Some chemists at wished to perpare a saturated solution of a silver compound and they wanted it to have the highest concentration of silver ion possible. Which of the following compound would they use ?
`K_(sp) (AgCI) = 1.8 xx 10^(-10), K_(sp) (AgBr) = 5.0 xx 10^(-13)`,
`K_(sp) (Ag_(2)CrO_(4)) = 2.4 xx 10^(-12) ["Use" 3sqrt(0.6) = 0.84]`

Options

A. `AgCI`
B. `AgBr`
C. `Ag_(2)CrO_(4)`
D. all of these

Explanation

To determine which silver compound will yield the highest concentration of silver ions in a saturated solution, we need to analyze the solubility of each compound based on their solubility product constants (Ksp). The compounds we are considering are AgCl, AgBr, and Ag2CrO4.

### Step-by-Step Solution:

1. **Identify the dissociation of each compound**:
- For **AgCl**:
\[ \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]
- For **AgBr**:
\[ \text{AgBr} \rightleftharpoons \text{Ag}^+ + \text{Br}^- \]
- For **Ag2CrO4**:
\[ \text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-} \]

2. **Write the expression for Ksp for each compound**:
- For **AgCl**:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S \cdot S = S^2 \]
Given \( K_{sp} = 1.8 \times 10^{-10} \), we have:
\[ S^2 = 1.8 \times 10^{-10} \implies S = \sqrt{1.8 \times 10^{-10}} = 0.134 \times 10^{-4} \text{ M} \]

- For **AgBr**:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] = S \cdot S = S^2 \]
Given \( K_{sp} = 5.0 \times 10^{-13} \), we have:
\[ S^2 = 5.0 \times 10^{-13} \implies S = \sqrt{5.0 \times 10^{-13}} = 0.007 \times 10^{-4} \text{ M} \]

- For **Ag2CrO4**:
\[ K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] = (2S)^2 \cdot S = 4S^3 \]
Given \( K_{sp} = 2.4 \times 10^{-12} \), we have:
\[ 4S^3 = 2.4 \times 10^{-12} \implies S^3 = \frac{2.4 \times 10^{-12}}{4} = 0.6 \times 10^{-12} \]
Thus,
\[ S = \sqrt[3]{0.6 \times 10^{-12}} \approx 0.84 \times 10^{-4} \text{ M} \]

3. **Compare the solubility of each compound**:
- **AgCl**: \( S = 0.134 \times 10^{-4} \text{ M} \)
- **AgBr**: \( S = 0.007 \times 10^{-4} \text{ M} \)
- **Ag2CrO4**: \( S = 0.84 \times 10^{-4} \text{ M} \)

4. **Conclusion**:
The compound with the highest solubility, and therefore the highest concentration of silver ions, is **Ag2CrO4**.

### Final Answer:
The chemists should use **Ag2CrO4** to prepare a saturated solution with the highest concentration of silver ions.