Let’s break this down step-by-step for each case: **distilled water, 0.05 M KCl, and 0.1 M MgCl₂**. We'll use the **Davies equation** for activity coefficients, and solve for the amount of CaCl₂ needed to achieve the target **calcium ion activity, {Ca²⁺} = 1.3 × 10⁻⁴ M**. --- ### **Step 1: Relate activity to concentration** The activity of Ca²⁺ is given by: \[ \{Ca^{2+}\} = \gamma_{Ca^{2+}} [Ca^{2+}] \] where: - \(\gamma_{Ca^{2+}}\) = activity coefficient of Ca²⁺ - \([Ca^{2+}]\) = molar concentration of Ca²⁺ To achieve the given activity, we need: \[ [Ca^{2+}] = \frac{\{Ca^{2+}\}}{\gamma_{Ca^{2+}}} \] --- ### **Step 2: Find the activity coefficient (\(\gamma_{Ca^{2+}}\)) using the Davies equation** The **Davies equation**: \[ \log_{10} \gamma_i = -0.51 z_i^2 \left( \frac{\sqrt{I}}{1+\sqrt{I}} - 0.3I \right) \] where: - \(z_i\) = charge of the ion (\(+2\) for Ca²⁺) - \(I\) = ionic strength of the solution **Ionic strength**: \[ I = \frac{1}{2} \sum c_i z_i^2 \] where: - \(c_i\) = molar concentration of each ion - \(z_i\) = charge of each ion We need to calculate \(I\) for each solution. --- #### **Case 1: Distilled Water (Negligible Ionic Strength)** - Only CaCl₂ is present. - Let \(x = [Ca^{2+}]\). - Each CaCl₂ gives 1 Ca²⁺ and 2 Cl⁻, so [Cl⁻] = 2x. \[ I = \frac{1}{2} (x \cdot 2^2 + 2x \cdot 1^2) = \frac{1}{2} (4x + 2x) = 3x \] --- #### **Case 2: 0.05 M KCl** - KCl is fully dissociated: [K⁺] = [Cl⁻] = 0.05 M - Add CaCl₂ to get [Ca²⁺] = x, [Cl⁻] increases by 2x. \[ I = \frac{1}{2} (0.05 \times 1^2 + 0.05 \times 1^2 + x \times 2^2 + 2x \times 1^2) \] \[ = \frac{1}{2} (0.05 + 0.05 + 4x + 2x) = \frac{1}{2} (0.10 + 6x) = 0.05 + 3x \] --- #### **Case 3: 0.1 M MgCl₂** - MgCl₂ dissociates to Mg²⁺ (0.1 M) and Cl⁻ (0.2 M) - Add CaCl₂: [Ca²⁺] = x, extra Cl⁻ = 2x \[ I = \frac{1}{2} (0.1 \times 2^2 + 0.2 \times 1^2 + x \times 2^2 + 2x \times 1^2) \] \[ = \frac{1}{2} (0.4 + 0.2 + 4x + 2x) = \frac{1}{2} (0.6 + 6x) = 0.3 + 3x \] --- ### **Step 3: Calculate [Ca²⁺], then mass of CaCl₂ needed** We need to **solve iteratively** because \(I\) depends on \(x = [Ca^{2+}]\), which in turn depends on \(\gamma_{Ca^{2+}}\), which itself depends on \(I\). But because the desired [Ca²⁺] is small compared to background salt in cases 2 and 3, we can **approximate** \(I\) using only the background electrolyte. --- #### **Case 1: Distilled Water** - \(I \approx 3x\) (since nothing else is present) - Initial guess: \(x = 1.3 \times 10^{-4}\) M. Calculate \(I\): \[ I = 3 \times 1.3 \times 10^{-4} = 3.9 \times 10^{-4} \] Calculate \(\gamma_{Ca^{2+}}\): \[ \sqrt{I} = \sqrt{3.9 \times 10^{-4}} \approx 0.0198 \] \[ \frac{\sqrt{I}}{1+\sqrt{I}} = \frac{0.0198}{1.0198} \approx 0.0194 \] \[ -0.51 \times 4 \times (0.0194 - 0.3 \times 3.9 \times 10^{-4}) \] \[ -0.51 \times 4 \times (0.0194 - 1.17 \times 10^{-4}) \approx -2.04 \times 0.01928 \approx -0.0394 \] \[ \gamma_{Ca^{2+}} = 10^{-0.0394} \approx 0.912 \] Now, \[ [Ca^{2+}] = \frac{1.3 \times 10^{-4}}{0.912} = 1.43 \times 10^{-4} \text{ M} \] Thus, we need \(1.43 \times 10^{-4}\) mol in 1 L. **CaCl₂ molar mass = 40.08 + 2×35.45 = 110.98 g/mol** \[ \text{Mass} = 1.43 \times 10^{-4} \text{ mol} \times 110.98 \text{ g/mol} = 0.0159 \text{ g} = 15.9 \text{ mg} \] **Explanation:** We calculated the ionic strength based only on CaCl₂, then used the Davies equation for the activity coefficient, then calculated the required concentration and mass. --- #### **Case 2: 0.05 M KCl** **Ionic strength:** \(I \approx 0.05\) (since 3x is negligible compared to 0.05) \[ \sqrt{0.05} = 0.224 \] \[ \frac{0.224}{1+0.224} = 0.183 \] \[ -0.51 \times 4 \times (0.183 - 0.3 \times 0.05) = -2.04 \times (0.183 - 0.015) = -2.04 \times 0.168 = -0.343 \] \[ \gamma_{Ca^{2+}} = 10^{-0.343} = 0.454 \] \[ [Ca^{2+}] = \frac{1.3 \times 10^{-4}}{0.454} = 2.87 \times 10^{-4} \text{ M} \] \[ \text{Mass} = 2.87 \times 10^{-4} \text{ mol} \times 110.98 \text{ g/mol} = 0.0319 \text{ g} = 31.9 \text{ mg} \] **Explanation:** In 0.05 M KCl, ionic strength is dominated by KCl, resulting in a lower activity coefficient (more deviation from ideality), requiring