20 ml of 0.2M `Al_2(SO_4)_3` is mixed with 20 ml of 6.6 M `BaCl_2` the concentration of `Cl^-` ion in solution is

Options

A. 0.2M
B. 6.6M
C. 0.02 M
D. 0.06 M

Answer

To find the concentration of Cl⁻ ions in the solution after mixing 20 mL of 0.2 M Al₂(SO₄)₃ with 20 mL of 6.6 M BaCl₂, we can follow these steps:

### Step 1: Calculate the moles of BaCl₂
First, we need to calculate the number of moles of BaCl₂ in the solution.

Given:
- Volume of BaCl₂ = 20 mL = 20/1000 L = 0.020 L
- Concentration of BaCl₂ = 6.6 M

Using the formula:
\[ \text{Moles of BaCl}_2 = \text{Concentration} \times \text{Volume} \]
\[ \text{Moles of BaCl}_2 = 6.6 \, \text{mol/L} \times 0.020 \, \text{L} = 0.132 \, \text{mol} \]

### Step 2: Determine the moles of Cl⁻ ions produced
BaCl₂ dissociates in solution as follows:
\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2 \text{Cl}^- \]

From the stoichiometry of the reaction, 1 mole of BaCl₂ produces 2 moles of Cl⁻ ions. Therefore, the moles of Cl⁻ ions produced are:
\[ \text{Moles of Cl}^- = 2 \times \text{Moles of BaCl}_2 = 2 \times 0.132 \, \text{mol} = 0.264 \, \text{mol} \]

### Step 3: Calculate the total volume of the solution
The total volume of the solution after mixing is:
\[ \text{Total Volume} = 20 \, \text{mL} + 20 \, \text{mL} = 40 \, \text{mL} = 40/1000 \, \text{L} = 0.040 \, \text{L} \]

### Step 4: Calculate the concentration of Cl⁻ ions
Now we can calculate the concentration of Cl⁻ ions in the solution using the formula:
\[ \text{Concentration of Cl}^- = \frac{\text{Moles of Cl}^-}{\text{Total Volume}} \]
\[ \text{Concentration of Cl}^- = \frac{0.264 \, \text{mol}}{0.040 \, \text{L}} = 6.6 \, \text{M} \]

### Final Answer
The concentration of Cl⁻ ions in the solution is **6.6 M**.

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