Let's break down the problem step by step: **Goal:** Prepare 1 L solutions with a calcium ion activity, \(\{Ca^{2+}\}\), of \(1.3 \times 10^{-4}\) M, using three different solvents: 1. Distilled water 2. 0.05 M KCl 3. 0.1 M MgCl\(_2\) We need to calculate the mass of CaCl\(_2\) to add in each case. ### **Step 1: Relate Activity to Concentration** The **activity** of an ion is given by: \[ \{Ca^{2+}\} = \gamma_{Ca^{2+}} [Ca^{2+}] \] Where: - \(\gamma_{Ca^{2+}}\) = activity coefficient for Ca\(^{2+}\) - \([Ca^{2+}]\) = molar concentration of Ca\(^{2+}\) So, \[ [Ca^{2+}] = \frac{\{Ca^{2+}\}}{\gamma_{Ca^{2+}}} \] **Explanation:** We need to determine \([Ca^{2+}]\) to know how much CaCl\(_2\) to add, but since the activity coefficient (\(\gamma\)) depends on the total ionic strength (which changes with each solution), we must calculate it for each case. --- ### **Step 2: Calculate Activity Coefficient (\(\gamma\)) Using the Davies Equation** The **Davies equation** for the activity coefficient of ion \(i\) (\(z_i\) = charge): \[ \log_{10} \gamma_i = -A z_i^2 \left( \frac{\sqrt{I}}{1 + \sqrt{I}} - 0.3I \right) \] Where: - \(A = 0.509\) at 25°C (for water) - \(z_i = +2\) for Ca\(^{2+}\) - \(I =\) ionic strength \(= \frac{1}{2} \sum c_j z_j^2\) (sum over all ions) Now, calculate \(I\) and \(\gamma_{Ca^{2+}}\) for each solution. #### **Case 1: Distilled Water (neglecting water's own ionic strength)** - Only CaCl\(_2\) is the source of ions. - CaCl\(_2\) → Ca\(^{2+}\) + 2Cl\(^{-}\) Let \(x = [Ca^{2+}]\) (to be determined). Thus, - \([Ca^{2+}] = x\) - \([Cl^{-}] = 2x\) \[ I = \frac{1}{2} \left( x \times 2^2 + 2x \times 1^2 \right) = \frac{1}{2} (4x + 2x) = 3x \] Now, we don't know \(x\) yet, but for small \(x\), \(I\) will be very small. Let's *assume* \(\gamma_{Ca^{2+}} \approx 1\) as a first approximation, then check. - \(\{Ca^{2+}\} = 1.3 \times 10^{-4}\) M - \(\gamma_{Ca^{2+}} \approx 1\) - \([Ca^{2+}] = \frac{1.3 \times 10^{-4}}{1} = 1.3 \times 10^{-4}\) M Now, \[ I = 3x = 3 \times 1.3 \times 10^{-4} = 3.9 \times 10^{-4} \, \text{M} \] \(\sqrt{I} = \sqrt{3.9 \times 10^{-4}} \approx 0.0198\) Now, plug into the Davies equation: \[ \log_{10} \gamma_{Ca^{2+}} = -0.509 \times 4 \left( \frac{0.0198}{1 + 0.0198} - 0.3 \times 3.9 \times 10^{-4} \right) \] \[ \frac{0.0198}{1.0198} \approx 0.0194 \] \[ 0.3 \times 3.9 \times 10^{-4} \approx 1.17 \times 10^{-4} \] \[ (0.0194 - 1.17 \times 10^{-4}) \approx 0.0193 \] \[ -0.509 \times 4 \times 0.0193 = -0.509 \times 0.0772 = -0.0393 \] \[ \gamma_{Ca^{2+}} = 10^{-0.0393} \approx 0.912 \] Now recalculate \([Ca^{2+}]\): \[ [Ca^{2+}] = \frac{1.3 \times 10^{-4}}{0.912} = 1.43 \times 10^{-4} \text{ M} \] Check new \(I\): \[ I = 3 \times 1.43 \times 10^{-4} = 4.29 \times 10^{-4} \] \(\sqrt{I} = 0.0207\) Repeat calculation with new \(I\): \[ \frac{0.0207}{1.0207} \approx 0.0203 \] \[ 0.3 \times 4.29 \times 10^{-4} \approx 1.29 \times 10^{-4} \] \[ 0.0203 - 0.000129 \approx 0.02017 \] \[ -0.509 \times 4 \times 0.02017 = -0.509 \times 0.08068 = -0.0411 \] \[ \gamma_{Ca^{2+}} = 10^{-0.0411} = 0.911 \] So, \([Ca^{2+}] = \frac{1.3 \times 10^{-4}}{0.911} = 1.43 \times 10^{-4}\) M **Explanation:** By