What would be solubility (in mole `L^(-1)`) of AgCl `(K_(sp) = 10^(-10))` in presence of `10^(-1)M AgNO_(3)` ?

Options

A. `10^(-5)`
B. `0.57 xx 10^(-5)`
C. `10^(-9)`
D. `0.83 xx 10^(-5)`

Explanation

To find the solubility of AgCl in the presence of 0.1 M AgNO3, we can follow these steps:

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

AgNO3 dissociates as:
\[ \text{AgNO}_3 (aq) \rightleftharpoons \text{Ag}^+ (aq) + \text{NO}_3^- (aq) \]

### Step 2: Define the solubility of AgCl
Let the solubility of AgCl in the solution be \( S \) (in moles per liter). Therefore, at equilibrium:
- The concentration of \( \text{Ag}^+ \) ions from AgCl will be \( S \).
- The concentration of \( \text{Cl}^- \) ions will also be \( S \).
- The concentration of \( \text{Ag}^+ \) ions from AgNO3 is given as \( 0.1 \, M \).

### Step 3: Calculate the total concentration of \( \text{Ag}^+ \)
The total concentration of \( \text{Ag}^+ \) ions in the solution will be:
\[ [\text{Ag}^+] = S + 0.1 \]

### Step 4: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Substituting the concentrations:
\[ K_{sp} = (S + 0.1)(S) \]

### Step 5: Substitute the value of \( K_{sp} \)
Given that \( K_{sp} = 10^{-10} \):
\[ 10^{-10} = (S + 0.1)(S) \]

### Step 6: Neglect \( S \) compared to \( 0.1 \)
Since \( S \) is expected to be very small compared to \( 0.1 \), we can approximate:
\[ S + 0.1 \approx 0.1 \]
Thus, the equation simplifies to:
\[ 10^{-10} = (0.1)(S) \]

### Step 7: Solve for \( S \)
Rearranging gives:
\[ S = \frac{10^{-10}}{0.1} = 10^{-9} \]

### Conclusion
The solubility of AgCl in the presence of \( 10^{-1} \, M \) AgNO3 is:
\[ S = 10^{-9} \, M \]