Problem 191 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(\mathrm{H}_{2} \mathrm{~S}\) is passed through an aqueous solution of an equilimolar mixture of \(\mathrm{Zn}^{2+}\) and \(\mathrm{Pb}^{2+}\) acidified with dilute acetic acid, \(\mathrm{ZnS}\) is not precipitated, because (a) \(\mathrm{K}_{\mathrm{sp}}(\mathrm{ZnS})<\mathrm{K}_{\text {?p }}(\mathrm{PbS})\) (b) \(\mathrm{K}_{\mathrm{s}}(\mathrm{ZnS})>\mathrm{K}_{\mathrm{s}}(\mathrm{PbS})\) (c) \(\mathrm{H}_{2} \mathrm{~S}\) decreases the \(\mathrm{K}_{\text {sp }}\) of \(\mathrm{ZnS}\) (d) \(\mathrm{H}_{2} \mathrm{~S}\) increases the \(\mathrm{K}_{\text {sp }}\) of \(\mathrm{PbS}\)

Step-by-Step Solution

Answer

(b) \(K_{sp}(ZnS)>K_{sp}(PbS)\).

Step 1: Understanding the Concept of Precipitation

When two ions in a solution form an insoluble compound, the substance precipitates out of the solution. The solubility product constant, \( K_{sp} \), helps predict this behavior. A compound precipitates when the product of the concentrations of its ions exceeds \( K_{sp} \).

Step 2: Comparing Solubility Product Constants

The solubility product constants for zinc sulfide (\( K_{sp}(ZnS) \)) and lead sulfide (\( K_{sp}(PbS) \)) are evaluated. According to typical solubility tables, \( K_{sp}(ZnS) \) is larger than \( K_{sp}(PbS) \), meaning \( ZnS \) is more soluble than \( PbS \).

Step 3: Impact of Acetic Acid on the Solution

The solution is acidified with dilute acetic acid. In an acidic medium, \( H_2S \) can produce \( S^{2-} \) ions, but the concentration of \( S^{2-} \) remains low due to low dissociation, which favors the precipitation of less soluble sulfides first.

Step 4: Predicting Which Compound Precipitates

Given that \( PbS \) is less soluble (lower \( K_{sp} \)) than \( ZnS \), in an acidic environment like this, only \( PbS \) will precipitate. The \( Zn^{2+} \) ions remain in solution because the concentration \( [S^{2-}] \) is insufficient to exceed \( K_{sp}(ZnS) \).

Key Concepts

Precipitation

Precipitation occurs when a solid forms from a solution due to chemical reactions, often between ions in the solution. This process happens when the product of the ion concentrations exceeds the solubility product constant, or \( K_{sp} \), of a compound. Let's break it down step by step.

• When ions in a solution form a less soluble compound, this compound falls out or precipitates.
• For a precipitation reaction to occur, the right conditions must be met, specifically those related to \( K_{sp} \).
• There are conditions where temperature, concentration, and presence of other ions, such as additional acids or bases, impact precipitation.

In the exercise problem, we have a mixture of \( \text{Zn}^{2+} \) and \( \text{Pb}^{2+} \) which react with \( \text{H}_2\text{S} \). The zinc sulfide \(( \text{ZnS})\) does not precipitate due to its higher \( K_{sp} \) compared to lead sulfide \((\text{PbS})\). The compound with the lower \( K_{sp} \), in this case, \(\text{PbS} \), forms a precipitate first. This understanding of precipitation is critical in predicting and determining which solid will fall out of solution under given conditions.

Solubility

Solubility describes how much of a substance can dissolve in a solvent before it reaches saturation. The solubility of a compound is often indicated by its solubility product constant, \( K_{sp} \).

Let's take a closer look:

• \( K_{sp} \) is a measure of the extent to which a compound dissociates into its ions in solution.
• A higher \( K_{sp} \) implies the compound is more soluble, while a lower \( K_{sp} \) signifies less solubility.
• Solubility can be influenced by various factors, such as temperature, common-ion effect, and pH levels of the solution.

In our exercise, \( K_{sp}(\text{ZnS}) \) is larger than \( K_{sp}(\text{PbS}) \), indicating that \( \text{ZnS} \) is more soluble than \( \text{PbS} \). Therefore, when \( \text{H}_2\text{S} \) is added, \( \text{PbS} \) with its lower solubility product, precipitates first, leaving \( \text{Zn}^{2+} \) ions in solution. Understanding the solubility and the corresponding \( K_{sp} \) values of compounds helps us predict and rationalize outcomes in chemical reactions.

Acetic acid impact

The presence of acetic acid in a chemical solution can significantly influence the behavior of the reaction, particularly its impact on solubility and precipitation. In this scenario, here's how acetic acid plays a role:

• Acetic acid, a weak acid, partially dissociates in water, maintaining the acidic environment necessary for certain reactions.
• In an acidic solution, such as one containing acetic acid, the dissociation of \( \text{H}_2\text{S} \) to produce \( \text{S}^{2-} \) ions is limited.
• This restriction means fewer sulfur ions are available to react with \( \text{Zn}^{2+} \), further preventing precipitation of \( \text{ZnS} \).

To summarize, the acidic environment created by acetic acid in the solution impacts the concentration of \( \text{S}^{2-} \), favoring the precipitation of sulfides that are less soluble, like \( \text{PbS} \), first. This is because their \( K_{sp} \) values are reached more swiftly compared to more soluble compounds. The presence and concentration of acetic acid thus play a pivotal role in directing the course of the reaction and understanding this can inform practical aspects of chemical applications and reactions.