The solubility product of AgCl is `1.44 xx 10^(-4)` at `100^(@)C`. The solubility of silver chloride in boiling water may be

Options

A. `0.72 xx 10^(-4) M`
B. `1.20 xx 10^(-2)M`
C. `0.72 xx 10^(-2) M`
D. `1.20 xx 10^(-4) M`

Explanation

To solve the problem regarding the solubility of silver chloride (AgCl) in boiling water, we will use the concept of solubility product (Ksp). Here are the steps to find the solubility:

### Step-by-Step Solution:

1. **Understanding the Dissociation of AgCl**:
- Silver chloride (AgCl) dissociates in water according to the following equation:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

2. **Defining Solubility (S)**:
- Let the solubility of AgCl in boiling water be \( S \) mol/L. This means that at equilibrium:
- The concentration of \(\text{Ag}^+\) ions = \( S \)
- The concentration of \(\text{Cl}^-\) ions = \( S \)

3. **Writing the Expression for Ksp**:
- The solubility product (Ksp) for AgCl can be expressed as:
\[ Ksp = [\text{Ag}^+][\text{Cl}^-] = S \cdot S = S^2 \]

4. **Substituting the Given Ksp Value**:
- We are given that the solubility product \( Ksp \) of AgCl at \( 100^\circ C \) is \( 1.44 \times 10^{-4} \).
- Therefore, we can set up the equation:
\[ S^2 = 1.44 \times 10^{-4} \]

5. **Calculating the Solubility (S)**:
- To find \( S \), we take the square root of both sides:
\[ S = \sqrt{1.44 \times 10^{-4}} \]
- Calculating this gives:
\[ S = 1.2 \times 10^{-2} \text{ mol/L} \]

6. **Conclusion**:
- The solubility of silver chloride (AgCl) in boiling water is \( 1.2 \times 10^{-2} \text{ mol/L} \).

### Final Answer:
The solubility of silver chloride in boiling water is \( 1.2 \times 10^{-2} \text{ mol/L} \).

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