Question

3.60 g KCl in 75.0 mL of 0.220 M CaCh solution. Assume that no volume change occurs when the salt is added to the solution.
Enter your answers, separated by commas, using three significant figures.

Answer 1

To determine the concentrations of ions in a solution containing 3.60 g of KCl added to 75.0 mL of a 0.220 M CaCl₂ solution, we need to follow these steps:

1. **Calculate the number of moles of KCl:**
- The molar mass of KCl (Potassium Chloride) is approximately \(39.10 \, \text{g/mol (K)} + 35.45 \, \text{g/mol (Cl)} \approx 74.55 \, \text{g/mol}\).
- Number of moles of KCl:
\[
\text{moles KCl} = \frac{3.60 \, \text{g}}{74.55 \, \text{g/mol}} \approx 0.0483 \, \text{mol KCl}
\]

2. **Calculate the total volume of the solution:**
- The volume of the CaCl₂ solution is 75.0 mL or 0.0750 L.
- Assuming no volume change, the total volume after adding KCl remains 0.0750 L.

3. **Calculate the number of moles of CaCl₂ in 0.220 M solution:**
- Moles of CaCl₂:
\[
\text{moles CaCl}_2 = 0.220 \, \text{mol/L} \times 0.0750 \, \text{L} = 0.0165 \, \text{mol CaCl}_2
\]

4. **Determine the concentrations of ions:**
- Each formula unit of CaCl₂ produces 1 Ca²⁺ and 2 Cl⁻ ions:
- Moles of Ca²⁺:
\[
\text{moles Ca}^{2+} = \text{moles CaCl}_2 = 0.0165 \, \text{mol}
\]
- Moles of Cl⁻ from CaCl₂:
\[
\text{moles Cl}^- (\text{from CaCl}_2) = 2 \times 0.0165 \, \text{mol} = 0.0330 \, \text{mol}
\]
- Moles of Cl⁻ from KCl:
\[
\text{moles Cl}^- (\text{from KCl}) = 0.0483 \, \text{mol}
\]
- Total moles of Cl⁻:
\[
\text{total moles Cl}^- = 0.0330 + 0.0483 = 0.0813 \, \text{mol}
\]

5. **Calculate the concentrations of ions in the total volume of solution:**
- Concentration of Ca²⁺:
\[
[\text{Ca}^{2+}] = \frac{0.0165 \, \text{mol}}{0.0750 \, \text{L}} \approx 0.220 \, \text{M}
\]
- Concentration of Cl⁻:
\[
[\text{Cl}^-] = \frac{0.0813 \, \text{mol}}{0.0750 \, \text{L}} \approx 1.084 \, \text{M}
\]

**Final concentrations with three significant figures:**
- [Ca²⁺], [Cl⁻]: 0.220 M, 1.08 M

Therefore, the final answer is:
**0.220, 1.08**