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 of finding the ionic product of calcium sulfate (CaSO₄) when equal volumes of 0.08 M calcium chloride (CaCl₂) and 0.02 M sodium sulfate (Na₂SO₄) are mixed, we can follow these steps:

### Step-by-Step Solution:

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

2. **Determine Initial Concentrations**: We have:
- Concentration of CaCl₂ = 0.08 M
- Concentration of Na₂SO₄ = 0.02 M

3. **Assume Equal Volumes**: Let the volume of each solution mixed be \( V \) liters. Therefore, the total volume after mixing is:
\[ V_{\text{total}} = V + V = 2V \]

4. **Calculate the Concentration of Ions After Mixing**:
- For calcium ions (Ca²⁺):
\[ [\text{Ca}^{2+}] = \frac{[\text{CaCl}_2] \times V}{V_{\text{total}}} = \frac{0.08 \, \text{mol/L} \times V}{2V} = \frac{0.08}{2} = 0.04 \, \text{M} \]
- For sulfate ions (SO₄²⁻):
\[ [\text{SO}_4^{2-}] = \frac{[\text{Na}_2\text{SO}_4] \times V}{V_{\text{total}}} = \frac{0.02 \, \text{mol/L} \times V}{2V} = \frac{0.02}{2} = 0.01 \, \text{M} \]

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

### Final Answer:
The ionic product of calcium sulfate (CaSO₄) is \( 4 \times 10^{-4} \, \text{M}^2 \).

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