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 solve the problem of finding the solubility of AgCl in the presence of 0.1 M AgNO3, 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
Let the solubility of AgCl in the solution be \( S \) (in moles per liter). When AgCl dissolves, it produces \( S \) moles of Ag\(^+\) and \( S \) moles of Cl\(^-\).

### Step 3: Consider the common ion effect
Since we have 0.1 M AgNO3 in the solution, it will dissociate completely to give:
\[ \text{AgNO}_3 (aq) \rightleftharpoons \text{Ag}^+ (aq) + \text{NO}_3^- (aq) \]
This means that the concentration of Ag\(^+\) in the solution due to AgNO3 is 0.1 M.

### Step 4: Set up the expression for Ksp
The solubility product constant \( K_{sp} \) for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Substituting the concentrations:
\[ K_{sp} = (0.1 + S)(S) \]
Given that \( K_{sp} = 10^{-10} \), we can write:
\[ 10^{-10} = (0.1 + S)(S) \]

### Step 5: Neglect the solubility \( S \)
Since the solubility \( S \) is expected to be very small compared to 0.1 M, we can approximate:
\[ 10^{-10} \approx (0.1)(S) \]

### Step 6: Solve for \( S \)
Rearranging the equation gives:
\[ S = \frac{10^{-10}}{0.1} = 10^{-9} \]

### Conclusion
Thus, the solubility of AgCl in the presence of 0.1 M AgNO3 is:
\[ S = 10^{-9} \, \text{mol L}^{-1} \]

### Final Answer
The solubility of AgCl in the presence of 0.1 M AgNO3 is \( 10^{-9} \, \text{mol L}^{-1} \).

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