If the solubility of AgCI (formula mass = 143 g/mole) in water at `25^(@)C` is `1.43 xx 10^(-4)` gm/100 ml of solution then the value of `K_(sp)` will be

Options

A. `1 xx 10^(-5)`
B. `2 xx 10^(-5)`
C. `1 xx 10^(-10)`
D. `2 xx 10^(-10)`

Answer

To find the solubility product constant (Ksp) for AgCl given its solubility in water, we can follow these steps:

### Step 1: Understand the Dissociation of AgCl
AgCl dissociates in water as follows:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

### Step 2: Write the Expression for Ksp
The solubility product constant (Ksp) for AgCl can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Since one mole of AgCl produces one mole of Ag\(^+\) and one mole of Cl\(^-\), if the solubility of AgCl is \( S \), then:
\[ K_{sp} = S \times S = S^2 \]

### Step 3: Convert Solubility from g/100 mL to moles/L
Given the solubility of AgCl is \( 1.43 \times 10^{-4} \) g/100 mL, we need to convert this to moles per liter (mol/L).

1. **Convert grams to moles**:
\[ \text{Moles of AgCl} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{1.43 \times 10^{-4} \text{ g}}{143 \text{ g/mol}} \]
\[ = 1.000 \times 10^{-6} \text{ moles} \]

2. **Convert 100 mL to liters**:
\[ 100 \text{ mL} = 0.1 \text{ L} \]

3. **Calculate molarity (moles per liter)**:
\[ S = \frac{1.000 \times 10^{-6} \text{ moles}}{0.1 \text{ L}} = 1.000 \times 10^{-5} \text{ mol/L} \]

### Step 4: Calculate Ksp
Now substitute \( S \) into the Ksp expression:
\[ K_{sp} = S^2 = (1.000 \times 10^{-5})^2 = 1.000 \times 10^{-10} \]

### Step 5: Conclusion
Thus, the value of \( K_{sp} \) for AgCl at 25°C is:
\[ K_{sp} = 1.0 \times 10^{-10} \]