A solution of 0.1 M in `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?
`[K_(sp)(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

Answer

To solve the problem, we need to determine the concentration of \( Cl^- \) ions when \( Ag_2CrO_4 \) begins to precipitate. We will use the solubility product constants (\( K_{sp} \)) for both \( AgCl \) and \( Ag_2CrO_4 \).

### Step-by-Step Solution:

1. **Identify the given concentrations and \( K_{sp} \) values**:
- Concentration of \( Cl^- = 0.1 \, M \)
- Concentration of \( CrO_4^{2-} = 10^{-4} \, M \)
- \( K_{sp}(AgCl) = 10^{-10} \, M^2 \)
- \( K_{sp}(Ag_2CrO_4) = 10^{-12} \, M^3 \)

2. **Determine the concentration of \( Ag^+ \) ions when \( AgCl \) starts to precipitate**:
- The dissociation of \( AgCl \) can be represented as:
\[ AgCl \rightleftharpoons Ag^+ + Cl^- \]
- The \( K_{sp} \) expression for \( AgCl \) is:
\[ K_{sp} = [Ag^+][Cl^-] \]
- Substituting the known values:
\[ 10^{-10} = [Ag^+](0.1) \]
- Rearranging to find \( [Ag^+] \):
\[ [Ag^+] = \frac{10^{-10}}{0.1} = 10^{-9} \, M \]

3. **Determine the concentration of \( Ag^+ \) ions when \( Ag_2CrO_4 \) starts to precipitate**:
- The dissociation of \( Ag_2CrO_4 \) can be represented as:
\[ Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-} \]
- The \( K_{sp} \) expression for \( Ag_2CrO_4 \) is:
\[ K_{sp} = [Ag^+]^2[CrO_4^{2-}] \]
- Substituting the known values:
\[ 10^{-12} = [Ag^+]^2(10^{-4}) \]
- Rearranging to find \( [Ag^+]^2 \):
\[ [Ag^+]^2 = \frac{10^{-12}}{10^{-4}} = 10^{-8} \]
- Taking the square root to find \( [Ag^+] \):
\[ [Ag^+] = 10^{-4} \, M \]

4. **Calculate the concentration of \( Cl^- \) when \( Ag_2CrO_4 \) begins to precipitate**:
- Using the \( K_{sp} \) expression for \( AgCl \) again:
\[ K_{sp} = [Ag^+][Cl^-] \]
- Substituting the \( [Ag^+] \) we found:
\[ 10^{-10} = (10^{-4})[Cl^-] \]
- Rearranging to find \( [Cl^-] \):
\[ [Cl^-] = \frac{10^{-10}}{10^{-4}} = 10^{-6} \, M \]

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

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