In which of the following solvents will `AgBr` has highest solubility?

Options

A. `10^(-3) M `NaBr
B. `10^(-3) M NH_(4)OH`
C. pure water
D. `10^(-3)` M HBr

Explanation

To determine in which solvent `AgBr` will have the highest solubility, we need to analyze the effect of each solvent on the dissociation of `AgBr` and the common ion effect.

### Step-by-Step Solution:

1. **Understanding the Dissociation of AgBr**:
- `AgBr` is a sparingly soluble salt that dissociates in water as follows:
\[ \text{AgBr (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Br}^- (aq) \]
- The solubility of `AgBr` is influenced by the concentration of its ions in solution.

2. **Analyzing the Solvents**:
- **Option 1: NaBr**:
- NaBr dissociates completely to give `Na^+` and `Br^-` ions.
- The increase in `Br^-` concentration due to NaBr will shift the equilibrium of `AgBr` to the left (backward direction) due to the common ion effect, thus decreasing the solubility of `AgBr`.

- **Option 2: NH4OH**:
- NH4OH dissociates to produce `NH4^+` and `OH^-` ions.
- The `OH^-` ions can react with `Ag^+` to form `AgOH`, effectively removing `Ag^+` ions from the solution.
- This decrease in `Ag^+` concentration will shift the equilibrium of `AgBr` to the right (forward direction), increasing its solubility.

- **Option 3: Pure Water**:
- Pure water has a very low concentration of ions and does not provide any common ions.
- While `AgBr` can dissolve in pure water, it will not be as effective as in NH4OH since there are no additional ions to affect the equilibrium.

- **Option 4: HBr**:
- HBr dissociates to produce `H^+` and `Br^-` ions.
- Similar to NaBr, the increase in `Br^-` concentration will shift the equilibrium of `AgBr` to the left, decreasing its solubility.

3. **Conclusion**:
- Among all the solvents analyzed, `NH4OH` provides a unique situation where the formation of `AgOH` decreases the concentration of `Ag^+` ions, thereby increasing the solubility of `AgBr`.
- Therefore, the solvent in which `AgBr` has the highest solubility is **Option 2: NH4OH**.