What are the units in which the solubility product of `Ca_(3)(PO_(4))_(2)` is expressed?

Options

A. `mol dm^(-3)`
B. `mol^(2) dm^(-6)`
C. `mol^(3) dm^(-9)`
D. `mol^(5) dm^(-15)`

Explanation

To determine the units in which the solubility product (Ksp) of \( Ca_3(PO_4)_2 \) is expressed, we can follow these steps:

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

### Step 2: Define the solubility
Let the solubility of \( Ca_3(PO_4)_2 \) be \( S \) (in moles per liter, mol/L).

### Step 3: Express concentrations of ions
From the dissociation equation:
- The concentration of \( Ca^{2+} \) ions will be \( 3S \).
- The concentration of \( PO_4^{3-} \) ions will be \( 2S \).

### Step 4: Write the expression for Ksp
The solubility product \( K_{sp} \) is given by the product of the concentrations of the ions raised to the power of their coefficients in the balanced equation:
\[ K_{sp} = [Ca^{2+}]^3 \times [PO_4^{3-}]^2 \]
Substituting the concentrations:
\[ K_{sp} = (3S)^3 \times (2S)^2 \]

### Step 5: Simplify the expression
Calculating the expression:
\[ K_{sp} = 27S^3 \times 4S^2 = 108S^5 \]

### Step 6: Determine the units of Ksp
Since \( S \) is in moles per liter (mol/L), we can express \( S^5 \) as:
\[ S^5 = (mol/L)^5 = mol^5/L^5 \]
Thus, the units of \( K_{sp} \) become:
\[ K_{sp} = 108 \times \frac{mol^5}{L^5} \]

### Step 7: Convert to standard units
In terms of standard units, we can express liters as \( dm^3 \):
\[ K_{sp} = 108 \times \frac{mol^5}{dm^{15}} \]

### Final Answer
The units in which the solubility product \( K_{sp} \) of \( Ca_3(PO_4)_2 \) is expressed are:
\[ mol^5 \cdot dm^{-15} \]