`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 the solubility product constant (Ksp) and how it behaves when additional ions are added to a saturated solution.

### Step-by-Step Solution:

1. **Understanding Ksp**: The solubility product constant (Ksp) for AgCl is given as \( K_{sp} = 1.8 \times 10^{-10} \). This value represents the equilibrium constant for the dissolution of AgCl in water, which can be expressed as:
\[ K_{sp} = [Ag^+][Cl^-] \]

2. **Initial Conditions**: In pure water, the solubility of AgCl is very low, and at equilibrium, we can assume that the concentrations of \( Ag^+ \) and \( Cl^- \) ions are equal. Let's denote the solubility of AgCl as \( S \). Therefore, at equilibrium:
\[ K_{sp} = S \times S = S^2 \]
From this, we can find the solubility \( S \):
\[ S^2 = 1.8 \times 10^{-10} \implies S = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \text{ M} \]

3. **Adding \( Ag^+ \) Ions**: Now, we are adding \( 10^{-5} \) moles of \( Ag^+ \) ions to the solution. This will increase the concentration of \( Ag^+ \) ions in the solution.

4. **New Concentration of \( Ag^+ \)**: The concentration of \( Ag^+ \) ions after adding \( 10^{-5} \) moles will be:
\[ [Ag^+] = S + 10^{-5} \approx 1.34 \times 10^{-5} + 1.0 \times 10^{-5} = 2.34 \times 10^{-5} \text{ M} \]

5. **Determining \( Cl^- \) Concentration**: Since \( K_{sp} \) is a constant at a given temperature, the addition of \( Ag^+ \) will shift the equilibrium to the left, causing some \( AgCl \) to precipitate out. The concentration of \( Cl^- \) will remain approximately equal to the solubility of AgCl, which we calculated as \( S \approx 1.34 \times 10^{-5} \text{ M} \).

6. **Calculating New Ksp**: The Ksp remains constant at a given temperature, regardless of the concentrations of the ions in the solution. Therefore, even after adding \( Ag^+ \), the value of \( K_{sp} \) will still be:
\[ K_{sp} = 1.8 \times 10^{-10} \]

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