Solubility of salt `A_(2)B_(3)` is `1 xx 10^(-4)`, its solubility product is

Options

A. `1.08 xx 10^(20)`
B. `1.08 xx 10^(18)`
C. `2.6 xx 10^(-18)`
D. `1.08 xx 10^(-18)`

Explanation

To find the solubility product (Ksp) of the salt \( A_2B_3 \) given its solubility, we can follow these steps:

### Step 1: Write the dissociation equation
The salt \( A_2B_3 \) dissociates in water according to the following equation:
\[ A_2B_3 (s) \rightleftharpoons 2A^{3+} (aq) + 3B^{2-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( A_2B_3 \) be \( S \). According to the problem, \( S = 1 \times 10^{-4} \).

### Step 3: Determine the concentrations of the ions at equilibrium
From the dissociation equation, we can see that:
- For every 1 mole of \( A_2B_3 \) that dissolves, 2 moles of \( A^{3+} \) and 3 moles of \( B^{2-} \) are produced.
- Therefore, at equilibrium:
- The concentration of \( A^{3+} \) will be \( 2S \)
- The concentration of \( B^{2-} \) will be \( 3S \)

Substituting \( S = 1 \times 10^{-4} \):
- Concentration of \( A^{3+} = 2S = 2 \times 1 \times 10^{-4} = 2 \times 10^{-4} \)
- Concentration of \( B^{2-} = 3S = 3 \times 1 \times 10^{-4} = 3 \times 10^{-4} \)

### Step 4: Write the expression for the solubility product (Ksp)
The solubility product \( Ksp \) is given by the expression:
\[ Ksp = [A^{3+}]^2 \cdot [B^{2-}]^3 \]

### Step 5: Substitute the equilibrium concentrations into the Ksp expression
Now substituting the concentrations:
\[ Ksp = (2S)^2 \cdot (3S)^3 \]
\[ Ksp = (2 \times 10^{-4})^2 \cdot (3 \times 10^{-4})^3 \]

### Step 6: Calculate Ksp
Calculating each part:
\[ (2 \times 10^{-4})^2 = 4 \times 10^{-8} \]
\[ (3 \times 10^{-4})^3 = 27 \times 10^{-12} \]

Now, multiply these two results:
\[ Ksp = 4 \times 10^{-8} \cdot 27 \times 10^{-12} = 108 \times 10^{-20} \]
\[ Ksp = 1.08 \times 10^{-18} \]

### Final Answer
Thus, the solubility product \( Ksp \) of salt \( A_2B_3 \) is:
\[ Ksp = 1.08 \times 10^{-18} \]