The volume of water needed to prepare a satured solution of `Ag^(o+)` having maximum `[Ag^(o+)]` ion by selecting one out of three slats form:
`AgC1(K_(sp) = 2.0 xx 10^(-10)), AgBr (K_(sp) = 5 xx 10^(-13))`, and `Ag_(2)CrO_(4)(K_(sp) = 2.4 xx 10^(-12))`. whcih compound should be used to have maximum `[Ag^(o+)]`?

Options

A. `AgC1`
B. `AgBr`
C. `Ag_(2)CrO_(4)`
D. Any one of them

Explanation

To determine which compound will yield the maximum concentration of Ag⁺ ions in a saturated solution, we will calculate the solubility of each compound based on their given Ksp values. The three compounds to consider are AgCl, AgBr, and Ag₂CrO₄.

### Step-by-Step Solution:

1. **Calculate the solubility of AgCl:**
- Dissociation: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
- Let the solubility of AgCl be S.
- Therefore, [Ag⁺] = S and [Cl⁻] = S.
- Ksp expression: Ksp = [Ag⁺][Cl⁻] = S × S = S².
- Given Ksp for AgCl = 2.0 × 10⁻¹⁰.
- Set up the equation: S² = 2.0 × 10⁻¹⁰.
- Solve for S:
\[ S = \sqrt{2.0 \times 10^{-10}} = 1.41 \times 10^{-5} \text{ M} \]

2. **Calculate the solubility of AgBr:**
- Dissociation: AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq)
- Let the solubility of AgBr be S.
- Therefore, [Ag⁺] = S and [Br⁻] = S.
- Ksp expression: Ksp = [Ag⁺][Br⁻] = S × S = S².
- Given Ksp for AgBr = 5.0 × 10⁻¹³.
- Set up the equation: S² = 5.0 × 10⁻¹³.
- Solve for S:
\[ S = \sqrt{5.0 \times 10^{-13}} = 7.07 \times 10^{-7} \text{ M} \]

3. **Calculate the solubility of Ag₂CrO₄:**
- Dissociation: Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)
- Let the solubility of Ag₂CrO₄ be S.
- Therefore, [Ag⁺] = 2S and [CrO₄²⁻] = S.
- Ksp expression: Ksp = [Ag⁺]²[CrO₄²⁻] = (2S)² × S = 4S³.
- Given Ksp for Ag₂CrO₄ = 2.4 × 10⁻¹².
- Set up the equation: 4S³ = 2.4 × 10⁻¹².
- Solve for S:
\[ S³ = \frac{2.4 \times 10^{-12}}{4} = 6.0 \times 10^{-13} \\ S = \sqrt[3]{6.0 \times 10^{-13}} \approx 8.43 \times 10^{-5} \text{ M} \]
- Therefore, [Ag⁺] = 2S = 2 × 8.43 × 10⁻⁵ = 1.686 × 10⁻⁴ M.

4. **Compare the concentrations of Ag⁺ ions:**
- For AgCl: [Ag⁺] = 1.41 × 10⁻⁵ M
- For AgBr: [Ag⁺] = 7.07 × 10⁻⁷ M
- For Ag₂CrO₄: [Ag⁺] = 1.686 × 10⁻⁴ M

5. **Conclusion:**
- The maximum concentration of Ag⁺ ions is from Ag₂CrO₄, which is 1.686 × 10⁻⁴ M. Therefore, to prepare a saturated solution with the maximum [Ag⁺], Ag₂CrO₄ should be used.