Let's solve the problem step-by-step for each case. ## **Given:** - Desired activity of Ca²⁺: \(\{Ca^{2+}\} = 1.3 \times 10^{-4}\) M - Need to find how much CaCl₂ to add to 1 L for: 1. Distilled water 2. 0.05 M KCl 3. 0.1 M MgCl₂ - All salts dissociate completely. - Use the **Davies equation** for activity coefficient (\(\gamma\)): \[ \log_{10} \gamma_i = -0.51 z_i^2 \left( \frac{\sqrt{I}}{1 + \sqrt{I}} - 0.3 I \right) \] where: - \(z_i\): charge of the ion - \(I\): ionic strength --- ## **General Approach** 1. **Calculate ionic strength \(I\)** for each solution. 2. **Calculate activity coefficient (\(\gamma_{Ca^{2+}}\))** for Ca²⁺. 3. **Relate activity and concentration:** \[ \{Ca^{2+}\} = \gamma_{Ca^{2+}} [Ca^{2+}] \] \[ [Ca^{2+}] = \frac{\{Ca^{2+}\}}{\gamma_{Ca^{2+}}} \] 4. **Determine how much CaCl₂ is needed** to get the required \([Ca^{2+}]\). --- ## **Step 1: Distilled Water** ### **Ionic Strength \(I\)** Initially, water has no ions. When you add CaCl₂: \[ CaCl_2 \rightarrow Ca^{2+} + 2Cl^- \] Let \(x = [Ca^{2+}]\), then \([Cl^-] = 2x\). \[ I = \frac{1}{2} \left( [Ca^{2+}](2^2) + [Cl^-](1^2) \right) = \frac{1}{2} \left( x \times 4 + 2x \times 1 \right) = \frac{1}{2} (4x + 2x) = 3x \] ### **Activity Coefficient (\(\gamma_{Ca^{2+}}\))** \[ \log_{10} \gamma_{Ca^{2+}} = -0.51 \times 4 \left( \frac{\sqrt{3x}}{1+\sqrt{3x}} - 0.3 \times 3x \right) \] But \(x\) is small, so we'll estimate \(\gamma_{Ca^{2+}}\) with \(x = 1.3 \times 10^{-4}\) M: \[ I = 3(1.3 \times 10^{-4}) = 3.9 \times 10^{-4} \text{ M} \] \[ \sqrt{I} = \sqrt{3.9 \times 10^{-4}} \approx 0.0198 \] \[ \frac{\sqrt{I}}{1+\sqrt{I}} = \frac{0.0198}{1.0198} \approx 0.0194 \] \[ \log_{10} \gamma_{Ca^{2+}} = -0.51 \times 4 \left(0.0194 - 0.3 \times 3.9 \times 10^{-4}\right) \] \[ 0.3 \times 3.9 \times 10^{-4} = 1.17 \times 10^{-4} \] \[ 0.0194 - 1.17 \times 10^{-4} \approx 0.0193 \] \[ -0.51 \times 4 \times 0.0193 = -0.0394 \] \[ \gamma_{Ca^{2+}} = 10^{-0.0394} \approx 0.91 \] ### **Calculate \([Ca^{2+}]\)** \[ [Ca^{2+}] = \frac{1.3 \times 10^{-4}}{0.91} = 1.43 \times 10^{-4} \text{ M} \] ### **Moles CaCl₂** Each mole CaCl₂ gives 1 mole Ca²⁺: \[ \text{moles CaCl}_2 = 1.43 \times 10^{-4} \text{ mol} \] ### **Grams CaCl₂** Molar mass CaCl₂ = 40.08 + 2×35.45 = 110.98 g/mol \[ \text{grams CaCl}_2 = 1.43 \times 10^{-4} \times 110.98 = 0.0159 \text{ g} = 15.9 \text{ mg} \] --- ## **Step 2: In 0.05 M KCl** - KCl dissociates: \(KCl \rightarrow K^+ + Cl^-\) - Initial: [K⁺] = [Cl⁻] = 0.05 