The solubility of a solution of AgCI(s) with solubility product `1.6 xx 10^(-10)` in 0.1 M NaCl solution would be :

Options

A. `1.26 xx 10^(-5)` M
B. `1.6 xx 10^(-9)` M
C. `1.6 xx 10^(-11)`M
D. zero

Answer

To find the solubility of AgCl in a 0.1 M NaCl solution, we can follow these steps:

### Step 1: Write the dissociation equation for AgCl
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

### Step 2: Define the solubility (S)
Let the solubility of AgCl in moles per liter be S. This means that when AgCl dissolves, it produces S moles of Ag⁺ and S moles of Cl⁻ ions.

### Step 3: Consider the contribution of Cl⁻ from NaCl
Since we have a 0.1 M NaCl solution, it will provide 0.1 M Cl⁻ ions. Therefore, the total concentration of Cl⁻ ions in the solution will be:
\[ \text{Total } [Cl⁻] = S + 0.1 \]

### Step 4: Write the expression for the solubility product (Ksp)
The solubility product (Ksp) expression for AgCl is given by:
\[ K_{sp} = [Ag⁺][Cl⁻] \]

Substituting the values we have:
\[ K_{sp} = S \cdot (S + 0.1) \]

### Step 5: Substitute the known Ksp value
We know that the Ksp for AgCl is \( 1.6 \times 10^{-10} \). Thus, we can write:
\[ 1.6 \times 10^{-10} = S \cdot (S + 0.1) \]

### Step 6: Make an approximation
Since the Ksp value is very small compared to the concentration of Cl⁻ from NaCl (0.1 M), we can assume that S is much smaller than 0.1 M. Therefore, we can neglect S in the expression:
\[ 1.6 \times 10^{-10} \approx S \cdot 0.1 \]

### Step 7: Solve for S
Now we can solve for S:
\[ S = \frac{1.6 \times 10^{-10}}{0.1} \]
\[ S = 1.6 \times 10^{-9} \text{ M} \]

### Conclusion
The solubility of AgCl in a 0.1 M NaCl solution is \( 1.6 \times 10^{-9} \) M.

### Final Answer
The correct option is B: \( 1.6 \times 10^{-9} \) M.

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