In a saturated aqueous solution of AgBr, concentration of `Ag^(+)` ion is `1xx10^(-6)` mol `L^(-1)` it `K_(sp)` for AgBr is `4xx10^(-13)`, then concentration of `Br^(-)` in solution is

Options

A. `1 xx 10^(-6)`
B. `2 xx 10^(-7)`
C. `4 xx 10^(-7)`
D. `2 xx 10^(-6)`

Answer

To find the concentration of \( Br^- \) in a saturated solution of \( AgBr \), we can follow these steps:

### Step 1: Write the dissociation equation for \( AgBr \)
The dissociation of silver bromide in water can be represented as:
\[ AgBr (s) \rightleftharpoons Ag^+ (aq) + Br^- (aq) \]

### Step 2: Define the solubility product constant (\( K_{sp} \))
The solubility product constant (\( K_{sp} \)) for the dissociation of \( AgBr \) can be expressed as:
\[ K_{sp} = [Ag^+][Br^-] \]
Where:
- \([Ag^+]\) is the concentration of silver ions.
- \([Br^-]\) is the concentration of bromide ions.

### Step 3: Substitute the known values into the \( K_{sp} \) expression
We are given:
- \( K_{sp} = 4 \times 10^{-13} \)
- \([Ag^+] = 1 \times 10^{-6} \, mol \, L^{-1}\)

Substituting these values into the \( K_{sp} \) expression:
\[ 4 \times 10^{-13} = (1 \times 10^{-6}) \times [Br^-] \]

### Step 4: Solve for \([Br^-]\)
To find \([Br^-]\), we rearrange the equation:
\[ [Br^-] = \frac{K_{sp}}{[Ag^+]} = \frac{4 \times 10^{-13}}{1 \times 10^{-6}} \]

### Step 5: Calculate \([Br^-]\)
Now, performing the calculation:
\[ [Br^-] = 4 \times 10^{-13} \times 10^{6} = 4 \times 10^{-7} \, mol \, L^{-1} \]

### Conclusion
The concentration of \( Br^- \) in the solution is:
\[ [Br^-] = 4 \times 10^{-7} \, mol \, L^{-1} \]