Which solid is less soluble in 0.5 M HCI than in water?

Options

A. `ZnCO_(3)`
B. `Hg_(2)CI_(2)`
C. `PbBr_(2)`
D. `AgI`

Answer

To determine which solid is less soluble in 0.5 M HCl than in water, we need to analyze the solubility of each of the given compounds in the presence of HCl, focusing on the common ion effect.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds given are ZnCO3, Hg2Cl2, PbBr2, and AgI.

2. **Understand the Common Ion Effect**: The common ion effect states that the solubility of a salt decreases in a solution that already contains one of the ions present in the salt. In this case, HCl provides H+ and Cl- ions.

3. **Analyze Each Compound**:
- **ZnCO3**: Dissolves in water to form Zn²⁺ and CO3²⁻. HCl will react with CO3²⁻ to form HCO3⁻ and water, but does not provide a common ion that would affect Zn²⁺ solubility.
- **Hg2Cl2**: Dissociates into Hg2²⁺ and Cl⁻ ions. The presence of Cl⁻ from HCl will shift the equilibrium to the left, reducing the solubility of Hg2Cl2 due to the common ion effect.
- **PbBr2**: Dissociates into Pb²⁺ and Br⁻ ions. HCl does not provide a common ion for Pb²⁺ or Br⁻, so its solubility remains unaffected.
- **AgI**: Dissociates into Ag⁺ and I⁻ ions. Similar to PbBr2, HCl does not provide a common ion for Ag⁺ or I⁻, so its solubility remains unaffected.

4. **Conclusion**: Among the compounds analyzed, Hg2Cl2 is the only one whose solubility decreases in the presence of 0.5 M HCl due to the common ion effect from the Cl⁻ ions.

### Final Answer:
**Hg2Cl2 is less soluble in 0.5 M HCl than in water.**

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