M When CaCl₂ is added: - [Ca²⁺] = \(x\) - [Cl⁻] = 0.05 + 2x (since each CaCl₂ adds 2 Cl⁻) \[ I = \frac{1}{2}([K^+](1^2) + [Cl^-](1^2) + [Ca^{2+}](2^2)) \] \[ = \frac{1}{2}(0.05 \times 1 + (0.05 + 2x) \times 1 + x \times 4) \] \[ = \frac{1}{2}(0.05 + 0.05 + 2x + 4x) = \frac{1}{2}(0.1 + 6x) = 0.05 + 3x \] Assume \(x\) is small, so \(I \approx 0.05\) M. \[ \sqrt{I} = \sqrt{0.05} = 0.224 \] \[ \frac{\sqrt{I}}{1+\sqrt{I}} = \frac{0.224}{1.224} = 0.183 \] \[ 0.3I = 0.015 \] \[ \log_{10} \gamma_{Ca^{2+}} = -0.51 \times 4 (0.183 - 0.015) = -0.51 \times 4 \times 0.168 = -0.342 \] \[ \gamma_{Ca^{2+}} = 10^{-0.342} = 0.456 \] \[ [Ca^{2+}] = \frac{1.3 \times 10^{-4}}{0.456} = 2.85 \times 10^{-4} \text{ M} \] \[ \text{moles CaCl}_2 = 2.85 \times 10^{-4} \text{ mol} \] \[ \text{grams CaCl}_2 = 2.85 \times 10^{-4} \times 110.98 = 0.0317 \text{ g} = 31.7 \text{ mg} \] --- ## **Step 3: In 0.1 M MgCl₂** - MgCl₂ dissociates: \(MgCl_2 \rightarrow Mg^{2+} + 2Cl^-\) - Initial: [Mg²⁺] = 0.1 M, [Cl⁻] = 0.2 M Add CaCl₂: - [Ca²⁺] = \(x\) - [Cl⁻] = 0.2 + 2x \[ I = \frac{1}{2}([Mg^{2+}](2^2) + [Cl^-](1^2) + [Ca^{2+}](2^2)) \] \[ = \frac{1}{2}(0.1 \times 4 + (0.2 + 2x) \times 1 + x \times 4) \] \[ = \frac{1}{2}(0.4 + 0.2 + 2x + 4x) = \frac{1}{2}(0.6 + 6x) = 0.3 + 3x \] Again, \(x\) is small, so \(I \approx 0.3\) M. \[ \sqrt{I} = \sqrt{0.3} = 0.548 \] \[ \frac{\sqrt{I}}{1+\sqrt{I}} = \frac{0.548}{1.548} = 0.354 \] \[ 0.3I = 0.09 \] \[ \log_{10} \gamma_{Ca^{2+}} = -0.51 \times 4 \times (0.354 - 0.09) = -0.51 \times 4 \times 0.264 = -0.538 \] \[ \gamma_{Ca^{2+}} = 10^{-0.538} = 0.290 \] \[ [Ca^{2+}] = \frac{1.3 \times 10^{-4}}{0.290} = 4.48 \times 10^{-4} \text{ M} \] \[ \text{moles CaCl}_2 = 4.48 \times 10^{-4} \text{ mol} \] \[ \text{grams CaCl}_2 = 4.48 \times 10^{-4} \times 110.98 = 0.0497 \text{ g} = 49.7 \text{ mg} \] --- ## **Summary Table** | Solution | \(\gamma_{Ca^{2+}}\) | [CaCl₂] (mol/L) | Grams CaCl₂ / 1 L | |-----------------------|----------------------|------------------|--------------------| | Distilled Water | 0.91 | 1.43×10⁻⁴ | **15.9 mg** | | 0.05 M KCl | 0.456 | 2.85×10⁻⁴ | **31.7 mg** | | 0.1 M MgCl₂ | 0.290 | 4.48×10⁻⁴ | **49.7 mg** | --- ## **Final Answers** **To make 1 liter of solution with a calcium ion activity of \(1.3 \times 10^{-4}\) M:** - **In distilled water:** Add **15.9 mg CaCl₂** - **In 0.05 M KCl:** Add **31.7 mg CaCl₂** - **In 0.1 M MgCl₂:** Add **49.7 mg CaCl₂** These amounts will give the desired Ca²⁺ activity in each solution.