more CaCl₂ to reach the same activity. --- #### **Case 3: 0.1 M MgCl₂** **Ionic strength:** \(I \approx 0.3\) (background from MgCl₂ dominates) \[ \sqrt{0.3} = 0.548 \] \[ \frac{0.548}{1+0.548} = 0.354 \] \[ -0.51 \times 4 \times (0.354 - 0.3 \times 0.3) = -2.04 \times (0.354 - 0.09) = -2.04 \times 0.264 = -0.538 \] \[ \gamma_{Ca^{2+}} = 10^{-0.538} = 0.290 \] \[ [Ca^{2+}] = \frac{1.3 \times 10^{-4}}{0.290} = 4.48 \times 10^{-4} \text{ M} \] \[ \text{Mass} = 4.48 \times 10^{-4} \text{ mol} \times 110.98 \text{ g/mol} = 0.0497 \text{ g} = 49.7 \text{ mg} \] **Explanation:** With high ionic strength from MgCl₂, the activity coefficient drops further, so you need even more CaCl₂ to reach the same Ca²⁺ activity. --- ## **Final Answers (rounded to 2 significant digits):** - **Distilled Water:** **16 mg CaCl₂** per 1 L - **0.05 M KCl:** **32 mg CaCl₂** per 1 L - **0.1 M MgCl₂:** **50 mg CaCl₂** per 1 L --- ### **Summary Table** | Solution | CaCl₂ to Add (mg/L) | |--------------------|--------------------| | Distilled Water | 16 | | 0.05 M KCl | 32 | | 0.1 M MgCl₂ | 50 | --- **Each calculation involved:** 1. Relating activity and concentration using the activity coefficient. 2. Calculating ionic strength and then the activity coefficient with the Davies equation. 3. Converting the required Ca²⁺ concentration to the needed mass of CaCl₂. Let me know if you need the calculations shown for more significant digits or want a step-by-step for each!

The approach to determining how much CaCl₂ to add involved understanding the relationship between ion activity and concentration, recognizing that activity coefficients modify this relationship due to ionic interactions in solution. We used the Davies equation to estimate activity coefficients (\(\gamma_{Ca^{2+}}\)), which depend on the ionic strength of the solution. Since ionic strength varies with the background electrolyte present (distilled water, KCl, MgCl₂), we calculated it based on the concentrations and charges of all ions in solution. For each scenario, we first estimated the ionic strength and then derived the activity coefficient for Ca²⁺. Using the target activity (\(1.3 \times 10^{-4}\) M), we calculated the necessary free ion concentration by dividing the activity by \(\gamma_{Ca^{2+}}\). Finally, we converted this concentration into the mass of CaCl₂ needed, considering its molar mass, to prepare 1 liter of solution with the desired calcium activity in each case.

Yes, the solution approach is correct and follows standard procedures for calculating ion activity in solutions with varying ionic strengths using the Davies equation. Here's a brief recheck of key points: 1. **Relationship between activity and concentration:** \(\{Ca^{2+}\} = \gamma_{Ca^{2+}} [Ca^{2+}]\). To achieve a specific activity (\(1.3 \times 10^{-4}\) M), we need to find the required free ion concentration considering the activity coefficient. 2. **Calculation of ionic strength:** For each solution, ionic strength \(I\) was appropriately estimated based on background salts and the added calcium chloride. 3. **Use of Davies equation:** The Davies equation was applied correctly to estimate \(\gamma_{Ca^{2+}}\) based on ionic strength. The calculations for \(\sqrt{I}\), the term \(\frac{\sqrt{I}}{1+\sqrt{I}}\), and then \(\log_{10}\gamma_{Ca^{2+}}\) were performed consistently. 4. **Iterative approximation:** Given the low target activity, the initial estimation of \(x\) (calcium concentration) and \(\gamma_{Ca^{2+}}\) is reasonable. For more precise results, iterative refinement could be performed, but for practical purposes, the approximation is adequate. 5. **Mass calculation:** The molar concentrations were converted to mass of CaCl₂ correctly, considering molar mass and volume. **In summary:** The methodology is sound, correctly applies the Davies equation, and accurately estimates the amount of CaCl₂ needed for each solution type to achieve the specified calcium activity. Minor refinements could be made with iterative calculations, but the provided answers are correct within typical experimental tolerances.