20mL of 0.2M `Al_(2)(SO_(4))_(3)` is mixed with 20mL of 0.6M `Bacl_(2)`.Concentration of `Al^(3+)`ion in the solution will be:

Options

A. 0.2M
B. 10.3M
C. 0.1M
D. 0.25M

Explanation

To find the concentration of \( Al^{3+} \) ions in the solution after mixing 20 mL of 0.2 M \( Al_2(SO_4)_3 \) with 20 mL of 0.6 M \( BaCl_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the moles of \( Al_2(SO_4)_3 \)**:
- The molarity (M) of \( Al_2(SO_4)_3 \) is 0.2 M and the volume (V) is 20 mL.
- Convert the volume from mL to L:
\[ V = 20 \, \text{mL} = 0.020 \, \text{L} \]
- Calculate the moles using the formula:
\[ \text{Moles of } Al_2(SO_4)_3 = M \times V = 0.2 \, \text{mol/L} \times 0.020 \, \text{L} = 0.004 \, \text{mol} \]

2. **Determine the moles of \( Al^{3+} \) produced**:
- Each mole of \( Al_2(SO_4)_3 \) produces 2 moles of \( Al^{3+} \):
\[ \text{Moles of } Al^{3+} = 0.004 \, \text{mol} \times 2 = 0.008 \, \text{mol} \]

3. **Calculate the total volume of the solution**:
- The total volume after mixing is:
\[ \text{Total Volume} = 20 \, \text{mL} + 20 \, \text{mL} = 40 \, \text{mL} = 0.040 \, \text{L} \]

4. **Calculate the concentration of \( Al^{3+} \)**:
- The concentration (C) of \( Al^{3+} \) can be calculated using the formula:
\[ C = \frac{\text{Moles of } Al^{3+}}{\text{Total Volume in L}} = \frac{0.008 \, \text{mol}}{0.040 \, \text{L}} = 0.2 \, \text{M} \]

### Final Answer:
The concentration of \( Al^{3+} \) ions in the solution is **0.2 M**.

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