Solubility of `AgBr` in water is `S_(1),` in `"0.01 M "CaBr_(2)` is `S_(2)` in `0.01 M NaBr` is `S_(3)` and in `0.05 M AgNO_(3)` is `S_(4)`. The correct order of these solubilities is `:`

Options

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

Explanation

To determine the correct order of solubility of `AgBr` in different solutions, we need to analyze the common ion effect and how it affects solubility.

### Step-by-step Solution:

1. **Understanding the Solubility of AgBr**:
- The solubility of `AgBr` in pure water (denoted as `S1`) is the highest because there are no common ions present to affect its solubility.

2. **Analyzing the Effect of Common Ions**:
- When `AgBr` dissolves, it dissociates into `Ag^+` and `Br^-` ions. The presence of either of these ions in the solution will decrease the solubility of `AgBr` due to the common ion effect.

3. **Solubility in 0.01 M CaBr2 (denoted as `S2`)**:
- `CaBr2` dissociates into `Ca^2+` and `2Br^-`. The concentration of `Br^-` ions from `CaBr2` is `2 x 0.01 M = 0.02 M`. This high concentration of `Br^-` will significantly reduce the solubility of `AgBr`.

4. **Solubility in 0.01 M NaBr (denoted as `S3`)**:
- `NaBr` dissociates into `Na^+` and `Br^-`. The concentration of `Br^-` ions from `NaBr` is `0.01 M`. This is lower than that from `CaBr2`, so the reduction in solubility of `AgBr` will be less compared to `S2`.

5. **Solubility in 0.05 M AgNO3 (denoted as `S4`)**:
- `AgNO3` dissociates into `Ag^+` and `NO3^-`. The concentration of `Ag^+` ions from `AgNO3` is `0.05 M`. This high concentration of `Ag^+` will also significantly reduce the solubility of `AgBr`.

6. **Comparing the Solubilities**:
- `S1` (in pure water) > `S3` (in 0.01 M NaBr) > `S2` (in 0.01 M CaBr2) > `S4` (in 0.05 M AgNO3).
- Therefore, the correct order of solubilities is:
\[ S1 > S3 > S2 > S4 \]

### Final Order of Solubility:
- The correct order of solubilities is: **S1 > S3 > S2 > S4**.