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. **Understanding the Dissociation of AgI**:
AgI dissociates in water as follows:
\[ \text{AgI (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \]
The solubility product (Ksp) expression for this equilibrium is:
\[ K_{sp} = [\text{Ag}^+][\text{I}^-] \]

2. **Setting Up the Problem**:
In a 10^(-4) N solution of KI, the concentration of I⁻ ions is 10^(-4) M because KI completely dissociates:
\[ \text{KI} \rightarrow \text{K}^+ + \text{I}^- \]
Therefore, we have:
\[ [\text{I}^-] = 10^{-4} \, \text{M} \]

3. **Substituting into the Ksp Expression**:
Let the solubility of AgI in this solution be \( s \) (in mol/L). Thus, the concentration of Ag⁺ ions will also be \( s \) since one mole of AgI produces one mole of Ag⁺. The Ksp expression becomes:
\[ K_{sp} = [\text{Ag}^+][\text{I}^-] = s \cdot (10^{-4}) \]

4. **Using the Value of Ksp**:
We know that \( K_{sp} = 1.0 \times 10^{-16} \). Therefore, we can set up the equation:
\[ 1.0 \times 10^{-16} = s \cdot (10^{-4}) \]

5. **Solving for s**:
Rearranging the equation to solve for \( s \):
\[ s = \frac{1.0 \times 10^{-16}}{10^{-4}} = 1.0 \times 10^{-12} \, \text{mol/L} \]

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

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