The solubility of AgCl in water at 298 K is `1.06xx10^(-5)` mole per litre. Calculate its solubility product at this temperature.

Explanation

To calculate the solubility product (Ksp) of AgCl at 298 K, we can follow these steps:

### Step 1: Understand the Dissociation of AgCl
AgCl is a sparingly soluble salt that dissociates in water according to the following equation:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

### Step 2: Define the Solubility
Let the solubility of AgCl in water be denoted as \( S \) (in moles per liter). According to the problem, \( S = 1.06 \times 10^{-5} \) mol/L.
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