A silver wire dipped in `0.1M HCI` solution saturated with `AgCI` develops a potential of `-0.25V`. If `E_(Ag//Ag^(+))^(@) =- 0.799V`, the `K_(sp)` of `AgCI` in pure water will be

Options

A. `2.95 xx 10^(-11)`
B. `5.1 xx 10^(-11)`
C. `3.95 xx 10^(-11)`
D. `1.95 xx 10^(-11)`

Explanation

To solve the problem, we need to determine the solubility product constant (Ksp) of AgCl in pure water using the given information. Here’s a step-by-step solution:

### Step 1: Understand the Reaction and Given Data
The dissolution of silver chloride (AgCl) can be represented as:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

From the question, we have:
- Concentration of Cl⁻ = 0.1 M (from HCl solution)
- Standard reduction potential for Ag⁺/Ag = -0.799 V
- The potential developed by the silver wire = -0.25 V

### Step 2: Use the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \]

For our reaction, the Nernst equation becomes:
\[ E = E^\circ - \frac{0.0591}{1} \log \frac{[\text{Ag}^+]}{1} \]

Where:
- \( E = -0.25 \, \text{V} \)
- \( E^\circ = -0.799 \, \text{V} \)
- \( n = 1 \) (since one electron is involved in the reaction)

### Step 3: Substitute Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ -0.25 = -0.799 - 0.0591 \log [\text{Ag}^+] \]

### Step 4: Rearranging the Equation
Rearranging the equation to isolate the logarithmic term:
\[ -0.25 + 0.799 = -0.0591 \log [\text{Ag}^+ \]
\[ 0.549 = -0.0591 \log [\text{Ag}^+] \]

### Step 5: Solve for \([\text{Ag}^+]\)
Now, divide both sides by -0.0591:
\[ \log [\text{Ag}^+] = \frac{0.549}{-0.0591} \]
Calculating the right-hand side:
\[ \log [\text{Ag}^+] \approx -9.3 \]
Now, converting from logarithmic form to concentration:
\[ [\text{Ag}^+] = 10^{-9.3} \approx 5.01 \times 10^{-10} \, \text{M} \]

### Step 6: Calculate Ksp
The Ksp expression for AgCl is:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Substituting the values:
\[ K_{sp} = (5.01 \times 10^{-10})(0.1) = 5.01 \times 10^{-11} \]

### Final Answer
Thus, the Ksp of AgCl in pure water is approximately:
\[ K_{sp} \approx 5.01 \times 10^{-11} \]