If the solubility products of AgCl and AgBr are `1.2 xx 10^(-10)` and `3.5 xx 10^(-13)` respectively, then the relation between the solubilities (denoted by the symbol'S') of these

Options

A. S of AgBr is less than that of AgCl
B. S of AgBr is greater than that of AgCl
C. S of AgBr is equal to that of AgCl
D. S of AgBr is `10^(6)` times greater than that of AgCl

Explanation

To determine the relationship between the solubilities of AgCl and AgBr based on their solubility products (Ksp), we can follow these steps:

### Step 1: Write the dissociation equations and Ksp expressions
For AgCl:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]
The solubility product expression is:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S^2 \]
where \( S \) is the solubility of AgCl.

For AgBr:
\[ \text{AgBr (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Br}^- (aq) \]
The solubility product expression is:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] = S^2 \]
where \( S \) is the solubility of AgBr.

### Step 2: Substitute the given Ksp values
Given:
- \( K_{sp} \) of AgCl = \( 1.2 \times 10^{-10} \)
- \( K_{sp} \) of AgBr = \( 3.5 \times 10^{-13} \)

For AgCl:
\[ S_{AgCl}^2 = 1.2 \times 10^{-10} \]
\[ S_{AgCl} = \sqrt{1.2 \times 10^{-10}} \]

For AgBr:
\[ S_{AgBr}^2 = 3.5 \times 10^{-13} \]
\[ S_{AgBr} = \sqrt{3.5 \times 10^{-13}} \]

### Step 3: Calculate the solubilities
Calculating \( S_{AgCl} \):
\[ S_{AgCl} = \sqrt{1.2 \times 10^{-10}} \approx 1.095 \times 10^{-5} \, \text{mol/L} \]

Calculating \( S_{AgBr} \):
\[ S_{AgBr} = \sqrt{3.5 \times 10^{-13}} \approx 5.916 \times 10^{-7} \, \text{mol/L} \]

### Step 4: Compare the solubilities
Now we compare the two solubilities:
- \( S_{AgCl} \approx 1.095 \times 10^{-5} \, \text{mol/L} \)
- \( S_{AgBr} \approx 5.916 \times 10^{-7} \, \text{mol/L} \)

Since \( S_{AgCl} > S_{AgBr} \), we conclude that:
\[ S_{AgBr} < S_{AgCl} \]

### Conclusion
The solubility of AgBr is less than the solubility of AgCl.