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 Al2(SO4)3` with `20 mL` of `0.6 M BaCl2`, we can follow these steps:

### Step 1: Calculate the number of moles of `Al2(SO4)3`
To find the number of moles of `Al2(SO4)3`, we can use the formula:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]

First, convert the volume from mL to L:
\[ 20 \, \text{mL} = 0.020 \, \text{L} \]

Now, calculate the moles:
\[ \text{Number of moles of } Al2(SO4)3 = 0.2 \, \text{M} \times 0.020 \, \text{L} = 0.004 \, \text{moles} \]

### Step 2: Determine the number of `Al^(3+)` ions produced
From the dissociation of `Al2(SO4)3`, we know that:
\[ 1 \, \text{mole of } Al2(SO4)3 \text{ produces } 2 \, \text{moles of } Al^{3+} \]

Thus, the number of moles of `Al^(3+)` produced is:
\[ 0.004 \, \text{moles of } Al2(SO4)3 \times 2 = 0.008 \, \text{moles of } Al^{3+} \]

### Step 3: Calculate the total volume of the solution
The total volume after mixing is:
\[ 20 \, \text{mL} + 20 \, \text{mL} = 40 \, \text{mL} = 0.040 \, \text{L} \]

### Step 4: Calculate the concentration of `Al^(3+)`
Now, we can find the concentration of `Al^(3+)` using the formula:
\[ \text{Concentration (M)} = \frac{\text{Number of moles}}{\text{Volume (in L)}} \]

Substituting the values:
\[ \text{Concentration of } Al^{3+} = \frac{0.008 \, \text{moles}}{0.040 \, \text{L}} = 0.2 \, \text{M} \]

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

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