The solubility of `AB_(2)` is `0.05` g per 100 mL at `25^@ C`.Calculate `K_(sp)` of `AB_(2)` at `25^(@)C`? [Atomic mass of A = 20 amu, atomic mass of B = 40 amu]

Options

A. `10^(3)`
B. `5 xx 10^(-7)`
C. `10^(-6)`
D. `5 xx 10^(-3)`

Answer

To calculate the solubility product constant \( K_{sp} \) of the salt \( AB_2 \) at \( 25^\circ C \), we will follow these steps:

### Step 1: Calculate the molecular weight of \( AB_2 \)
The molecular weight of \( AB_2 \) can be calculated using the atomic masses provided:
- Atomic mass of \( A = 20 \, \text{amu} \)
- Atomic mass of \( B = 40 \, \text{amu} \)

The molecular weight of \( AB_2 \) is given by:
\[ \text{Molecular weight of } AB_2 = \text{mass of } A + 2 \times \text{mass of } B = 20 + 2 \times 40 = 20 + 80 = 100 \, \text{amu} \]

### Step 2: Convert solubility from grams per 100 mL to moles per liter
The solubility of \( AB_2 \) is given as \( 0.05 \, \text{g} \) per \( 100 \, \text{mL} \). To convert this to moles per liter (mol/L), we use the formula:
\[ \text{Solubility (mol/L)} = \frac{\text{Weight (g)}}{\text{Molecular Weight (g/mol)} \times \text{Volume (L)}} \]
First, convert \( 100 \, \text{mL} \) to liters:
\[ 100 \, \text{mL} = 0.1 \, \text{L} \]
Now, substituting the values:
\[ \text{Solubility} = \frac{0.05 \, \text{g}}{100 \, \text{g/mol} \times 0.1 \, \text{L}} = \frac{0.05}{10} = 0.005 \, \text{mol/L} = 5 \times 10^{-3} \, \text{mol/L} \]

### Step 3: Write the dissociation equation
The dissociation of \( AB_2 \) in water can be represented as:
\[ AB_2 (s) \rightleftharpoons A^{2+} (aq) + 2B^{-} (aq) \]

### Step 4: Set up the expression for \( K_{sp} \)
From the dissociation equation, we can see that:
- For every 1 mole of \( AB_2 \) that dissolves, it produces 1 mole of \( A^{2+} \) and 2 moles of \( B^{-} \).
- If \( s \) is the solubility of \( AB_2 \), then at equilibrium:
- Concentration of \( A^{2+} = s \)
- Concentration of \( B^{-} = 2s \)

Thus, the expression for \( K_{sp} \) is:
\[ K_{sp} = [A^{2+}][B^{-}]^2 = s \cdot (2s)^2 = s \cdot 4s^2 = 4s^3 \]

### Step 5: Substitute the value of \( s \) into the \( K_{sp} \) expression
Now, substituting \( s = 5 \times 10^{-3} \):
\[ K_{sp} = 4 \cdot (5 \times 10^{-3})^3 \]
Calculating \( (5 \times 10^{-3})^3 \):
\[ (5 \times 10^{-3})^3 = 125 \times 10^{-9} = 1.25 \times 10^{-7} \]
Now substituting back:
\[ K_{sp} = 4 \cdot 1.25 \times 10^{-7} = 5 \times 10^{-7} \]

### Final Answer
Thus, the solubility product constant \( K_{sp} \) of \( AB_2 \) at \( 25^\circ C \) is:
\[ \boxed{5 \times 10^{-7}} \]