The solubility of a springly soluble salt `AB_(2)` in water is `1.0xx10^(-5) mol L^(-1)`. Its solubility product is:

Options

A. `1.0xx10^(-10)M^(3)`
B. `4xx10^(-15)M^(3)`
C. `4xx10^(-10)M^(3)`
D. `1.0xx10^(-15)M^(3)`

Explanation

To find the solubility product (Ksp) of the sparingly soluble salt \( AB_2 \) given its solubility in water, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the salt \( AB_2 \) in water can be represented as:
\[ AB_2 (s) \rightleftharpoons A^{2+} (aq) + 2B^{-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( AB_2 \) be \( S \). According to the problem, \( S = 1.0 \times 10^{-5} \, \text{mol L}^{-1} \).

### Step 3: Determine the concentrations of ions
From the dissociation equation, we can see that:
- For every 1 mole of \( AB_2 \) that dissolves, 1 mole of \( A^{2+} \) and 2 moles of \( B^{-} \) are produced.
- Therefore, at equilibrium:
- The concentration of \( A^{2+} \) will be \( S \).
- The concentration of \( B^{-} \) will be \( 2S \).

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

### Step 5: Substitute the values
Now, substituting \( S = 1.0 \times 10^{-5} \):
\[ Ksp = (1.0 \times 10^{-5}) \times (2 \times 1.0 \times 10^{-5})^2 \]
Calculating \( (2S)^2 \):
\[ (2 \times 1.0 \times 10^{-5})^2 = 4 \times (1.0 \times 10^{-5})^2 = 4 \times 1.0 \times 10^{-10} = 4.0 \times 10^{-10} \]

### Step 6: Calculate Ksp
Now substituting back into the Ksp expression:
\[ Ksp = (1.0 \times 10^{-5}) \times (4.0 \times 10^{-10}) = 4.0 \times 10^{-15} \]

### Final Answer
Thus, the solubility product \( Ksp \) of the salt \( AB_2 \) is:
\[ Ksp = 4.0 \times 10^{-15} \, \text{mol}^3 \text{L}^{-3} \]