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 (\(K_{sp}\)) for silver chloride (AgCl) given its solubility, we can follow these steps:

### Step 1: Understand the Dissociation of AgCl
When AgCl dissolves in water, it dissociates into silver ions (\(Ag^+\)) and chloride ions (\(Cl^-\)):
\[ AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq) \]
Let the solubility of AgCl be \(S\) moles per liter. Therefore, at equilibrium:
- The concentration of \(Ag^+\) = \(S\)
- The concentration of \(Cl^-\) = \(S\)

### Step 2: Write the Expression for \(K_{sp}\)
The solubility product constant (\(K_{sp}\)) is given by:
\[ K_{sp} = [Ag^+][Cl^-] = S \times S = S^2 \]

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

1. Convert grams to moles using the molar mass of AgCl:
- Molar mass of AgCl = 143 g/mol
- Moles of AgCl in 100 ml:
\[ \text{Moles} = \frac{1.43 \times 10^{-4} \text{ g}}{143 \text{ g/mol}} = \frac{1.43 \times 10^{-4}}{143} \]
\[ = 1.000 \times 10^{-6} \text{ mol} \]

2. Since this is for 100 ml, to find the concentration in 1 L (1000 ml), multiply by 10:
\[ S = 1.000 \times 10^{-6} \text{ mol} \times 10 = 1.000 \times 10^{-5} \text{ mol/L} \]

### Step 4: Calculate \(K_{sp}\)
Now that we have the solubility \(S\) in mol/L, we can find \(K_{sp}\):
\[ K_{sp} = S^2 = (1.000 \times 10^{-5})^2 = 1.000 \times 10^{-10} \]

### Conclusion
Thus, the value of \(K_{sp}\) for AgCl at 25°C is:
\[ K_{sp} = 1.00 \times 10^{-10} \]