Calculate the solubility of silver chloride in water at room temperature if the solubility product of AgCl is `1.6xx10^(-10)`.

Answer

To calculate the solubility of silver chloride (AgCl) in water at room temperature using the given solubility product (Ksp), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Dissociation of AgCl**:
Silver chloride (AgCl) is a sparingly soluble salt that dissociates in water according to the following equation:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]
From this equation, we can see that for every mole of AgCl that dissolves, one mole of Ag\(^+\) ions and one mole of Cl\(^-\) ions are produced.

2. **Define Solubility**:
Let the solubility of AgCl in water be represented by \( S \). This means that at equilibrium:
- The concentration of Ag\(^+\) ions will be \( S \) moles per liter.
- The concentration of Cl\(^-\) ions will also be \( S \) moles per liter.

3. **Write the Expression for Ksp**:
The solubility product (Ksp) expression for AgCl can be written as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Substituting the concentrations from step 2, we have:
\[ K_{sp} = S \cdot S = S^2 \]

4. **Substitute the Given Ksp Value**:
We are given that \( K_{sp} = 1.6 \times 10^{-10} \). Therefore, we can set up the equation:
\[ S^2 = 1.6 \times 10^{-10} \]

5. **Solve for S**:
To find the solubility \( S \), we take the square root of both sides:
\[ S = \sqrt{1.6 \times 10^{-10}} \]

6. **Calculate the Square Root**:
Performing the calculation:
\[ S \approx 1.26 \times 10^{-5} \text{ moles per liter} \]

### Final Answer:
The solubility of silver chloride (AgCl) in water at room temperature is approximately:
\[ S \approx 1.26 \times 10^{-5} \text{ moles per liter} \]