The solubility product of `AgCl` is `1.8xx10^(-10)` at `18^(@)C`. The solubility of `AgCl` in `0.1 M` solution of sodium chloride would be
(a)`1.26xx10^(-5)M`
(b)`1.8xx10^(-9)M`
(c)`1.6xx10^(-11)M`
(d)zero

Options

A. `1.26xx10^(-5)M`
B. `1.8xx10^(-9)M`
C. `1.6xx10^(-11)M`
D. zero

Explanation

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

### Step 1: Write the dissociation equation for AgCl
AgCl dissociates in water as follows:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

### Step 2: Define the solubility product (Ksp)
The solubility product (Ksp) expression for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Given that \( K_{sp} = 1.8 \times 10^{-10} \) at 18°C.

### Step 3: Set up the concentrations in the presence of NaCl
In a 0.1 M NaCl solution, NaCl completely dissociates into Na\(^+\) and Cl\(^-\) ions:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
This means that the concentration of Cl\(^-\) from NaCl is 0.1 M.

Let \( s \) be the solubility of AgCl in this solution. The concentration of Ag\(^+\) will be \( s \), and the concentration of Cl\(^-\) will be \( s + 0.1 \) (since we have 0.1 M from NaCl).

### Step 4: Substitute into the Ksp expression
Substituting the concentrations into the Ksp expression:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s(s + 0.1) \]
Since \( s \) is expected to be very small compared to 0.1, we can simplify this to:
\[ K_{sp} \approx s(0.1) \]

### Step 5: Solve for s
Now we can substitute the value of Ksp:
\[ 1.8 \times 10^{-10} = s(0.1) \]
\[ s = \frac{1.8 \times 10^{-10}}{0.1} \]
\[ s = 1.8 \times 10^{-9} \, M \]

### Conclusion
The solubility of AgCl in a 0.1 M NaCl solution is:
\[ s = 1.8 \times 10^{-9} \, M \]

Thus, the correct answer is (b) \( 1.8 \times 10^{-9} M \).

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