The solubility of `A_2X_2` is y mol/dm`""^3.` Its solubility product is

Options

A. `6y^4`
B. `64y^4`
C. `36y^5`
D. `108y^5`

Answer

To find the solubility product (Ksp) of the salt \( A_2X_2 \) given its solubility \( y \) mol/dm³, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
The dissociation of the salt \( A_2X_2 \) in water can be represented as:
\[ A_2X_2 (s) \rightleftharpoons 2A^+ (aq) + 2X^{2-} (aq) \]

2. **Determine the Concentrations**:
If the solubility of \( A_2X_2 \) is \( y \) mol/dm³, then when it dissolves:
- The concentration of \( A^+ \) ions will be \( 2y \) (since there are 2 moles of \( A^+ \) produced for every mole of \( A_2X_2 \)).
- The concentration of \( X^{2-} \) ions will be \( 2y \) (since there are also 2 moles of \( X^{2-} \) produced for every mole of \( A_2X_2 \)).

3. **Write the Expression for Ksp**:
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [A^+]^2 [X^{2-}]^2 \]

4. **Substitute the Concentrations into the Ksp Expression**:
Substitute the concentrations we found:
\[ K_{sp} = (2y)^2 (2y)^2 \]

5. **Calculate Ksp**:
Simplifying the expression:
\[ K_{sp} = (4y^2)(4y^2) = 16y^4 \]

### Final Answer:
Thus, the solubility product \( K_{sp} \) of \( A_2X_2 \) is:
\[ K_{sp} = 16y^4 \]