A solution contains `0.1 M` is `Cl^(-)` and `10^(-4) M CrO_(4)^(2-)`. If solid `AgNO_(3)` is gradually added to this solution, what will be the concentration of `Cl^(-)` when `Ag_(2)CrO_(4)` begins to precipitate?
`(Ksp (AgCl) = 10^(-10) M^(2)`,
`K_(sp) (Ag_(2)CrO_(4)) = 10^(-12)M^(3))`

Options

A. `10^(-6)M`
B. `10^(-4) M`
C. `10^(-5) M`
D. `10^(-9) M`

Explanation

To solve the problem, we need to determine the concentration of \( Cl^- \) when \( Ag_2CrO_4 \) begins to precipitate after adding \( AgNO_3 \) to the solution containing \( 0.1 M \, Cl^- \) and \( 10^{-4} M \, CrO_4^{2-} \).

### Step-by-Step Solution:

1. **Identify the Ksp Values**:
- For \( AgCl \): \( K_{sp} = 10^{-10} \, M^2 \)
- For \( Ag_2CrO_4 \): \( K_{sp} = 10^{-12} \, M^3 \)

2. **Determine the Concentration of \( Ag^+ \) for \( Ag_2CrO_4 \) Precipitation**:
- The solubility product expression for \( Ag_2CrO_4 \) is:
\[ K_{sp} = [Ag^+]^2 \cdot [CrO_4^{2-}] \]
- Rearranging gives:
\[ [Ag^+]^2 = \frac{K_{sp}}{[CrO_4^{2-}]} \]
- Substituting the values:
\[ [Ag^+]^2 = \frac{10^{-12}}{10^{-4}} = 10^{-8} \]
- Taking the square root:
\[ [Ag^+] = 10^{-4} \, M \]

3. **Determine the Concentration of \( Cl^- \) at the Precipitation Point of \( AgCl \)**:
- The solubility product expression for \( AgCl \) is:
\[ K_{sp} = [Ag^+] \cdot [Cl^-] \]
- Rearranging gives:
\[ [Cl^-] = \frac{K_{sp}}{[Ag^+]} \]
- Substituting the values:
\[ [Cl^-] = \frac{10^{-10}}{10^{-4}} = 10^{-6} \, M \]

4. **Conclusion**:
- The concentration of \( Cl^- \) when \( Ag_2CrO_4 \) begins to precipitate is \( 10^{-6} \, M \).

### Final Answer:
The concentration of \( Cl^- \) when \( Ag_2CrO_4 \) begins to precipitate is \( 10^{-6} \, M \).