iterating, we see the value stabilizes quickly. The required [Ca\(^{2+}\)] is **1.43 × 10\(^{-4}\) M**. --- #### **Case 2: In 0.05 M KCl** - KCl → K\(^+\) + Cl\(^-\) - Add CaCl\(_2\) (unknown x M): gives x M Ca\(^{2+}\) and 2x M Cl\(^-\) - Total [K\(^+\)] = 0.05 M - Total [Cl\(^-\)] = 0.05 M (from KCl) + 2x (from CaCl\(_2\)) Ionic strength: \[ I = \frac{1}{2} \left( x \times 4 + 0.05 \times 1 + 2x \times 1 \right) \] But sum for all ions: \[ I = \frac{1}{2} \left( x \times 4 + 0.05 \times 1 + (0.05 + 2x) \times 1 \right) \] Wait: KCl gives 0.05 M K\(^+\) and 0.05 M Cl\(^-\). CaCl\(_2\) gives x M Ca\(^{2+}\) and 2x M Cl\(^-\). So, - [K\(^+\)] = 0.05 M - [Ca\(^{2+}\)] = x - [Cl\(^-\)] = 0.05 + 2x \[ I = \frac{1}{2}\left( [K^+] \times 1^2 + [Ca^{2+}] \times 2^2 + [Cl^-] \times 1^2 \right) = \frac{1}{2}(0.05 \times 1 + x \times 4 + (0.05+2x) \times 1) \] \[ = \frac{1}{2}(0.05 + 4x + 0.05 + 2x) = \frac{1}{2}(0.1 + 6x) = 0.05 + 3x \] First, let's assume x is much smaller than 0.05 (which it is), so \(I \approx 0.05\). \(\sqrt{I} = \sqrt{0.05} = 0.2236\) \[ \frac{0.2236}{1.2236} = 0.1827 \] \[ 0.3 \times 0.05 = 0.015 \] \[ 0.1827 - 0.015 = 0.1677 \] \[ -0.509 \times 4 \times 0.1677 = -0.509 \times 0.6708 = -0.3415 \] \[ \gamma_{Ca^{2+}} = 10^{-0.3415} = 0.455 \] Now, \[ [Ca^{2+}] = \frac{1.3 \times 10^{-4}}{0.455} = 2.86 \times 10^{-4} \text{ M} \] Check new \(I = 0.05 + 3 \times 2.86 \times 10^{-4} = 0.0509\), which is nearly unchanged. **Explanation:** The ionic strength is dominated by the background KCl, so the correction from the CaCl\(_2\) is minimal. The required [Ca\(^{2+}\)] is **2.86 × 10\(^{-4}\) M**. --- #### **Case 3: In 0.1 M MgCl\(_2\)** MgCl\(_2\) → Mg\(^{2+}\) + 2Cl\(^{-}\) - [Mg\(^{2+}\)] = 0.1 M - [Cl\(^-\)] = 0.2 M (from MgCl\(_2\)) + 2x (from CaCl\(_2\)) - [Ca\(^{2+}\)] = x \[ I = \frac{1}{2}([Mg^{2+}] \times 2^2 + [Ca^{2+}] \times 2^2 + [Cl^-] \times 1^2) \] \[ = \frac{1}{2}(0.1 \times 4 + x \times 4 + (0.2 + 2x) \times 1) = \frac{1}{2}(0.4 + 4x + 0.2 + 2x) = \frac{1}{2}(0.6 + 6x) = 0.3 + 3x \] Assume x is small, so \(I \approx 0.3\). \(\sqrt{I} = \sqrt{0.3} = 0.5477\) \[ \frac{0.5477}{1.5477} = 0.3539 \] \[ 0.3 \times 0.3 = 0.09 \] \[ 0.3539 - 0.09 = 0.2639 \] \[ -0.509 \times 4 \times 0.2639 = -0.509 \times 1.0556 = -0.5376 \] \[ \gamma_{Ca^{2+}} = 10^{-0.5376} = 0.290 \] \[ [Ca^{2+}] = \frac{1.3 \times 10^{-4}}{0.290} = 4.48 \times 10^{-4} \text{ M} \] Now, \(I = 0.3 + 3 \times 4.48 \times 10^{-4} = 0.3013\), unchanged. **Explanation:** The ionic strength is dominated by the background MgCl\(_2\), so the correction from CaCl\(_2\) is negligible. The required [Ca\(^{2+}\)] is **4.48 × 10\(^{-4}\) M**. --- ### **Step 3: Calculate the mass of CaCl\(_2\) to add** CaCl\(_2\) (anhydrous) molar mass = 40.08 (Ca) + 2 × 35.45 (Cl) = **110.98 g/mol** For 1 L: \[ \text{moles CaCl}_2 = [Ca^{2+}] \] \[ \text{mass CaCl}_2 = \text{moles} \times 110.98 \text{ g/mol} \] #### **Case 1: Distilled water** \[ \text{mass} = 1.43 \times 10^{-4} \times 110.98 = 0.0159 \text{ g} = 15.9 \text{ mg} \] #### **Case 2: 0.05 M KCl** \[ \text{mass} = 2.86 \times 10^{-4} \times 110.98 = 0.0318 \text{ g} = 31.8 \text{ mg} \] ####