Find the `Cl^(-)` concentration in solution which is obtained by mixing one mole each of `BaCl_(2),` `NaCl` and HCl in 500 ml of water.

Answer

To find the concentration of \( Cl^- \) ions in the solution obtained by mixing one mole each of \( BaCl_2 \), \( NaCl \), and \( HCl \) in 500 ml of water, we can follow these steps:

### Step 1: Determine the contribution of \( Cl^- \) ions from each compound.

1. **From \( BaCl_2 \)**:
- Each formula unit of \( BaCl_2 \) produces 2 \( Cl^- \) ions.
- Therefore, from 1 mole of \( BaCl_2 \), we get:
\[ 1 \, \text{mole of } BaCl_2 \times 2 \, \text{moles of } Cl^- = 2 \, \text{moles of } Cl^- \]

2. **From \( NaCl \)**:
- Each formula unit of \( NaCl \) produces 1 \( Cl^- \) ion.
- Therefore, from 1 mole of \( NaCl \), we get:
\[ 1 \, \text{mole of } NaCl \times 1 \, \text{mole of } Cl^- = 1 \, \text{mole of } Cl^- \]

3. **From \( HCl \)**:
- Each formula unit of \( HCl \) also produces 1 \( Cl^- \) ion.
- Therefore, from 1 mole of \( HCl \), we get:
\[ 1 \, \text{mole of } HCl \times 1 \, \text{mole of } Cl^- = 1 \, \text{mole of } Cl^- \]

### Step 2: Calculate the total moles of \( Cl^- \).

Now, we add up the contributions from all three compounds:
\[ \text{Total } Cl^- = 2 \, \text{moles from } BaCl_2 + 1 \, \text{mole from } NaCl + 1 \, \text{mole from } HCl = 4 \, \text{moles of } Cl^- \]

### Step 3: Calculate the molarity of \( Cl^- \).

Molarity (M) is defined as the number of moles of solute per liter of solution. We have 4 moles of \( Cl^- \) in a total volume of 500 ml (which is 0.5 liters).

Using the formula for molarity:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Substituting the values:
\[ \text{Molarity of } Cl^- = \frac{4 \, \text{moles}}{0.5 \, \text{liters}} = 8 \, \text{M} \]

### Final Answer:
The concentration of \( Cl^- \) in the solution is \( 8 \, \text{M} \).

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