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) \(\mathrm{K}_{\mathrm{s}}(\mathrm{ZnS})<\mathrm{K}_{\mathrm{si}}(\mathrm{PbS})\) (b) \(\mathrm{K}_{\mathrm{ss}}(\mathrm{ZnS})>\mathrm{K}_{s p}(\mathrm{PbS})\) (c) \(\mathrm{H}_{2} \mathrm{~S}\) decreases the \(\mathrm{K}_{\text {sp }}\) of \(\mathrm{ZnS}\) (d) \(\mathrm{H}_{2} \mathrm{~S}\) increases the \(\mathrm{K}_{\text {p }}\) of \(\mathrm{PbS}\)

Step-by-Step Solution

Answer

Option (b) is correct; \( K_{\text{ss}}(\mathrm{ZnS}) > K_{\text{sp}}(\mathrm{PbS}) \).

Step 1: Understanding the Problem

We have a solution of an equimolar mixture of \( \mathrm{Zn}^{2+} \) and \( \mathrm{Pb}^{2+} \) ions in dilute acetic acid. When \( \mathrm{H}_2 \mathrm{S} \) gas is passed through it, \( \mathrm{ZnS} \) is not precipitated. We need to determine the reason for this based on the given options.

Step 2: Analyzing Solubility Product (Ksp) Concept

The solubility product (\( K_{\text{sp}} \)) of a compound represents its degree of solubility in a solution. A smaller \( K_{\text{sp}} \) means the compound is less soluble and more likely to precipitate. Conversely, a larger \( K_{\text{sp}} \) implies greater solubility and less tendency to precipitate.

Step 3: Comparing Ksp Values

\( \mathrm{ZnS} \) and \( \mathrm{PbS} \) have different \( K_{\text{sp}} \) values. \( K_{\text{sp}}(\mathrm{ZnS}) \) is typically larger than \( K_{\text{sp}}(\mathrm{PbS}) \), meaning \( \mathrm{ZnS} \) is more soluble in the solution than \( \mathrm{PbS} \).

Step 4: Evaluating Option (b)

Option (b) states \( K_{\text{ss}}(\mathrm{ZnS}) > K_{\text{sp}}(\mathrm{PbS}) \). This implies that \( \mathrm{ZnS} \) has higher solubility than \( \mathrm{PbS} \) in the given conditions, hence it does not precipitate, while \( \mathrm{PbS} \) does.

Step 5: Concluding the Correct Option

Based on the comparison and analysis, it is clear that the solubility of \( \mathrm{ZnS} \) is indeed greater than that of \( \mathrm{PbS} \). Thus, option (b) is the correct reason why \( \mathrm{ZnS} \) does not precipitate when \( \mathrm{H}_2 \mathrm{S} \) is passed through the solution.

Key Concepts

Solubility Product (Ksp)

The solubility product, often represented as \( K_{\text{sp}} \), is a vital concept in understanding how substances dissolve in solutions. It provides insight into the solubility of ionic compounds, especially those that are sparingly soluble. The \( K_{\text{sp}} \) value is unique to each compound and depends on the temperature and the nature of the solvent.
If you have a lower \( K_{\text{sp}} \), that generally indicates the compound is less soluble and more prone to form a precipitate.

• A lower \( K_{\text{sp}} \) means a substance will precipitate out of a solution more readily compared to a substance with a higher \( K_{\text{sp}} \).
• Larger \( K_{\text{sp}} \) values suggest greater solubility.

For example, if we compare \( \mathrm{ZnS} \) and \( \mathrm{PbS} \), \( \mathrm{ZnS} \) typically has a larger \( K_{\text{sp}} \) than \( \mathrm{PbS} \), which explains why \( \mathrm{ZnS} \) remains dissolved in the solution.

Precipitation Reactions

Precipitation reactions occur when two solutions containing soluble salts are mixed, resulting in the formation of an insoluble solid or precipitate. The formula of the precipitate can be predicted by considering the ions involved and their respective solubility products.
A key factor determining whether a precipitate will form is the concentration of the ions and the comparison of the product of these concentrations to the \( K_{\text{sp}} \).
If the product exceeds the \( K_{\text{sp}} \), precipitation occurs.

• In our exercise, \( \mathrm{PbS} \) precipitates first because its \( K_{\text{sp}} \) is smaller, meaning it's less soluble.
• This leaves \( \mathrm{ZnS} \) in solution because its solubility is greater, given its higher \( K_{\text{sp}} \).

Precipitation plays a crucial role not only in chemistry experiments but also in many industrial and environmental processes.

Metal Sulfides

Metal sulfides such as \( \mathrm{ZnS} \) and \( \mathrm{PbS} \) are salts that include a metal cation and the sulfide anion \( \mathrm{S}^{2-} \). These compounds are typically formed through precipitation reactions.
The \( K_{\text{sp}} \) values of metal sulfides vary, depending on their nature and the conditions, such as the pH of the environment. In acidified solutions, like the one mentioned in the exercise, the solubility can be influenced by shifts in equilibrium.

• For \( \mathrm{PbS} \), its lower \( K_{\text{sp}} \) means it is less soluble, hence more prone to precipitate first the moment \( \mathrm{H}_{2} \mathrm{~S} \) is introduced.
• In contrast, \( \mathrm{ZnS} \) with higher solubility remains dissolved.

These principles of metal sulfides are vital in understanding processes from mining to water treatment.

Acid-Base Chemistry

Acid-base chemistry plays a role in determining the solubility of metal sulfides in a solution. The presence of acids can alter the solubility behavior by affecting the ion concentrations. Acetic acid, for instance, can impact solubility because it is capable of adjusting the overall pH of the solution.
Acids can provide additional protons, \( \mathrm{H}^+ \), which may interact with sulfide ions to shift the equilibrium.

• In our scenario, dilute acetic acid does not fully dissociate, providing a slightly acidic environment.
• This can suppress the precipitation of sulfides like \( \mathrm{ZnS} \), which require a higher concentration of \( \mathrm{S}^{2-} \) to precipitate out.

Understanding acid-base interactions is crucial because they are intertwined with solubility and precipitation, influencing various chemical systems.