Will a precipitate of `CaSO_4`, form if
(i) equal volumes of 0.02 M `CaCl_2` and 0.0004 M `Na_2SO_4` solutions are mixed?
(ii) equal volumes of 0.08 M `CaCl_2` and 0.02 M `Na_2SO_4` are mixed? `K_(sp)(CaSO_4) =24 xx 10^(-5)`.

Answer

To determine whether a precipitate of \( \text{CaSO}_4 \) will form when mixing the given solutions, we need to calculate the ionic product and compare it with the solubility product constant \( K_{sp} \) of \( \text{CaSO}_4 \).

### Given Data:
- \( K_{sp}(\text{CaSO}_4) = 24 \times 10^{-5} \)
- For part (i):
- \( [\text{CaCl}_2] = 0.02 \, M \)
- \( [\text{Na}_2\text{SO}_4] = 0.0004 \, M \)
- For part (ii):
- \( [\text{CaCl}_2] = 0.08 \, M \)
- \( [\text{Na}_2\text{SO}_4] = 0.02 \, M \)

### Step-by-Step Solution:

#### Part (i):
1. **Calculate the new concentrations after mixing equal volumes:**
- When equal volumes of \( 0.02 \, M \) \( \text{CaCl}_2 \) and \( 0.0004 \, M \) \( \text{Na}_2\text{SO}_4 \) are mixed, the concentrations will be halved.
- New concentration of \( \text{Ca}^{2+} \):
\[ [\text{Ca}^{2+}] = \frac{0.02}{2} = 0.01 \, M \]
- New concentration of \( \text{SO}_4^{2-} \):
\[ [\text{SO}_4^{2-}] = \frac{0.0004}{2} = 0.0002 \, M = 2 \times 10^{-4} \, M \]

2. **Calculate the ionic product \( Q_{sp} \):**
\[ Q_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = (0.01)(2 \times 10^{-4}) = 2 \times 10^{-6} \]

3. **Compare \( Q_{sp} \) with \( K_{sp} \):**
- \( K_{sp} = 24 \times 10^{-5} = 2.4 \times 10^{-4} \)
- Since \( Q_{sp} = 2 \times 10^{-6} < K_{sp} \), no precipitate will form.

#### Part (ii):
1. **Calculate the new concentrations after mixing equal volumes:**
- When equal volumes of \( 0.08 \, M \) \( \text{CaCl}_2 \) and \( 0.02 \, M \) \( \text{Na}_2\text{SO}_4 \) are mixed, the concentrations will also be halved.
- New concentration of \( \text{Ca}^{2+} \):
\[ [\text{Ca}^{2+}] = \frac{0.08}{2} = 0.04 \, M \]
- New concentration of \( \text{SO}_4^{2-} \):
\[ [\text{SO}_4^{2-}] = \frac{0.02}{2} = 0.01 \, M \]

2. **Calculate the ionic product \( Q_{sp} \):**
\[ Q_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = (0.04)(0.01) = 0.0004 = 4 \times 10^{-4} \]

3. **Compare \( Q_{sp} \) with \( K_{sp} \):**
- \( K_{sp} = 24 \times 10^{-5} = 2.4 \times 10^{-4} \)
- Since \( Q_{sp} = 4 \times 10^{-4} > K_{sp} \), a precipitate will form.

### Summary of Results:
- (i) No precipitate forms.
- (ii) A precipitate forms.