{2+}\) and \(\mathrm{Ni}^{2+}\) gives yellow precipitate of CdS on passing hydrogen sulphide. Statement : Solubility product of \(\mathrm{CdS}\) is more than that of NiS. (a) If both assertion and statement are correct and statement is an explanation of assertion. (b) If assertion is correct and statement is wrong, statement is not an explanation of assertion. (c) If assertion is wrong and statement is correct, statement is not an explanation of assertion. (d) If both assertion and statement are wrong and statement is not explanation of assertion.

Step-by-Step Solution

Answer

(b) Assertion is correct but the statement is wrong and not an explanation.

Step 1: Understanding the assertion

The assertion states that when a very dilute acidic solution containing \(\mathrm{Cd}^{2+}\) and \(\mathrm{Ni}^{2+}\) is treated with hydrogen sulfide \(\text{H}_2\text{S}\), a yellow precipitate of \(\mathrm{CdS}\) forms. This implies that \(\mathrm{CdS}\) is less soluble than \(\mathrm{NiS}\) under these conditions, hence precipitates out first.

Step 2: Analyzing the statement

The statement claims that the solubility product \(K_{sp}\) of \(\mathrm{CdS}\) is more than that of \(\mathrm{NiS}\). Solubility product \(K_{sp}\) is related to the solubility of a compound: the smaller the \(K_{sp}\), the less soluble the compound is. Therefore, if \(\mathrm{CdS}\) precipitates first, its solubility product should be lower than that of \(\mathrm{NiS}\).

Step 3: Compare Assertion and Statement

In comparing the assertion and statement, the assertion correctly identifies that \(\mathrm{CdS}\) precipitates first. However, the statement that \(\mathrm{CdS}\) has a higher solubility product than \(\mathrm{NiS}\) contradicts the implications needed for \(\mathrm{CdS}\) to precipitate first. The statement is therefore incorrect.

Step 4: Choosing the correct option

Since the assertion is correct but the statement is incorrect and does not support or explain the assertion, the correct choice is (b): If the assertion is correct and statement is wrong, statement is not an explanation of assertion.

Key Concepts

Solubility Product

The solubility product, often abbreviated as \( K_{sp} \), is a crucial concept in chemistry that helps us understand the solubility of compounds. It refers to a specific type of equilibrium constant employed for solids dissolving in water. This constant quantifies the extent of solubility of a substance.
For a compound like cadmium sulfide (\( \text{CdS} \)), represented as \( \text{Cd}^{2+} + \text{S}^{2-} \rightleftharpoons \text{CdS} \), the solubility product expression is \( K_{sp} = [\text{Cd}^{2+}][\text{S}^{2-}] \).

• If \( K_{sp} \) is small, the compound is less soluble in water, meaning it will more readily form a precipitate.
• A larger \( K_{sp} \) indicates higher solubility, meaning more of the substance stays in solution.

Understanding \( K_{sp} \) allows us to predict when a precipitate will form during a chemical reaction. This is particularly useful in processes like the hydrogen sulfide reaction, where the relative solubility of compounds dictates their order of precipitation.

Hydrogen Sulfide Reaction

The hydrogen sulfide reaction is an intriguing process where hydrogen sulfide gas (\( \text{H}_2\text{S} \)) is introduced to a solution containing metal ions. When \( \text{H}_2\text{S} \) is passed through such a solution, it can lead to the formation of metal sulfides, which often appear as precipitates.

• In acidic solutions, certain metal ions, like \( \text{Cd}^{2+} \), readily form insoluble sulfides like \( \text{CdS} \).
• The formation of these precipitates relies on the solubility product of each potential metal sulfide.

This selectivity is critical when dealing with a solution containing multiple metal ions, such as \( \text{Cd}^{2+} \) and \( \text{Ni}^{2+} \). The metal with the lower \( K_{sp} \) will precipitate first because its sulfide form is less soluble. Thus, the hydrogen sulfide reaction is a valuable tool for separating metal ions based on their sulfide solubility.

CdS Precipitation

The precipitation of cadmium sulfide (\( \text{CdS} \)) is a practical demonstration of the relationship between solubility products and chemical reactions. When cadmium ions (\( \text{Cd}^{2+} \)) are present in a solution and exposed to hydrogen sulfide gas, they form cadmium sulfide. This compound appears as a distinct yellow precipitate. Key factors leading to \( \text{CdS} \) precipitation include:

• The solubility product of \( \text{CdS} \) is small, meaning the compound is less soluble in water.
• The solution conditions, such as pH, can affect the solubility and thus the extent of precipitation.

In the presence of other metal ions, such as \( \text{Ni}^{2+} \), \( \text{CdS} \) precipitates first due to its smaller \( K_{sp} \), confirming its lower solubility compared to nickel sulfide (\( \text{NiS} \)). This behavior serves as a foundational concept in understanding how to manipulate solutions to selectively precipitate desired compounds.