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:
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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: \[ \text{moles KCl} = \frac{3.60 , \text{g}}{74.55 , \text{g/mol}} \approx 0.0483 , \text{mol KCl} \]
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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.
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Calculate the number of moles of CaCl₂ in 0.220 M solution:
- Moles of CaCl₂: \[ \text{moles CaCl}_2 = 0.220 , \text{mol/L} \times 0.0750 , \text{L} = 0.0165 , \text{mol CaCl}_2 \]
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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⁻: \[ \text{total moles Cl}^- = 0.0330 + 0.0483 = 0.0813 , \text{mol} \]
- Each formula unit of CaCl₂ produces 1 Ca²⁺ and 2 Cl⁻ ions:
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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