`K_(sp)` of AgCl is `1xx10^(-10)`. Its solubility in 0.1 M `KNO_(3)` will be :

Options

A. `10^(-5)` moles/litre
B. `gt10^(-5)` moles/litre
C. `lt10^(-5)` moles/litre
D. None of these

Answer

To solve the problem of finding the solubility of AgCl in a 0.1 M KNO₃ solution, we can follow these steps:

### Step 1: Understanding the Given Information
We are given the solubility product constant (Ksp) of AgCl, which is \( K_{sp} = 1 \times 10^{-10} \). We also know that KNO₃ is a salt of a strong acid (HNO₃) and a strong base (KOH), and it does not provide any common ions with AgCl.

**Hint:** Identify the components of the solutions and their properties (strong acid/base vs. weak acid/base).

### Step 2: Dissociation of AgCl
AgCl dissociates in water according to the following equilibrium:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

Let the solubility of AgCl in the solution be \( S \) (in moles per liter). Therefore, at equilibrium:
- The concentration of \( \text{Ag}^+ \) ions = \( S \)
- The concentration of \( \text{Cl}^- \) ions = \( S \)

**Hint:** Write the dissociation equation for the salt and express the concentrations of the ions in terms of solubility.

### Step 3: Writing the Expression for Ksp
The Ksp expression for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Substituting the values we have:
\[ K_{sp} = S \times S = S^2 \]

**Hint:** Remember that Ksp is the product of the concentrations of the ions at equilibrium.

### Step 4: Substituting the Ksp Value
We know that:
\[ K_{sp} = 1 \times 10^{-10} \]
Thus, we can set up the equation:
\[ S^2 = 1 \times 10^{-10} \]

**Hint:** Set the Ksp equal to the expression derived from solubility.

### Step 5: Solving for Solubility (S)
To find the solubility \( S \), we take the square root of both sides:
\[ S = \sqrt{1 \times 10^{-10}} \]
Calculating this gives:
\[ S = 10^{-5} \text{ moles per liter} \]

**Hint:** Remember to take the square root to find the solubility.

### Step 6: Conclusion
The solubility of AgCl in 0.1 M KNO₃ is \( 10^{-5} \) moles per liter.

**Final Answer:** The solubility of AgCl in 0.1 M KNO₃ is \( 10^{-5} \) moles per liter.

---