The solubility of `A_(2)B_(3)` is "x mol dm"^(-3)`. Its `K_(sp)` is

Options

A. `6x^(4)`
B. `64x^(4)`
C. `36x^(5)`
D. `108x^(5)`

Explanation

To find the solubility product constant (Ksp) for the salt \( A_2B_3 \) given its solubility \( x \) mol/dm³, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( A_2B_3 \) in water can be expressed as:
\[ A_2B_3 (s) \rightleftharpoons 2 A^{3+} (aq) + 3 B^{2-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( A_2B_3 \) be \( x \) mol/dm³. This means that when \( A_2B_3 \) dissolves, it produces:
- 2 moles of \( A^{3+} \) ions for every mole of \( A_2B_3 \) that dissolves.
- 3 moles of \( B^{2-} \) ions for every mole of \( A_2B_3 \) that dissolves.

### Step 3: Calculate the concentrations of the ions
From the dissociation, we can express the concentrations of the ions in terms of \( x \):
- The concentration of \( A^{3+} \) ions will be \( 2x \) mol/dm³.
- The concentration of \( B^{2-} \) ions will be \( 3x \) mol/dm³.

### Step 4: Write the expression for Ksp
The solubility product constant \( K_{sp} \) is given by the expression:
\[ K_{sp} = [A^{3+}]^2 [B^{2-}]^3 \]
Substituting the concentrations we found:
\[ K_{sp} = (2x)^2 (3x)^3 \]

### Step 5: Simplify the expression
Now we can simplify this expression:
\[ K_{sp} = (4x^2) \cdot (27x^3) = 108x^5 \]

### Conclusion
Thus, the solubility product \( K_{sp} \) for \( A_2B_3 \) is:
\[ K_{sp} = 108x^5 \]