To solve the problem, we need to analyze the situation step by step.
### Step 1: Understand the Initial Concentrations
We have a solution containing both KI and KCl, each at a concentration of 0.001 M (or 10^(-3) M). When these solutions are mixed, we need to consider the dissociation of these salts:
- **KI dissociates into:**
\[
K^+ + I^-
\]
- **KCl dissociates into:**
\[
K^+ + Cl^-
\]
### Step 2: Calculate the Total Volume After Mixing
When 20 mL of the KI and KCl solution is added to 20 mL of a saturated AgI solution, the total volume becomes:
\[
20 \, \text{mL} + 20 \, \text{mL} = 40 \, \text{mL}
\]
### Step 3: Calculate the New Concentrations After Mixing
Since the volume has doubled, the concentrations of KI and KCl will be halved:
- New concentration of \( K^+ \) and \( I^- \) from KI:
\[
\frac{0.001 \, \text{M}}{2} = 0.0005 \, \text{M} \, (5 \times 10^{-4} \, \text{M})
\]
- New concentration of \( Cl^- \) from KCl:
\[
\frac{0.001 \, \text{M}}{2} = 0.0005 \, \text{M} \, (5 \times 10^{-4} \, \text{M})
\]
### Step 4: Determine the Solubility of AgI
The solubility product (\( K_{sp} \)) of AgI is given as \( 10^{-16} \). If we let the solubility of AgI be \( S \), then:
\[
K_{sp} = [Ag^+][I^-] = S^2
\]
\[
S^2 = 10^{-16} \implies S = 10^{-8} \, \text{M}
\]
### Step 5: Calculate the Concentration of Iodide Ion
After mixing, the concentration of \( I^- \) will include contributions from both the saturated AgI solution and the KI solution:
- From AgI:
\[
\frac{10^{-8}}{2} = 5 \times 10^{-9} \, \text{M}
\]
- From KI:
\[
5 \times 10^{-4} \, \text{M}
\]
- Total \( I^- \) concentration:
\[
I^- = 5 \times 10^{-9} + 5 \times 10^{-4} = 5.005 \times 10^{-4} \, \text{M}
\]
### Step 6: Calculate the Ionic Product for AgI
The ionic product (\( IP \)) for AgI after mixing is:
\[
IP = [Ag^+][I^-] = \left(\frac{10^{-8}}{2}\right) \times \left(5.005 \times 10^{-4}\right)
\]
Calculating:
\[
IP = 5 \times 10^{-9} \times 5.005 \times 10^{-4} = 2.5025 \times 10^{-12}
\]
### Step 7: Compare Ionic Product with Ksp
Since \( K_{sp} \) for AgI is \( 10^{-16} \):
\[
IP (2.5025 \times 10^{-12}) > K_{sp} (10^{-16})
\]
This means that AgI will precipitate.
### Step 8: Calculate the Ionic Product for AgCl
Now, we need to check if AgCl will precipitate:
- The \( K_{sp} \) of AgCl is \( 10^{-10} \).
- The ionic product for AgCl is:
\[
IP = [Ag^+][Cl^-] = \left(\frac{10^{-8}}{2}\right) \times \left(5 \times 10^{-4}\right)
\]
Calculating:
\[
IP = 5 \times 10^{-9} \times 5 \times 10^{-4} = 2.5 \times 10^{-12}
\]
### Step 9: Compare Ionic Product with Ksp for AgCl
Since \( IP (2.5 \times 10^{-12}) < K_{sp} (10^{-10}) \):
This means that AgCl will not precipitate.
### Conclusion
The only precipitate that will form is AgI.
### Final Answer
**AgI will precipitate, but AgCl will not.**
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