A solution is 0.01 M Kl and 0.1 M KCl. If solid `AgNO_(3)` is added to the solution, what is the `[I^(-)]` when AgCl begins to precipitate?
`[K_(SP)(Agl)=1.5xx10^(-16),K_(SP)(AgCl)=1.8xx10^(-10)]`

Options

A. `3.5xx10^(-7)`
B. `6.1xx10^(-8)`
C. `2.2xx10^(-7)`
D. `8.3xx10^(-8)`

Answer

To solve the problem, we need to determine the concentration of iodide ions \([I^-]\) when silver chloride \((AgCl)\) begins to precipitate after adding solid silver nitrate \((AgNO_3)\) to a solution containing potassium iodide \((KI)\) and potassium chloride \((KCl)\).

### Step-by-Step Solution:

1. **Identify the Initial Concentrations**:
- The solution contains:
- \([KI] = 0.01 \, M\) (which provides \([I^-] = 0.01 \, M\))
- \([KCl] = 0.1 \, M\) (which provides \([Cl^-] = 0.1 \, M\))

2. **Dissociation of Added Silver Nitrate**:
- When \(AgNO_3\) is added, it dissociates into:
- \(Ag^+ + NO_3^-\)

3. **Precipitation of Silver Chloride**:
- \(Ag^+\) ions can react with \(Cl^-\) ions to form \(AgCl\) precipitate. The solubility product constant (\(K_{sp}\)) for \(AgCl\) is given as:
\[ K_{sp}(AgCl) = [Ag^+][Cl^-] = 1.8 \times 10^{-10} \]

4. **Calculate the Concentration of \(Ag^+\) at Precipitation**:
- At the point of precipitation, we can express the \(K_{sp}\) in terms of the known concentration of \(Cl^-\):
\[ [Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \, M \]

5. **Precipitation of Silver Iodide**:
- Now, we need to find the concentration of \(I^-\) when \(AgI\) begins to precipitate. The \(K_{sp}\) for \(AgI\) is given as:
\[ K_{sp}(AgI) = [Ag^+][I^-] = 1.5 \times 10^{-16} \]

6. **Calculate the Concentration of \(I^-\)**:
- Using the previously calculated concentration of \(Ag^+\):
\[ [I^-] = \frac{K_{sp}}{[Ag^+]} = \frac{1.5 \times 10^{-16}}{1.8 \times 10^{-9}} \approx 8.33 \times 10^{-8} \, M \]

### Final Answer:
The concentration of iodide ions \([I^-]\) when \(AgCl\) begins to precipitate is approximately:
\[ [I^-] \approx 8.33 \times 10^{-8} \, M \]