The expression for the solubility product of `Ca_(3)(PO_(4))_(2)` will be `K_(sp)=`......... .

Answer

To find the expression for the solubility product (Ksp) of calcium phosphate, \( \text{Ca}_3(\text{PO}_4)_2 \), we will follow these steps:

### Step 1: Write the Dissociation Equation
The first step is to write the dissociation equation for the sparingly soluble salt \( \text{Ca}_3(\text{PO}_4)_2 \).

\[ \text{Ca}_3(\text{PO}_4)_2 (s) \rightleftharpoons 3 \text{Ca}^{2+} (aq) + 2 \text{PO}_4^{3-} (aq) \]

### Step 2: Define Solubility
Let the solubility of \( \text{Ca}_3(\text{PO}_4)_2 \) in mol/L be \( S \). This means that when \( \text{Ca}_3(\text{PO}_4)_2 \) dissolves, it produces ions in the following amounts:
- For every 1 mole of \( \text{Ca}_3(\text{PO}_4)_2 \) that dissolves, it produces 3 moles of \( \text{Ca}^{2+} \) ions.
- It produces 2 moles of \( \text{PO}_4^{3-} \) ions.

Thus, at equilibrium:
- The concentration of \( \text{Ca}^{2+} \) ions = \( 3S \)
- The concentration of \( \text{PO}_4^{3-} \) ions = \( 2S \)

### Step 3: Write the Ksp Expression
The solubility product \( K_{sp} \) is defined as the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced equation.

\[ K_{sp} = [\text{Ca}^{2+}]^3 \times [\text{PO}_4^{3-}]^2 \]

Substituting the concentrations from Step 2:

\[ K_{sp} = (3S)^3 \times (2S)^2 \]

### Step 4: Simplify the Expression
Now we will simplify the expression:

\[ K_{sp} = (27S^3) \times (4S^2) = 108S^5 \]

### Final Answer
Thus, the expression for the solubility product \( K_{sp} \) of \( \text{Ca}_3(\text{PO}_4)_2 \) is:

\[ K_{sp} = 108S^5 \]

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