Determine the solubility of AgCl (in mole/litre) in water. `K_(sp) (AgCl)=1.8 xx 10^(-10)`.

Answer

To determine the solubility of AgCl in water, we need to use the solubility product constant (Ksp) given for AgCl. The Ksp expression for AgCl can be written as:

\[ K_{sp} = [Ag^+] [Cl^-] \]

1. **Write the Dissolution Equation**:
When AgCl dissolves in water, it dissociates into its ions:
\[ AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq) \]

2. **Define the Solubility**:
Let the solubility of AgCl in water be \( s \) moles per liter. Therefore, at equilibrium:
\[ [Ag^+] = s \]
\[ [Cl^-] = s \]

3. **Substitute into the Ksp Expression**:
Now, substitute these concentrations into the Ksp expression:
\[ K_{sp} = [Ag^+] [Cl^-] = s \cdot s = s^2 \]

4. **Set Up the Equation**:
We know that \( K_{sp} = 1.8 \times 10^{-10} \). So we can set up the equation:
\[ s^2 = 1.8 \times 10^{-10} \]

5. **Solve for Solubility (s)**:
To find \( s \), take the square root of both sides:
\[ s = \sqrt{1.8 \times 10^{-10}} \]

6. **Calculate the Value**:
Now, calculate the value:
\[ s \approx 1.34 \times 10^{-5} \, \text{moles/liter} \]

Thus, the solubility of AgCl in water is approximately \( 1.34 \times 10^{-5} \) moles per liter.