`Hg_(2)Cl_(2)` aprecipitate if (a) ____is greater than `K_(sp) (Hg_(2)Cl_(2))`.

Answer

To determine the condition under which the precipitate of mercury(I) chloride (Hg₂Cl₂) forms, we need to understand the relationship between the ionic product and the solubility product (Ksp) of the salt.

### Step-by-Step Solution:

1. **Understand the Concept of Ksp**:
- Ksp (solubility product) is a constant that represents the maximum concentration of ions in a saturated solution of a sparingly soluble salt. For Hg₂Cl₂, the Ksp value is specific to its dissolution in water.

2. **Write the Dissociation Equation**:
- When Hg₂Cl₂ dissolves in water, it dissociates into its ions:
\[ \text{Hg}_2\text{Cl}_2 (s) \rightleftharpoons \text{Hg}_2^{2+} (aq) + 2 \text{Cl}^- (aq) \]

3. **Write the Expression for Ionic Product (IP)**:
- The ionic product (IP) is calculated using the concentrations of the ions in solution:
\[ \text{IP} = [\text{Hg}_2^{2+}] \times [\text{Cl}^-]^2 \]

4. **Condition for Precipitation**:
- A precipitate forms when the ionic product exceeds the solubility product:
\[ \text{IP} > Ksp \]
- Therefore, for Hg₂Cl₂ to precipitate, the condition is:
\[ [\text{Hg}_2^{2+}] \times [\text{Cl}^-]^2 > Ksp(\text{Hg}_2\text{Cl}_2) \]

5. **Conclusion**:
- Thus, Hg₂Cl₂ will precipitate if the ionic product, which is the product of the concentration of Hg₂²⁺ ions and the square of the concentration of Cl⁻ ions, is greater than the Ksp of Hg₂Cl₂.

### Final Answer:
Hg₂Cl₂ precipitates if the ionic product is greater than Ksp(Hg₂Cl₂).

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