0.03 mole of `Ca^(2+)` ions is added to a litre of 0.01 M `SO_(4)^(2-)` solution. Will it cause precipitation of `CaSO_(4)` ? `K_(sp) "for" CaSO_(4)=2.4xx10^(-5)`.

Explanation

To determine whether the addition of 0.03 moles of \( \text{Ca}^{2+} \) ions to a 1-liter solution of 0.01 M \( \text{SO}_4^{2-} \) will cause precipitation of \( \text{CaSO}_4 \), we need to compare the ionic product (IP) of the solution to the solubility product constant (\( K_{sp} \)) of \( \text{CaSO}_4 \).

### Step-by-Step Solution:

1. **Identify the concentrations of ions:**
- The concentration of \( \text{Ca}^{2+} \) ions after adding 0.03 moles to 1 liter of solution is:
\[ [\text{Ca}^{2+}] = \frac{0.03 \text{ moles}}{1 \text{ L}} = 0.03 \text{ M} \]
- The concentration of \( \text{SO}_4^{2-} \) ions is given as:
\[ [\text{SO}_4^{2-}] = 0.01 \text{ M} \]

2. **Calculate the ionic product (IP):**
- The ionic product for the precipitation of \( \text{CaSO}_4 \) is given by:
\[ IP = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = (0.03)(0.01) = 0.0003 = 3.0 \times 10^{-4} \]

3. **Compare the ionic product with \( K_{sp} \):**
- The solubility product constant for \( \text{CaSO}_4 \) is given as:
\[ K_{sp} = 2.4 \times 10^{-5} \]
- Now, we compare the ionic product with the \( K_{sp} \):
\[ IP = 3.0 \times 10^{-4} \quad \text{and} \quad K_{sp} = 2.4 \times 10^{-5} \]

4. **Determine if precipitation occurs:**
- Since \( IP > K_{sp} \), this indicates that the solution is supersaturated with respect to \( \text{CaSO}_4 \), and thus precipitation will occur.

### Conclusion:
Yes, the addition of 0.03 moles of \( \text{Ca}^{2+} \) ions to the solution will cause precipitation of \( \text{CaSO}_4 \).