In which of the following the solubility of AgCl will be minimum ?

Options

A. `0.1 M NaNO_(3)`
B. Water
C. `0.1 M NaCl`
D. `0.1 M NaBr`

Explanation

To determine in which solution the solubility of AgCl will be minimum, we need to consider the common ion effect. The common ion effect states that the solubility of a salt is reduced in a solution that contains a common ion.

### Step-by-Step Solution:

1. **Identify the Dissociation of AgCl**:
AgCl dissociates in water as follows:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

2. **Consider the Solutions Given**:
We need to analyze the dissociation of the potential solutions:
- **NaNO3**: Dissociates into Na\(^+\) and NO3\(^-\)
- **Water**: Dissociates into H\(^+\) and OH\(^-\)
- **NaCl**: Dissociates into Na\(^+\) and Cl\(^-\)
- **NaBr**: Dissociates into Na\(^+\) and Br\(^-\)

3. **Identify Common Ions**:
- In NaCl, the Cl\(^-\) ion is a common ion with AgCl.
- In NaNO3 and water, there are no common ions with AgCl.
- NaBr does not provide a common ion for AgCl either.

4. **Apply the Common Ion Effect**:
The presence of Cl\(^-\) from NaCl will shift the equilibrium of the AgCl dissociation to the left, reducing the solubility of AgCl due to the common ion effect.

5. **Conclusion**:
Therefore, the solubility of AgCl will be minimum in the presence of NaCl because it provides the common ion Cl\(^-\).

### Final Answer:
The solubility of AgCl will be minimum in **NaCl**.

---