The solubility product of `AgI` at `25^(@)C` is `1.0xx10^(-16) mol^(2) L^(-2)`. The solubility of `AgI` in `10^(-4) N` solution of `KI` at `25^(@)C` is approximately ( in `mol L^(-1)`)

Options

A. `1.0xx10^(-10)`
B. `1.0xx10^(-8)`
C. `1.0xx10^(-16)`
D. `1.0xx10^(12)`

Explanation

To solve the problem, we need to find the solubility of AgI in a 10^(-4) N solution of KI, given that the solubility product (Ksp) of AgI is 1.0 x 10^(-16) mol² L^(-2).

### Step-by-Step Solution:

1. **Understand the Dissociation of KI:**
KI dissociates in solution to give K⁺ and I⁻ ions:
\[ KI \rightarrow K^+ + I^- \]
Since the concentration of KI is 10^(-4) N, we have:
\[ [I^-] = 10^{-4} \text{ mol L}^{-1} \]

2. **Dissociation of AgI:**
AgI dissociates in solution to give Ag⁺ and I⁻ ions:
\[ AgI \rightleftharpoons Ag^+ + I^- \]
Let the solubility of AgI in this solution be \( x \) mol L^(-1). Therefore, at equilibrium:
\[ [Ag^+] = x \text{ mol L}^{-1} \]
\[ [I^-] = 10^{-4} + x \text{ mol L}^{-1} \]

3. **Setting Up the Solubility Product Expression:**
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [Ag^+][I^-] \]
Substituting the values we have:
\[ K_{sp} = x(10^{-4} + x) \]
Given \( K_{sp} = 1.0 \times 10^{-16} \):
\[ 1.0 \times 10^{-16} = x(10^{-4} + x) \]

4. **Assuming x is Small:**
Since \( x \) (the solubility of AgI) is expected to be very small compared to \( 10^{-4} \), we can approximate:
\[ 10^{-4} + x \approx 10^{-4} \]
Thus, the equation simplifies to:
\[ 1.0 \times 10^{-16} = x \cdot 10^{-4} \]

5. **Solving for x:**
Rearranging the equation gives:
\[ x = \frac{1.0 \times 10^{-16}}{10^{-4}} = 1.0 \times 10^{-12} \text{ mol L}^{-1} \]

### Final Answer:
The solubility of AgI in a 10^(-4) N solution of KI at 25°C is approximately:
\[ \boxed{1.0 \times 10^{-12} \text{ mol L}^{-1}} \]