Let the solubilities of `AgCI` in `H_(2)O`, and in `0.01M CaCI_(2), 0.01M NaCI`, and `0.05M AgNO_(3)` be `S_(1),S_(2),S_(3),S_(4)`, respectively. What is the correct relationship between these quantites.

Options

A. `S_(1) gt S_(2) gt S_(3) gt S_(4)`
B. `S_(1) gt S_(2) = S_(3) gt S_(4)`
C. `S_(1) gt S_(3) gt S_(2) gtS_(4)`
D. `S_(4) gt S_(2) gt S_(3) gt S_(1)`

Explanation

To determine the correct relationship between the solubilities of AgCl in different solutions, we will analyze the solubility in pure water and in the presence of other salts.

### Step-by-Step Solution:

1. **Understanding the Dissociation of AgCl**:
- AgCl dissociates in water as follows:
\[ \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]
- Let the solubility of AgCl in pure water be \( S_1 \). Thus, at equilibrium:
\[ [\text{Ag}^+] = S_1 \quad \text{and} \quad [\text{Cl}^-] = S_1 \]
- The solubility product \( K_{sp} \) for AgCl can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S_1^2 \]
- Therefore, we have:
\[ S_1 = \sqrt{K_{sp}} \]

2. **Solubility in 0.01 M CaCl2**:
- CaCl2 dissociates in water as follows:
\[ \text{CaCl}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{Cl}^- \]
- The concentration of Cl⁻ from CaCl2 will be \( 0.02 \, \text{M} \) (since it produces 2 moles of Cl⁻ per mole of CaCl2).
- The total concentration of Cl⁻ becomes:
\[ [\text{Cl}^-] = S_2 + 0.02 \approx 0.02 \quad (\text{since } S_2 \text{ is negligible}) \]
- The solubility product expression becomes:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-]^2 = S_2 (0.02)^2 \]
- Thus, we can express \( S_2 \) as:
\[ S_2 = \frac{K_{sp}}{(0.02)^2} \]

3. **Solubility in 0.01 M NaCl**:
- NaCl dissociates as:
\[ \text{NaCl} \rightleftharpoons \text{Na}^+ + \text{Cl}^- \]
- The concentration of Cl⁻ from NaCl is \( 0.01 \, \text{M} \).
- The total concentration of Cl⁻ becomes:
\[ [\text{Cl}^-] = S_3 + 0.01 \approx 0.01 \]
- The solubility product expression becomes:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-]^2 = S_3 (0.01)^2 \]
- Thus, we can express \( S_3 \) as:
\[ S_3 = \frac{K_{sp}}{(0.01)^2} \]

4. **Solubility in 0.05 M AgNO3**:
- AgNO3 dissociates as:
\[ \text{AgNO}_3 \rightleftharpoons \text{Ag}^+ + \text{NO}_3^- \]
- The concentration of Ag⁺ from AgNO3 is \( 0.05 \, \text{M} \).
- The total concentration of Ag⁺ becomes:
\[ [\text{Ag}^+] = S_4 + 0.05 \approx 0.05 \]
- The solubility product expression becomes:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-]^2 = (0.05)(S_4)^2 \]
- Thus, we can express \( S_4 \) as:
\[ S_4 = \frac{K_{sp}}{(0.05)} \]

5. **Comparing Solubilities**:
- From the expressions derived:
- \( S_1 = \sqrt{K_{sp}} \)
- \( S_2 = \frac{K_{sp}}{(0.02)^2} \)
- \( S_3 = \frac{K_{sp}}{(0.01)^2} \)
- \( S_4 = \frac{K_{sp}}{0.05} \)

- Since \( (0.01)^2 < (0.02)^2 < 0.05 \), we can conclude:
\[ S_1 > S_3 > S_2 > S_4 \]

### Final Relationship:
The correct relationship is:
\[ S_1 > S_3 > S_2 > S_4 \]