The concentration of Kl and KCl in a certain solution containing both is 0.001 M each . If 20 mL of this solution is added to 20 mL of a saturated solution of Agl in water . What will happen ?
`K_(sp)` of `AgCl = 10 ^(-10) , K_(sq)` of `AgI = 10^(-16)`

Options

A. Agl will be precipitated
B. AgCl will be precipitated
C. Both AgCl and Agl will be precipitated
D. There will be no precipitate

Answer

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. When these salts dissolve, they dissociate into their respective ions:
- KI → K⁺ + I⁻
- KCl → K⁺ + Cl⁻

Thus, the concentrations of the ions are:
- [I⁻] = 0.001 M
- [Cl⁻] = 0.001 M

### Step 2: Mixing the Solutions
When we mix 20 mL of the KI and KCl solution with 20 mL of a saturated solution of AgI, the total volume becomes 40 mL. The concentrations of the ions will be halved because the volume has doubled:
- [I⁻] after mixing = 0.001 M / 2 = 0.0005 M
- [Cl⁻] after mixing = 0.001 M / 2 = 0.0005 M

### Step 3: Determine the Saturated Concentration of AgI
The solubility product (Ksp) of AgI is given as \(10^{-16}\). The equilibrium expression for AgI is:
\[ K_{sp} = [Ag^+][I^-] \]

Let the solubility of AgI be \(S\). Therefore,
\[ K_{sp} = S \cdot S = S^2 \]
\[ S^2 = 10^{-16} \]
\[ S = 10^{-8} \text{ M} \]

This means in a saturated solution of AgI, the concentration of Ag⁺ is \(10^{-8}\) M and the concentration of I⁻ is also \(10^{-8}\) M.

### Step 4: Calculate the Ion Concentrations After Mixing
After mixing, we have:
- [I⁻] from KI and KCl = 0.0005 M
- [Ag⁺] from the saturated AgI = \(10^{-8}\) M

### Step 5: Calculate the Ion Product for AgI
The ion product (IP) for AgI after mixing can be calculated as:
\[ IP = [Ag^+][I^-] \]
\[ IP = (10^{-8})(0.0005) = 5 \times 10^{-12} \]

### Step 6: Compare IP with Ksp of AgI
Since \(K_{sp}\) of AgI is \(10^{-16}\) and \(IP = 5 \times 10^{-12}\):
\[ IP > K_{sp} \]
This indicates that AgI will precipitate out of the solution.

### Step 7: Analyze the Situation for AgCl
Now we check for AgCl, whose \(K_{sp}\) is \(10^{-10}\):
\[ K_{sp} = [Ag^+][Cl^-] \]
Using the same concentration for Ag⁺ as before (\(10^{-8}\) M) and [Cl⁻] = 0.0005 M:
\[ IP_{AgCl} = (10^{-8})(0.0005) = 5 \times 10^{-12} \]

### Step 8: Compare IP with Ksp of AgCl
Since \(K_{sp}\) of AgCl is \(10^{-10}\) and \(IP_{AgCl} = 5 \times 10^{-12}\):
\[ IP < K_{sp} \]
This indicates that AgCl will not precipitate.

### Conclusion
In conclusion, when 20 mL of the solution containing KI and KCl is mixed with 20 mL of a saturated solution of AgI, AgI will precipitate out of the solution, while AgCl will not.