The correct order of increasing solubility of AgCI in
(A) water (B) 0.1 M NaCI
(C ) 0.1 M `BaCI_2` (D) 0.1 M `NH_3` is

Options

A. `A lt B lt C lt D`
B. `B lt C lt A lt D`
C. `C lt B lt D lt A`
D. `C lt B lt A lt D`

Answer

To determine the correct order of increasing solubility of AgCl in the given solutions, we need to analyze how the presence of each solution affects the solubility of AgCl.

### Step-by-Step Solution:

1. **Understanding AgCl Solubility**:
- AgCl is a sparingly soluble salt. Its solubility in pure water is limited and can be represented by the solubility product constant (Ksp). The solubility (S) of AgCl in pure water can be expressed as:
\[ K_{sp} = [Ag^+][Cl^-] = S^2 \]
- Therefore, in pure water, the solubility is S.

2. **Solubility in 0.1 M NaCl**:
- In a solution of NaCl, the concentration of Cl⁻ ions is already increased due to the presence of NaCl. This will shift the equilibrium of the dissolution of AgCl according to Le Chatelier's principle, reducing its solubility.
- The Ksp expression becomes:
\[ K_{sp} = S \times [0.1] \implies S = \frac{K_{sp}}{0.1} \]
- Thus, the solubility of AgCl in 0.1 M NaCl is less than in pure water.

3. **Solubility in 0.1 M BaCl₂**:
- BaCl₂ dissociates into Ba²⁺ and 2Cl⁻ ions. Therefore, the concentration of Cl⁻ ions in this solution is 0.2 M.
- The Ksp expression becomes:
\[ K_{sp} = S \times [0.2] \implies S = \frac{K_{sp}}{0.2} \]
- This means the solubility of AgCl in 0.1 M BaCl₂ is even lower than in 0.1 M NaCl due to the higher concentration of Cl⁻ ions.

4. **Solubility in 0.1 M NH₃**:
- Ammonia (NH₃) can form a complex with Ag⁺ ions, which reduces the concentration of free Ag⁺ ions in solution. This shifts the equilibrium to the right, increasing the solubility of AgCl.
- The Ksp expression remains the same, but due to the formation of the complex, the effective concentration of Ag⁺ decreases, leading to increased solubility of AgCl.

5. **Comparing Solubilities**:
- From the analysis, we have:
- Solubility in water (S)
- Solubility in 0.1 M NaCl \(\left(\frac{K_{sp}}{0.1}\right)\)
- Solubility in 0.1 M BaCl₂ \(\left(\frac{K_{sp}}{0.2}\right)\)
- Solubility in 0.1 M NH₃ (increased due to complexation)
- The order of increasing solubility is:
\[ \text{BaCl}_2 < \text{NaCl} < \text{Water} < \text{NH}_3 \]

### Final Order:
Thus, the correct order of increasing solubility of AgCl is:
**(C) 0.1 M BaCl₂ < (B) 0.1 M NaCl < (A) Water < (D) 0.1 M NH₃**