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

Options

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

Answer

To solve the problem, we need to calculate the solubility of AgI in a \(10^{-4} \, \text{M}\) solution of KI, given that the solubility product (\(K_{sp}\)) of AgI is \(1.0 \times 10^{-16} \, \text{mol}^2 \, \text{L}^{-2}\).

### Step-by-Step Solution:

1. **Understanding the Dissociation of AgI**:
AgI is a sparingly soluble salt that dissociates in water as follows:
\[ \text{AgI (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \]
Let the solubility of AgI in pure water be \(S\) (in mol/L). At equilibrium, the concentration of \(\text{Ag}^+\) and \(\text{I}^-\) will both be \(S\).

2. **Writing the Expression for \(K_{sp}\)**:
The solubility product \(K_{sp}\) is given by:
\[ K_{sp} = [\text{Ag}^+][\text{I}^-] \]
Therefore, in pure water:
\[ K_{sp} = S \cdot S = S^2 \]
Given \(K_{sp} = 1.0 \times 10^{-16}\):
\[ S^2 = 1.0 \times 10^{-16} \]
\[ S = \sqrt{1.0 \times 10^{-16}} = 1.0 \times 10^{-8} \, \text{mol/L} \]

3. **Considering the Effect of KI**:
When we add \(10^{-4} \, \text{M}\) KI to the solution, it dissociates completely into \(K^+\) and \(I^-\):
\[ \text{KI (s)} \rightarrow \text{K}^+ (aq) + \text{I}^- (aq) \]
Thus, the concentration of \(\text{I}^-\) from KI is \(10^{-4} \, \text{M}\).

4. **Setting Up the New Equilibrium**:
In the presence of \(10^{-4} \, \text{M}\) \(\text{I}^-\), the total concentration of \(\text{I}^-\) becomes:
\[ [\text{I}^-] = S + 10^{-4} \approx 10^{-4} \, \text{M} \quad (\text{since } S \text{ is very small}) \]
Let \(S'\) be the new solubility of AgI in the \(10^{-4} \, \text{M}\) KI solution. At equilibrium:
\[ [\text{Ag}^+] = S' \quad \text{and} \quad [\text{I}^-] = 10^{-4} \, \text{M} \]

5. **Applying the \(K_{sp}\) Expression**:
Now we can write the \(K_{sp}\) expression again:
\[ K_{sp} = [\text{Ag}^+][\text{I}^-] = S' \cdot (10^{-4}) \]
Substituting the value of \(K_{sp}\):
\[ 1.0 \times 10^{-16} = S' \cdot (10^{-4}) \]

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

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