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 will follow these steps:

### Step 1: Understand the given data
We are given:
- The potential of the silver wire in a saturated AgCl solution: \( E = -0.25 \, \text{V} \)
- The standard electrode potential for the Ag/Ag\(^+\) half-cell: \( E^\circ_{Ag/Ag^+} = -0.799 \, \text{V} \)
- The concentration of Cl\(^-\) ions in the solution: \( [Cl^-] = 0.1 \, \text{M} \)

### Step 2: Use the Nernst equation
The Nernst equation relates the cell potential to the concentrations of the ions involved in the half-reaction:
\[ E = E^\circ - \frac{0.0592}{n} \log \left( \frac{1}{[Ag^+]} \right) \]
where \( n \) is the number of electrons transferred (which is 1 for silver).

Substituting the known values into the Nernst equation:
\[ -0.25 = -0.799 - \frac{0.0592}{1} \log \left( [Ag^+] \right) \]

### Step 3: Rearranging the equation
Rearranging the equation to isolate the logarithmic term:
\[ -0.25 + 0.799 = -0.0592 \log \left( [Ag^+] \right) \]
\[ 0.549 = -0.0592 \log \left( [Ag^+] \right) \]

### Step 4: Solve for \([Ag^+]\)
Now, divide both sides by -0.0592:
\[ \log \left( [Ag^+] \right) = \frac{0.549}{-0.0592} \]
Calculating the right-hand side:
\[ \log \left( [Ag^+] \right) = -9.273 \]

Now, to find \([Ag^+]\), we take the antilogarithm:
\[ [Ag^+] = 10^{-9.273} \approx 5.33 \times 10^{-10} \, \text{M} \]

### Step 5: Calculate \( K_{sp} \)
The solubility product constant \( K_{sp} \) for AgCl is given by:
\[ K_{sp} = [Ag^+][Cl^-] \]
Substituting the values we have:
\[ K_{sp} = (5.33 \times 10^{-10})(0.1) = 5.33 \times 10^{-11} \]

### Final Answer
Thus, the \( K_{sp} \) of AgCl in pure water is:
\[ K_{sp} = 5.33 \times 10^{-11} \]

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