Silver ions are added to a solution with `[Br^(-)]=[Cl^(-)]=[CO_(3)^(2-)]=[AsO_(4)^(3-)]`=0.1M. Which compound will precipitate with lowest `[Ag^(+)]`?

Options

A. `AgBr(K_(sp)=5xx10^(-13))`
B. `AgCl(K_(sp)=1.8xx10^(-10))`
C. `Ag2CO_(3)(K_(sp)=8.1xx10^(-12))`
D. `Ag_(3)AsO_(4)(K_(sp)=1xx10^(-22))`

Answer

To determine which compound will precipitate with the lowest concentration of silver ions \([Ag^+]\), we need to analyze the solubility products (\(K_{sp}\)) of the silver salts formed with the given anions: bromide \((Br^-)\), chloride \((Cl^-)\), carbonate \((CO_3^{2-})\), and arsenate \((AsO_4^{3-})\).

### Step-by-Step Solution:

1. **Identify the Compounds**:
- The compounds formed with silver ions are:
- Silver bromide \((AgBr)\)
- Silver chloride \((AgCl)\)
- Silver carbonate \((Ag_2CO_3)\)
- Silver arsenate \((Ag_3AsO_4)\)

2. **Write the Dissociation Equations**:
- For \(AgBr\):
\[ AgBr \rightleftharpoons Ag^+ + Br^- \]
- For \(AgCl\):
\[ AgCl \rightleftharpoons Ag^+ + Cl^- \]
- For \(Ag_2CO_3\):
\[ Ag_2CO_3 \rightleftharpoons 2Ag^+ + CO_3^{2-} \]
- For \(Ag_3AsO_4\):
\[ Ag_3AsO_4 \rightleftharpoons 3Ag^+ + AsO_4^{3-} \]

3. **Set Up the \(K_{sp}\) Expressions**:
- For \(AgBr\):
\[ K_{sp} = [Ag^+][Br^-] \quad (K_{sp} = 5 \times 10^{-13}) \]
- For \(AgCl\):
\[ K_{sp} = [Ag^+][Cl^-] \quad (K_{sp} = 1.8 \times 10^{-10}) \]
- For \(Ag_2CO_3\):
\[ K_{sp} = [Ag^+]^2[CO_3^{2-}] \quad (K_{sp} = 8.1 \times 10^{-12}) \]
- For \(Ag_3AsO_4\):
\[ K_{sp} = [Ag^+]^3[AsO_4^{3-}] \quad (K_{sp} = 1.0 \times 10^{-22}) \]

4. **Calculate the Required \([Ag^+]\) Concentrations**:
- For \(AgBr\):
\[ [Ag^+] = \frac{K_{sp}}{[Br^-]} = \frac{5 \times 10^{-13}}{0.1} = 5 \times 10^{-12} \, M \]
- For \(AgCl\):
\[ [Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \, M \]
- For \(Ag_2CO_3\):
\[ K_{sp} = [Ag^+]^2[CO_3^{2-}] \implies [Ag^+]^2 = \frac{K_{sp}}{[CO_3^{2-}]} = \frac{8.1 \times 10^{-12}}{0.1} = 8.1 \times 10^{-11} \]
\[ [Ag^+] = \sqrt{8.1 \times 10^{-11}} \approx 9 \times 10^{-6} \, M \]
- For \(Ag_3AsO_4\):
\[ K_{sp} = [Ag^+]^3[AsO_4^{3-}] \implies [Ag^+]^3 = \frac{K_{sp}}{[AsO_4^{3-}]} = \frac{1.0 \times 10^{-22}}{0.1} = 1.0 \times 10^{-21} \]
\[ [Ag^+] = \sqrt[3]{1.0 \times 10^{-21}} \approx 1.0 \times 10^{-7} \, M \]

5. **Compare the \([Ag^+]\) Concentrations**:
- \(AgBr\): \(5 \times 10^{-12} \, M\)
- \(AgCl\): \(1.8 \times 10^{-9} \, M\)
- \(Ag_2CO_3\): \(9 \times 10^{-6} \, M\)
- \(Ag_3AsO_4\): \(1.0 \times 10^{-7} \, M\)

6. **Conclusion**:
The compound that will precipitate with the lowest \([Ag^+]\) concentration is **AgBr** with a concentration of \(5 \times 10^{-12} \, M\).

### Final Answer:
**AgBr** will precipitate with the lowest \([Ag^+]\).