If the solubility product of `AgBrO_(3)`, and `AgSO_(4)` are `5.5 xx 10^(-5)` and `2 xx 10^(-5)` respectively, the relationship between the solubility of these salts can be correctly represented as

Options

A. `S_(AgBrO_(3)) gt S_(Ag_(2)SO_(4))`
B. `S_(AgBrO_(3)) = (S_(Ag_(2))SO_(4))`
C. `S_(AgBrO_(3)) lt S_(Ag_(2)SO_(4))`
D. Unpredictable

Answer

To solve the problem, we need to determine the solubility of two salts, AgBrO₃ and Ag₂SO₄, based on their solubility products (Ksp). Let's break down the steps:

### Step 1: Write the dissociation equations for both salts.

1. **For AgBrO₃:**
\[ \text{AgBrO}_3 (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{BrO}_3^- (aq) \]

2. **For Ag₂SO₄:**
\[ \text{Ag}_2\text{SO}_4 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{SO}_4^{2-} (aq) \]

### Step 2: Write the expressions for the solubility products.

1. **For AgBrO₃:**
\[ K_{sp} = [\text{Ag}^+][\text{BrO}_3^-] = S \cdot S = S^2 \]
Given \( K_{sp} = 5.5 \times 10^{-5} \), we have:
\[ S^2 = 5.5 \times 10^{-5} \]

2. **For Ag₂SO₄:**
\[ K_{sp} = [\text{Ag}^+]^2[\text{SO}_4^{2-}] = (2S)^2 \cdot S = 4S^3 \]
Given \( K_{sp} = 2 \times 10^{-5} \), we have:
\[ 4S^3 = 2 \times 10^{-5} \]

### Step 3: Calculate the solubility (S) for each salt.

1. **For AgBrO₃:**
\[ S = \sqrt{5.5 \times 10^{-5}} = \sqrt{5.5} \times 10^{-2.5} \approx 0.0074 \, \text{mol/L} \]

2. **For Ag₂SO₄:**
\[ S^3 = \frac{2 \times 10^{-5}}{4} = 0.5 \times 10^{-5} = 5 \times 10^{-6} \]
\[ S = \sqrt[3]{5 \times 10^{-6}} \approx 0.017 \, \text{mol/L} \]

### Step 4: Compare the solubility of both salts.

- Solubility of AgBrO₃ = 0.0074 mol/L
- Solubility of Ag₂SO₄ = 0.017 mol/L

Since \( 0.017 > 0.0074 \), we conclude that the solubility of Ag₂SO₄ is greater than that of AgBrO₃.

### Final Conclusion:
The relationship between the solubility of these salts can be correctly represented as:
\[ \text{Solubility of } Ag_2SO_4 > \text{Solubility of } AgBrO_3 \]