The solubility product of `CaSO_(4)` is `2.4xx10^(-5)`. When 100 mL of `0.01M CaCl_(2)` and 100 mL of `0.002M Na_(2)SO_(4)` are mixed, then :

Options

A. `Na_(2)SO_(4)` will precipitate
B. Both will precipitate
C. `CaSO_(4)` will precipitate
D. None will precipitate

Explanation

To solve the problem step by step, we will determine the concentrations of the ions in the mixed solution and then calculate the ionic product (Q) to compare it with the solubility product (Ksp) of \(CaSO_4\).

### Step 1: Calculate the concentration of \(Ca^{2+}\) ions in the mixture

When we mix 100 mL of \(0.01 M\) \(CaCl_2\) with 100 mL of \(0.002 M\) \(Na_2SO_4\), the total volume of the solution becomes \(200 mL\).

The concentration of \(Ca^{2+}\) ions can be calculated using the formula:

\[ \text{Concentration of } Ca^{2+} = \frac{C_1 \times V_1}{V_f} \]

Where:
- \(C_1 = 0.01 M\) (initial concentration of \(CaCl_2\))
- \(V_1 = 100 mL\) (volume of \(CaCl_2\))
- \(V_f = 200 mL\) (final volume after mixing)

Calculating:

\[ \text{Concentration of } Ca^{2+} = \frac{0.01 \, \text{mol/L} \times 100 \, \text{mL}}{200 \, \text{mL}} = \frac{0.01 \times 100}{200} = 0.005 \, M \]

### Step 2: Calculate the concentration of \(SO_4^{2-}\) ions in the mixture

Similarly, we can calculate the concentration of \(SO_4^{2-}\) ions:

\[ \text{Concentration of } SO_4^{2-} = \frac{C_2 \times V_2}{V_f} \]

Where:
- \(C_2 = 0.002 M\) (initial concentration of \(Na_2SO_4\))
- \(V_2 = 100 mL\) (volume of \(Na_2SO_4\))
- \(V_f = 200 mL\) (final volume after mixing)

Calculating:

\[ \text{Concentration of } SO_4^{2-} = \frac{0.002 \, \text{mol/L} \times 100 \, \text{mL}}{200 \, \text{mL}} = \frac{0.002 \times 100}{200} = 0.001 \, M \]

### Step 3: Calculate the ionic product (Q)

The ionic product \(Q\) for the dissolution of \(CaSO_4\) is given by:

\[ Q = [Ca^{2+}][SO_4^{2-}] \]

Substituting the concentrations we found:

\[ Q = (0.005)(0.001) = 5 \times 10^{-6} \]

### Step 4: Compare Q with Ksp

The given solubility product \(K_{sp}\) of \(CaSO_4\) is:

\[ K_{sp} = 2.4 \times 10^{-5} \]

Now we compare \(Q\) with \(K_{sp}\):

- \(Q = 5 \times 10^{-6}\)
- \(K_{sp} = 2.4 \times 10^{-5}\)

Since \(Q < K_{sp}\), the solution is unsaturated, and no precipitation will occur.

### Conclusion

Since \(Q < K_{sp}\), \(CaSO_4\) will not precipitate.