You are given a `10^(-5)` M NaCl solution and `10^(-8)M AgNO_3` solution . They are mixed in 1:1 volume ratio . `K_(sp)(AgCl)=10^(-5) M^(2)` Choose the correct statement.

Options

A. Precipitation will take place
B. Precipitation will not take place
C. Cannot be determined
D. After 1 hour precipitation will take place

Answer

To solve the problem step by step, we need to analyze the given solutions and their concentrations after mixing.

### Step 1: Understand the initial concentrations
We have two solutions:
- NaCl solution: \( [NaCl] = 10^{-5} \, M \)
- AgNO₃ solution: \( [AgNO_3] = 10^{-8} \, M \)

### Step 2: Mixing the solutions
When we mix equal volumes of both solutions, the total volume doubles. Therefore, the concentration of each ion will be halved.

### Step 3: Calculate the new concentrations after mixing
- For NaCl:
\[ [Na^+] = [Cl^-] = \frac{10^{-5}}{2} = 5 \times 10^{-6} \, M \]

- For AgNO₃:
\[ [Ag^+] = \frac{10^{-8}}{2} = 5 \times 10^{-9} \, M \]

### Step 4: Calculate the ionic product (IP) for AgCl
The ionic product \( IP \) for the precipitation of AgCl can be calculated as follows:
\[ IP = [Ag^+][Cl^-] = (5 \times 10^{-9})(5 \times 10^{-6}) = 25 \times 10^{-15} \]

### Step 5: Compare the ionic product with Ksp
The solubility product constant (\( K_{sp} \)) for AgCl is given as:
\[ K_{sp} = 10^{-5} \, M^2 \]

Now we compare the ionic product with the \( K_{sp} \):
\[ IP = 25 \times 10^{-15} \quad \text{and} \quad K_{sp} = 10^{-5} = 10 \times 10^{-6} \]

Since \( 25 \times 10^{-15} < 10^{-5} \), we conclude that:
\[ IP < K_{sp} \]

### Step 6: Conclusion
Since the ionic product is less than the solubility product, no precipitation of AgCl will occur.

### Final Answer
The correct statement is: **Precipitation will not take place.**

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