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)`.
Calculate the `[CI^(Theta)]` in the equilibrium solution.

Options

A. `5 xx 10^(-5)M`
B. `2.5 xx 10^(-5)`
C. `5 xx 10^(-3)M`
D. `2.5 xx 10^(-3)M`

Explanation

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

### Step 1: Calculate the initial concentrations after mixing
When we mix 500 mL of 0.01 M AgNO₃ with 500 mL of a solution that is 0.01 M in both NaCl and NaBr, the total volume becomes 1 L (1000 mL).

- The initial concentration of Ag⁺ from AgNO₃:
\[ [Ag^+] = \frac{0.01 \, \text{mol/L} \times 0.5 \, \text{L}}{1 \, \text{L}} = 0.005 \, \text{mol/L} = 5 \times 10^{-3} \, \text{mol/L} \]

- The initial concentrations of Cl⁻ and Br⁻ from NaCl and NaBr:
\[ [Cl^-] = [Br^-] = \frac{0.01 \, \text{mol/L} \times 0.5 \, \text{L}}{1 \, \text{L}} = 0.005 \, \text{mol/L} = 5 \times 10^{-3} \, \text{mol/L} \]

### Step 2: Determine the solubility product (Ksp) and precipitation
We need to check which salt will precipitate first based on their Ksp values:

- For AgCl:
\[ K_{sp} = [Ag^+][Cl^-] = 10^{-10} \]

- For AgBr:
\[ K_{sp} = [Ag^+][Br^-] = 5 \times 10^{-13} \]

### Step 3: Calculate the concentration of Cl⁻ at the point of precipitation
To find out if AgCl will precipitate, we set up the expression for Ksp:

Using the initial concentrations:
\[ [Ag^+] = 5 \times 10^{-3} \, \text{mol/L}, \quad [Cl^-] = 5 \times 10^{-3} \, \text{mol/L} \]

Calculating the product:
\[ [Ag^+][Cl^-] = (5 \times 10^{-3})(5 \times 10^{-3}) = 25 \times 10^{-6} = 2.5 \times 10^{-5} \]

Since \(2.5 \times 10^{-5} > 10^{-10}\), AgCl will precipitate.

### Step 4: Determine the equilibrium concentration of Cl⁻
When AgCl precipitates, it will consume Cl⁻ ions. We need to find the equilibrium concentration of Cl⁻ after precipitation occurs.

Let \(x\) be the amount of Cl⁻ that precipitates:
\[ K_{sp} = [Ag^+][Cl^-] \implies 10^{-10} = (5 \times 10^{-3})(5 \times 10^{-3} - x) \]

Assuming \(x\) is small compared to \(5 \times 10^{-3}\):
\[ 10^{-10} = (5 \times 10^{-3})(5 \times 10^{-3}) - (5 \times 10^{-3})x \]
\[ 10^{-10} = 2.5 \times 10^{-5} - (5 \times 10^{-3})x \]
\[ (5 \times 10^{-3})x = 2.5 \times 10^{-5} - 10^{-10} \]
\[ (5 \times 10^{-3})x \approx 2.5 \times 10^{-5} \]
\[ x \approx \frac{2.5 \times 10^{-5}}{5 \times 10^{-3}} = 5 \times 10^{-3} \]

Thus, the equilibrium concentration of Cl⁻:
\[ [Cl^-]_{eq} = 5 \times 10^{-3} - x \approx 0 \]

However, since AgBr has a higher Ksp, it will not precipitate until all Cl⁻ is consumed, and thus:
\[ [Cl^-]_{eq} = 5 \times 10^{-3} \, \text{mol/L} \]

### Final Answer
The concentration of \([Cl^-]\) in the equilibrium solution is:
\[ \boxed{5 \times 10^{-3} \, \text{mol/L}} \]