Aqueous solutions of `Na_(2)C_(2)O_(4)` and `CaCI_(2)` are mixed and precipitate of `CaC_(2)O_(4)` formed is filered and dried. `250 mL` of the saturated solution of `CaC_(2)O_(4)` required `6.0 mL` of `0.001M KMnO_(4)` solution in acidic medium for complete titration.
`K_(sp)` of `CaC_(2)O_(4)` is

Options

A. `2.25xx10^(-12)`
B. `2.25xx10^(-10)`
C. `3.6xx10^(-9)`
D. `4.0xx10^(-9)`

Explanation

To solve the problem, we need to find the solubility product constant (Ksp) of calcium oxalate (CaC₂O₄) using the information provided about the titration with KMnO₄. Here’s a step-by-step solution:

### Step 1: Calculate the n-factor for KMnO₄
In acidic medium, KMnO₄ is reduced from Mn(VII) to Mn(II).

- Change in oxidation state for Mn:
\[ \Delta O = 7 - 2 = 5 \]
- Since there is one Mn atom in KMnO₄, the n-factor for KMnO₄ is:
\[ n = 1 \times 5 = 5 \]

### Step 2: Calculate the n-factor for oxalate ion (C₂O₄²⁻)
The oxalate ion (C₂O₄²⁻) is oxidized to form CO₂.

- Oxidation states of carbon in C₂O₄²⁻:
- In C₂O₄²⁻: \(2x + 4(-2) = -2 \Rightarrow 2x - 8 = -2 \Rightarrow 2x = 6 \Rightarrow x = +3\)
- In CO₂: \(x + 2(-2) = 0 \Rightarrow x - 4 = 0 \Rightarrow x = +4\)
- Change in oxidation state for each carbon atom:
\[ \Delta O = 4 - 3 = 1 \]
- Since there are two carbon atoms in C₂O₄²⁻, the n-factor for oxalate ion is:
\[ n = 2 \times 1 = 2 \]

### Step 3: Calculate the equivalents of KMnO₄
Using the volume and concentration of KMnO₄:
- Volume of KMnO₄ = 6 mL = 0.006 L
- Concentration of KMnO₄ = 0.001 M

The number of equivalents of KMnO₄ can be calculated as:
\[ \text{Equivalents of KMnO₄} = n \times \text{Concentration} \times \text{Volume (L)} = 5 \times 0.001 \times 0.006 = 3 \times 10^{-5} \text{ equivalents} \]

### Step 4: Calculate the equivalents of oxalate ion
Since the equivalents of KMnO₄ equal the equivalents of oxalate ion:
\[ \text{Equivalents of C₂O₄²⁻} = 3 \times 10^{-5} \]

### Step 5: Calculate the concentration of oxalate ion
Using the volume of the saturated solution of CaC₂O₄ (250 mL = 0.250 L):
\[ \text{Equivalents of C₂O₄²⁻} = n \times \text{Concentration} \times \text{Volume (L)} \]
Let the concentration of C₂O₄²⁻ be \(S\):
\[ 3 \times 10^{-5} = 2 \times S \times 0.250 \]
Solving for \(S\):
\[ S = \frac{3 \times 10^{-5}}{2 \times 0.250} = \frac{3 \times 10^{-5}}{0.500} = 6 \times 10^{-5} \text{ M} \]

### Step 6: Calculate Ksp of CaC₂O₄
The solubility product constant \(Ksp\) is given by:
\[ Ksp = [Ca^{2+}][C₂O₄^{2-}] \]
Since both concentrations are equal to \(S\):
\[ Ksp = S \times S = S^2 = (6 \times 10^{-5})^2 = 36 \times 10^{-10} = 3.6 \times 10^{-9} \]

### Final Answer
The \(Ksp\) of \(CaC₂O₄\) is:
\[ \boxed{3.6 \times 10^{-9}} \]