A saturated solution of `Ca_(3)(PO_(4))_(2)` has `[Ca^(2+)]=2xx10^(-8)` M and `[PO_(4)^(3-)]=1.6xx10^(-5)` M `K_(sp)` of `Ca_(3)(PO_(4))_(2)` is :

Options

A. `3.2xx10^(-13)`
B. `3.2xx10^(-34)`
C. `2.048xx10^(-33)`
D. None of these

Answer

To find the solubility product constant (Ksp) for the saturated solution of calcium phosphate, \( Ca_3(PO_4)_2 \), we can 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: Identify the Stoichiometric Ratios
From the dissociation equation, we can see that:
- For every 1 mole of \( Ca_3(PO_4)_2 \) that dissolves, 3 moles of \( Ca^{2+} \) and 2 moles of \( PO_4^{3-} \) are produced.

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

### Step 4: Substitute the Given Concentrations
We are given:
- \([Ca^{2+}] = 2 \times 10^{-8} \, M\)
- \([PO_4^{3-}] = 1.6 \times 10^{-5} \, M\)

Now, substituting these values into the \( Ksp \) expression:
\[ Ksp = (2 \times 10^{-8})^3 \times (1.6 \times 10^{-5})^2 \]

### Step 5: Calculate Each Part
First, calculate \( (2 \times 10^{-8})^3 \):
\[ (2 \times 10^{-8})^3 = 8 \times 10^{-24} \]

Next, calculate \( (1.6 \times 10^{-5})^2 \):
\[ (1.6 \times 10^{-5})^2 = 2.56 \times 10^{-10} \]

### Step 6: Combine the Results
Now, multiply the two results together:
\[ Ksp = 8 \times 10^{-24} \times 2.56 \times 10^{-10} \]

Calculating this gives:
\[ Ksp = 20.48 \times 10^{-34} \]

### Step 7: Express in Standard Form
Finally, we can express this in standard form:
\[ Ksp = 2.048 \times 10^{-33} \]

### Final Answer
The solubility product constant \( Ksp \) of \( Ca_3(PO_4)_2 \) is:
\[ Ksp = 2.048 \times 10^{-33} \]

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