The solubility of `Ca_(3)(PO_(4))_(2)` in water is y moles `//` litre. Its solubility product is

Options

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

Explanation

To find the solubility product (Ksp) of calcium phosphate, \( Ca_3(PO_4)_2 \), we will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of calcium phosphate in water can be represented as:
\[ Ca_3(PO_4)_2 (s) \rightleftharpoons 3Ca^{2+} (aq) + 2PO_4^{3-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( Ca_3(PO_4)_2 \) in water be \( y \) moles per liter. This means that at equilibrium:
- The concentration of \( Ca^{2+} \) ions will be \( 3y \) (since 3 moles of \( Ca^{2+} \) are produced for every mole of \( Ca_3(PO_4)_2 \)).
- The concentration of \( PO_4^{3-} \) ions will be \( 2y \) (since 2 moles of \( PO_4^{3-} \) are produced for every mole of \( Ca_3(PO_4)_2 \)).

### Step 3: Write the expression for Ksp
The solubility product \( Ksp \) is given by the formula:
\[ Ksp = [Ca^{2+}]^3 \times [PO_4^{3-}]^2 \]
Substituting the concentrations from Step 2:
\[ Ksp = (3y)^3 \times (2y)^2 \]

### Step 4: Simplify the expression
Now, we will simplify the expression:
\[ Ksp = (27y^3) \times (4y^2) = 108y^5 \]

### Step 5: Final result
Thus, the solubility product \( Ksp \) of \( Ca_3(PO_4)_2 \) is:
\[ Ksp = 108y^5 \]