The solubility of AgCl is `1xx10^(-5)mol//L`. Its solubility in 0.1 molar sodium chloride solution is

Options

A. `1 xx 10^(-10)`
B. `1 xx 10^(-5)`
C. `1xx 10^(-9)`
D. `1 xx 10^(-4)`

Answer

To find the solubility of AgCl in a 0.1 M sodium chloride solution, 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 (S) of AgCl
Let \( S \) be the solubility of AgCl in moles per liter (mol/L). Therefore, at equilibrium:
- The concentration of \( \text{Ag}^+ \) ions will be \( S \).
- The concentration of \( \text{Cl}^- \) ions will also be \( S \).

### Step 3: Write the expression for the solubility product constant (Ksp)
The solubility product constant \( K_{sp} \) for AgCl can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S \times S = S^2 \]

### Step 4: Substitute the given solubility into the Ksp expression
Given that the solubility of AgCl in pure water is \( 1 \times 10^{-5} \) mol/L, we can calculate \( K_{sp} \):
\[ K_{sp} = (1 \times 10^{-5})^2 = 1 \times 10^{-10} \]

### Step 5: Consider the effect of the common ion (Cl⁻) from NaCl
When AgCl is added to a 0.1 M NaCl solution, the concentration of \( \text{Cl}^- \) ions is already 0.1 M due to the dissociation of NaCl:
\[ \text{NaCl} \rightleftharpoons \text{Na}^+ + \text{Cl}^- \]

### Step 6: Set up the new equilibrium expression
Let \( x \) be the solubility of AgCl in the 0.1 M NaCl solution. At equilibrium:
- The concentration of \( \text{Ag}^+ \) will be \( x \).
- The concentration of \( \text{Cl}^- \) will be \( 0.1 + x \) (but since \( x \) is very small compared to 0.1, we can approximate this as 0.1).

Thus, the Ksp expression becomes:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = x \times (0.1) \]

### Step 7: Substitute Ksp into the equation
We know that \( K_{sp} = 1 \times 10^{-10} \), so we can set up the equation:
\[ 1 \times 10^{-10} = x \times 0.1 \]

### Step 8: Solve for x
Rearranging the equation gives:
\[ x = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9} \]

### Conclusion
The solubility of AgCl in a 0.1 M sodium chloride solution is:
\[ \text{Solubility} = 1 \times 10^{-9} \text{ mol/L} \]

### Final Answer
The correct option is \( 1 \times 10^{-9} \).

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