What volume (in ml) of 0.8M `AlCl_(3)` solution should be mixed with
50 ml of 0.2 M `CaCl_(2)` solution to get solution of chloride ion concentration equal to 0.6 M ?

Answer

To solve the problem, we need to find the volume of 0.8 M `AlCl_(3)` solution that should be mixed with 50 ml of 0.2 M `CaCl_(2)` solution to achieve a final chloride ion concentration of 0.6 M.

### Step-by-Step Solution:

1. **Determine the contribution of chloride ions from `AlCl_(3)`**:
- `AlCl_(3)` dissociates into one `Al^(3+)` ion and three `Cl^(-)` ions.
- Therefore, the concentration of chloride ions from `AlCl_(3)` is:
\[ \text{Chloride concentration from } AlCl_3 = 3 \times 0.8 \text{ M} = 2.4 \text{ M} \]

2. **Determine the contribution of chloride ions from `CaCl_(2)`**:
- `CaCl_(2)` dissociates into one `Ca^(2+)` ion and two `Cl^(-)` ions.
- Therefore, the concentration of chloride ions from `CaCl_(2)` is:
\[ \text{Chloride concentration from } CaCl_2 = 2 \times 0.2 \text{ M} = 0.4 \text{ M} \]

3. **Let the volume of `AlCl_(3)` solution be `x` ml**:
- The total volume of the mixed solution will be \( x + 50 \) ml.

4. **Set up the equation for final chloride ion concentration**:
- The total moles of chloride ions from both solutions must equal the final concentration of chloride ions multiplied by the total volume:
\[ \text{Final concentration} = \frac{\text{Moles of } Cl^-}{\text{Total volume}} \]
- Thus, we have:
\[ 0.6 = \frac{(2.4 \cdot x) + (0.4 \cdot 50)}{x + 50} \]

5. **Substitute the values into the equation**:
- Substitute \(0.4 \cdot 50 = 20\):
\[ 0.6 = \frac{(2.4x + 20)}{x + 50} \]

6. **Cross-multiply to eliminate the fraction**:
\[ 0.6(x + 50) = 2.4x + 20 \]
- Expanding the left side:
\[ 0.6x + 30 = 2.4x + 20 \]

7. **Rearranging the equation**:
- Move all terms involving \(x\) to one side:
\[ 30 - 20 = 2.4x - 0.6x \]
\[ 10 = 1.8x \]

8. **Solve for \(x\)**:
\[ x = \frac{10}{1.8} \approx 5.56 \text{ ml} \]

### Final Answer:
The volume of 0.8 M `AlCl_(3)` solution that should be mixed is approximately **5.56 ml**.

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