Solubility of `AgCl` at `20^(@)C` is `1.435xx10^(-3) g per litre`. The solubility product of `AgCl` is

Options

A. `1xx10^(-5)`
B. `1xx10^(-10)`
C. `1.435xx10^(-5)`
D. `108xx10^(-3)`

Explanation

To find the solubility product (Ksp) of AgCl given its solubility, we will follow these steps:

### Step 1: Convert the solubility from grams per liter to moles per liter
The solubility of AgCl at 20°C is given as \(1.435 \times 10^{-3}\) g/L. We need to convert this to moles per liter (mol/L).

**Molar mass of AgCl:**
- Silver (Ag) = 107.87 g/mol
- Chlorine (Cl) = 35.45 g/mol
- Molar mass of AgCl = 107.87 + 35.45 = 143.32 g/mol

**Calculation of moles:**
\[ \text{Moles of AgCl} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ \text{Moles of AgCl} = \frac{1.435 \times 10^{-3} \text{ g}}{143.32 \text{ g/mol}} \approx 1.00 \times 10^{-5} \text{ mol/L} \]

### Step 2: Write the dissociation equation for AgCl
When AgCl dissolves in water, it dissociates into its ions:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

### Step 3: Determine the concentration of ions
From the dissociation, we see that for every mole of AgCl that dissolves, it produces one mole of Ag\(^+\) and one mole of Cl\(^-\). Therefore, if the solubility (S) of AgCl is \(1.00 \times 10^{-5}\) mol/L, then:
\[ [\text{Ag}^+] = S = 1.00 \times 10^{-5} \text{ mol/L} \]
\[ [\text{Cl}^-] = S = 1.00 \times 10^{-5} \text{ mol/L} \]

### Step 4: Write the expression for the solubility product (Ksp)
The solubility product constant (Ksp) for AgCl is given by:
\[ Ksp = [\text{Ag}^+][\text{Cl}^-] \]
Substituting the concentrations:
\[ Ksp = (1.00 \times 10^{-5})(1.00 \times 10^{-5}) = 1.00 \times 10^{-10} \]

### Final Answer
The solubility product (Ksp) of AgCl at 20°C is:
\[ \boxed{1.00 \times 10^{-10}} \]

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