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 `[Ag^(o+)]` 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 will follow these steps:

### Step 1: Determine the initial concentrations after mixing the solutions
When we mix 500 mL of 0.01 M AgNO₃ with 500 mL of a solution containing both 0.01 M NaCl and 0.01 M NaBr, the total volume becomes 1000 mL (500 mL + 500 mL).

- The concentration of Ag⁺ ions from AgNO₃:
\[ \text{Initial concentration of Ag}^+ = \frac{0.01 \, \text{mol/L} \times 500 \, \text{mL}}{1000 \, \text{mL}} = 0.005 \, \text{M} = 5 \times 10^{-3} \, \text{M} \]

- The concentration of Cl⁻ ions from NaCl:
\[ \text{Initial concentration of Cl}^- = \frac{0.01 \, \text{mol/L} \times 500 \, \text{mL}}{1000 \, \text{mL}} = 0.005 \, \text{M} = 5 \times 10^{-3} \, \text{M} \]

- The concentration of Br⁻ ions from NaBr:
\[ \text{Initial concentration of Br}^- = \frac{0.01 \, \text{mol/L} \times 500 \, \text{mL}}{1000 \, \text{mL}} = 0.005 \, \text{M} = 5 \times 10^{-3} \, \text{M} \]

### Step 2: Calculate the solubility product (Ksp) expressions
For the precipitation reactions, we can use the Ksp values given:

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

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

### Step 3: Set up the equilibrium expressions
Let \( x \) be the concentration of Ag⁺ that remains in solution at equilibrium after some precipitate forms.

For AgCl:
\[ K_{sp} = x \cdot (5 \times 10^{-3}) = 10^{-10} \]
\[ x = \frac{10^{-10}}{5 \times 10^{-3}} = 2 \times 10^{-8} \, \text{M} \]

For AgBr:
\[ K_{sp} = x \cdot (5 \times 10^{-3}) = 5 \times 10^{-13} \]
\[ x = \frac{5 \times 10^{-13}}{5 \times 10^{-3}} = 1 \times 10^{-10} \, \text{M} \]

### Step 4: Determine the limiting factor
Since AgBr has a lower solubility product than AgCl, it will precipitate first, and thus the concentration of Ag⁺ will be limited by the solubility of AgBr.

### Step 5: Conclusion
The concentration of Ag⁺ at equilibrium will be determined by the Ksp of AgBr, which is:
\[ [Ag^+] = 1 \times 10^{-10} \, \text{M} \]

### Final Answer:
The concentration of Ag⁺ in the equilibrium solution is \( 2 \times 10^{-10} \, \text{M} \).

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