The following solutions are mixed: `500mL of 0.01 M AgNO_(3)` and `500mL` solution that was both `0.01M` in `NaCI` and `0.01M` in `NaBr`. Given `K_(sp) AGCI = 10^(-10), K_(sp) AgBr = 5 xx 10^(-13)`,`[Ag^(+)] = 2 xx 10^(-8)`.
Calculate the `[Br^(Theta)]` in the equilibrium solution.

Options

A. `2.0 xx10^(-8)M`
B. `2.0 xx 10^(-10)M`
C. `2.5 xx 10^(-5)M`
D. `2.5 xx 10^(-8)M`

Explanation

To solve the problem, we need to find the concentration of bromide ions \([Br^-]\) in the equilibrium solution after mixing the given solutions. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Mixing of Solutions
We are mixing:
- 500 mL of 0.01 M AgNO₃
- 500 mL of a solution that is 0.01 M in both NaCl and NaBr

### Step 2: Calculate the Moles of Ag⁺
First, we calculate the moles of Ag⁺ from AgNO₃:
\[ \text{Moles of Ag}^+ = \text{Concentration} \times \text{Volume} = 0.01 \, \text{M} \times 0.5 \, \text{L} = 0.005 \, \text{moles} \]

### Step 3: Calculate the Total Volume of the Mixture
The total volume after mixing the two solutions is:
\[ \text{Total Volume} = 500 \, \text{mL} + 500 \, \text{mL} = 1000 \, \text{mL} = 1 \, \text{L} \]

### Step 4: Calculate the Concentration of Ag⁺ in the Mixture
The concentration of Ag⁺ after mixing is:
\[ [\text{Ag}^+] = \frac{\text{Moles of Ag}^+}{\text{Total Volume}} = \frac{0.005 \, \text{moles}}{1 \, \text{L}} = 0.005 \, \text{M} = 5 \times 10^{-3} \, \text{M} \]

### Step 5: Write the Expression for Ksp of AgBr
The solubility product constant \(K_{sp}\) for AgBr is given by:
\[ K_{sp} = [Ag^+][Br^-] \]
Given \(K_{sp} = 5 \times 10^{-13}\).

### Step 6: Substitute the Known Values
We know that \([Ag^+] = 2 \times 10^{-8} \, \text{M}\) (as given in the problem). We can substitute this into the \(K_{sp}\) expression:
\[ 5 \times 10^{-13} = (2 \times 10^{-8})[Br^-] \]

### Step 7: Solve for [Br⁻]
Rearranging the equation to find \([Br^-]\):
\[ [Br^-] = \frac{5 \times 10^{-13}}{2 \times 10^{-8}} = 2.5 \times 10^{-5} \, \text{M} \]

### Final Answer
The concentration of bromide ions \([Br^-]\) in the equilibrium solution is:
\[ \boxed{2.5 \times 10^{-5} \, \text{M}} \]