The solubility product of silver chloride is `1.8 xx 10^(-10)` at 298 K. The solubility of AgCl in 0.01 M HCl solution in mol `// dm^(3)` is

Options

A. `2.4 xx 10^(-9)`
B. `3.6 xx 10^(-8)`
C. `0.9 xx 10^(-10)`
D. `1.8 xx 10^(-8)`

Explanation

To find the solubility of silver chloride (AgCl) in a 0.01 M HCl solution, we can follow these steps:

### Step 1: Understand the solubility product (Ksp)
The solubility product (Ksp) of AgCl is given as \(1.8 \times 10^{-10}\) at 298 K. The dissociation of AgCl in water can be represented as:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]
The Ksp expression for this equilibrium is:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]

### Step 2: Determine the concentration of Cl⁻ in the solution
In a 0.01 M HCl solution, HCl fully dissociates to provide Cl⁻ ions. Therefore, the concentration of Cl⁻ in the solution is:
\[ [\text{Cl}^-] = 0.01 \, \text{M} \]

### Step 3: Set up the expression for Ksp
Let the solubility of AgCl in the 0.01 M HCl solution be \(X\) mol/dm³. This means that the concentration of Ag⁺ ions will be \(X\) mol/dm³, and the concentration of Cl⁻ ions will be:
\[ [\text{Cl}^-] = 0.01 + X \approx 0.01 \, \text{M} \quad (\text{since } X \text{ is very small compared to } 0.01) \]

### Step 4: Substitute into the Ksp expression
Substituting into the Ksp expression:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = X \cdot 0.01 \]
Setting this equal to the Ksp value:
\[ 1.8 \times 10^{-10} = X \cdot 0.01 \]

### Step 5: Solve for X
Now, solve for \(X\):
\[ X = \frac{1.8 \times 10^{-10}}{0.01} = 1.8 \times 10^{-8} \, \text{mol/dm}^3 \]

### Final Answer
The solubility of AgCl in 0.01 M HCl solution is:
\[ \boxed{1.8 \times 10^{-8} \, \text{mol/dm}^3} \]

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