Equal volume of 0.08 M `CaCI_(2)` and 0.02 M sodium sulphate are mixed. What is ionic product of `CaSO_(4)`?

Options

A. `10 xx 10^(-4)`
B. `8 xx 10^(-4)`
C. `4 xx 10^(-4)`
D. `2 xx 10^(-4)`

Answer

To solve the problem, we need to find the ionic product of calcium sulfate (CaSO₄) when equal volumes of 0.08 M CaCl₂ and 0.02 M Na₂SO₄ are mixed.

### Step-by-Step Solution:

1. **Identify the Reaction**:
When CaCl₂ and Na₂SO₄ are mixed, they react to form calcium sulfate (CaSO₄) and sodium chloride (NaCl):
\[ \text{CaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{NaCl} \]

2. **Determine the Initial Molarities**:
- Molarity of CaCl₂ (M₁) = 0.08 M
- Molarity of Na₂SO₄ (M₂) = 0.02 M

3. **Assume Equal Volume**:
Let the volume of each solution be \( V \) liters. Therefore, the total volume after mixing will be \( V + V = 2V \).

4. **Calculate the Concentration of Calcium Ions (Ca²⁺)**:
The concentration of calcium ions after mixing can be calculated using the formula:
\[ [\text{Ca}^{2+}] = \frac{M_1 \cdot V}{V + V} = \frac{0.08 \cdot V}{2V} = \frac{0.08}{2} = 0.04 \, \text{M} \]

5. **Calculate the Concentration of Sulfate Ions (SO₄²⁻)**:
The concentration of sulfate ions after mixing can be calculated similarly:
\[ [\text{SO}_4^{2-}] = \frac{M_2 \cdot V}{V + V} = \frac{0.02 \cdot V}{2V} = \frac{0.02}{2} = 0.01 \, \text{M} \]

6. **Calculate the Ionic Product (Ksp) of CaSO₄**:
The ionic product (IP) of calcium sulfate is given by the product of the concentrations of its ions:
\[ \text{IP} = [\text{Ca}^{2+}] \cdot [\text{SO}_4^{2-}] = 0.04 \cdot 0.01 = 0.0004 \]
\[ \text{IP} = 4 \times 10^{-4} \]

### Final Answer:
The ionic product of CaSO₄ is \( 4 \times 10^{-4} \).

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