For the sparingly soluble salts, an equilibrium is established between the undissolved solid unit and ions of the dissolved salt. For a solution of the salt like `A_x B_y`.
`A_x B_y =xA^(y+)+yB^(x-)`
`K_(sp)=[A^(y+)]^x [B^(x-)]^y`
For precipitation to occur , the ionic product must be greater than `K_(sp)`.
The solubility of `A_2X_3` is y mol `dm^(-3)` . Its solubility product is

Options

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

Explanation

To find the solubility product (Ksp) of the sparingly soluble salt \( A_2X_3 \), we will follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of the salt \( A_2X_3 \) in water can be represented as:
\[ A_2X_3 (s) \rightleftharpoons 2A^{3+} (aq) + 3X^{2-} (aq) \]

### Step 2: Determine the Molar Solubility
Let the solubility of \( A_2X_3 \) be \( y \) mol/dm³. This means that when \( A_2X_3 \) dissolves, it produces:
- \( 2y \) mol/dm³ of \( A^{3+} \)
- \( 3y \) mol/dm³ of \( X^{2-} \)

### Step 3: Write the Expression for Ksp
The solubility product \( K_{sp} \) is defined as:
\[ K_{sp} = [A^{3+}]^2 [X^{2-}]^3 \]
Substituting the concentrations from the dissociation:
\[ K_{sp} = (2y)^2 (3y)^3 \]

### Step 4: Simplify the Expression
Calculating the expression:
\[ K_{sp} = (2y)^2 \cdot (3y)^3 = 4y^2 \cdot 27y^3 = 108y^5 \]

### Step 5: Final Result
Thus, the solubility product \( K_{sp} \) of \( A_2X_3 \) is:
\[ K_{sp} = 108y^5 \]