`K_(sp)` of AgCl in water at `25^(@)C` is `1.8xx10^(-10)`. If `10^(-5)` mole of `Ag^(+)` ions are added to this solution. `K_(sp)` will be:

Options

A. `1.8xx10^(-15)`
B. `1.8xx10^(-5)`
C. `1.8xx10^(-10)`
D. none of these

Explanation

To solve the problem, we need to understand the concept of solubility product constant (\(K_{sp}\)) and how it behaves when additional ions are introduced into a saturated solution.

### Step-by-step Solution:

1. **Understand the Dissociation of AgCl**:
- When silver chloride (AgCl) dissolves in water, it dissociates into silver ions (\(Ag^+\)) and chloride ions (\(Cl^-\)):
\[ AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq) \]

2. **Write the Expression for \(K_{sp}\)**:
- The solubility product constant (\(K_{sp}\)) for AgCl is given by:
\[ K_{sp} = [Ag^+][Cl^-] \]
- It is given that \(K_{sp} = 1.8 \times 10^{-10}\) at \(25^{\circ}C\).

3. **Initial Concentrations**:
- Let the solubility of AgCl in water be \(S\). Therefore, at equilibrium:
\[ [Ag^+] = S \quad \text{and} \quad [Cl^-] = S \]
- Thus, we can express \(K_{sp}\) as:
\[ K_{sp} = S \times S = S^2 \]
- From this, we can find \(S\):
\[ S^2 = 1.8 \times 10^{-10} \implies S = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \text{ M} \]

4. **Adding \(Ag^+\) Ions**:
- We are adding \(10^{-5}\) moles of \(Ag^+\) ions to the solution. This will increase the concentration of \(Ag^+\) ions in the solution.
- The new concentration of \(Ag^+\) will be:
\[ [Ag^+]_{new} = S + 10^{-5} \text{ M} \]

5. **Effect on \(K_{sp}\)**:
- It is important to note that \(K_{sp}\) is a constant at a given temperature and does not change with the addition of \(Ag^+\) ions. Instead, the equilibrium will shift to maintain the \(K_{sp}\) value.
- Therefore, the addition of \(Ag^+\) ions will cause the equilibrium to shift left, leading to the precipitation of AgCl until the product of the concentrations of the ions equals the \(K_{sp}\).

6. **Conclusion**:
- Since \(K_{sp}\) remains constant regardless of the concentration changes due to the addition of \(Ag^+\) ions, the final \(K_{sp}\) value remains:
\[ K_{sp} = 1.8 \times 10^{-10} \]

### Final Answer:
The \(K_{sp}\) of AgCl after adding \(10^{-5}\) moles of \(Ag^+\) ions remains \(1.8 \times 10^{-10}\).