✨ VIPSolutions

Let's break down and solve the given liquid-liquid extraction problem step by step for **Case (1)**. --- ### **Given Data** - **Feed**: 100 kg solution (25 wt% acetic acid, 75 wt% water) - Acetic acid: \( 0.25 \times 100 = 25 \) kg - Water: \( 0.75 \times 100 = 75 \) kg - **Solvent (isopropyl ether)**: 100 kg - **Temperature**: 20°C - **Operation**: Single-stage, all 100 kg ether contacted with the feed at once --- ### **Step 1: System Overview** - **Components**: Acetic acid (A), Water (W), Isopropyl ether (S) - **Streams**: - **Feed (F)**: 100 kg (A + W) - **Solvent (S)**: 100 kg (pure isopropyl ether) - **Raffinate (R)**: Aqueous phase after extraction - **Extract (E)**: Ether phase after extraction --- ### **Step 2: Draw the System** \[ \text{Feed (100 kg, 25% acid, 75% water)} + \text{Solvent (100 kg ether)} \rightarrow \text{Raffinate (R)} + \text{Extract (E)} \] --- ### **Step 3: Material Balances** Let: - \( R \) = mass of raffinate phase (aqueous) - \( E \) = mass of extract phase (ether) - \( x_A, x_W, x_S \) = mass fractions of acetic acid, water, and ether in raffinate - \( y_A, y_W, y_S \) = mass fractions of acetic acid, water, and ether in extract **Overall balance:** \[ F + S = R + E \\ 100 + 100 = R + E \implies R + E = 200 \text{ kg} \] **Component balances:** - Acetic acid: \( 25 = R x_A + E y_A \) - Water: \( 75 = R x_W + E y_W \) - Ether: \( 100 = R x_S + E y_S \) --- ### **Step 4: Use Equilibrium Data (Graphical Method)** For liquid-liquid extraction, the **graphical method** involves using a ternary phase diagram for (water, acetic acid, isopropyl ether) at 20°C. If you do not have the diagram, follow this outline: #### **A. Plot the overall mixture point** - **Feed composition** (before mixing with ether): 25% acetic acid, 75% water, 0% ether. - **Solvent composition**: 0% acetic acid, 0% water, 100% ether. - **Overall mixture**: - Acetic acid: \( 25 \) kg - Water: \( 75 \) kg - Ether: \( 100 \) kg - **Total:** 200 kg - **Overall mass fractions:** - Acetic acid: \( \frac{25}{200} = 0.125 \) - Water: \( \frac{75}{200} = 0.375 \) - Ether: \( \frac{100}{200} = 0.5 \) #### **B. Tie Line and Lever Rule** 1. **Locate overall composition** on ternary diagram. 2. **Draw tie line** passing through the overall composition, connecting equilibrium compositions of raffinate and extract. 3. **Determine raffinate and extract compositions** at ends of tie line. 4. **Apply lever rule** to calculate the amounts of each phase. --- ### **Step 5: Sample Solution (Assuming Graphical Data)** Suppose from the ternary diagram at 20°C: - **Raffinate composition** (aqueous phase): \( x_A = 0.07, x_W = 0.90, x_S = 0.03 \) - **Extract composition** (ether phase): \( y_A = 0.13, y_W = 0.02, y_S = 0.85 \) #### **A. Set up equations** Let \( R \) = mass of raffinate, \( E = 200 - R \). - Acetic acid: \( 25 = R(0.07) + (200 - R)(0.13) \) - Water: \( 75 = R(0.90) + (200 - R)(0.02) \) - Ether: \( 100 = R(0.03) + (200 - R)(0.85) \) #### **B. Solve for R** Using acetic acid balance: \[ 25 = 0.07R + 0.13(200 - R) \\ 25 = 0.07R + 26 - 0.13R \\ 25 - 26 = (0.07 - 0.13)R \\ -1 = -0.06R \\ R = \frac{-1}{-0.06} = 16.67 \text{ kg} \] \[ E = 200 - 16.67 = 183.33 \text{ kg} \] #### **C. Check with Water Balance** \[ 75 = 0.90 \times 16.67 + 0.02 \times 183.33 \\ 75 = 15 + 3.67 = 18.67 \text{ (not matching 75, so compositions from the diagram need to be adjusted)} \] #### **D. General Procedure** - Use the correct compositions from the ternary diagram for accurate results. - Apply the lever rule: \[ \frac{R}{E} = \frac{|P - E|}{|R - P|} \] where \( P \) is the overall mixture point, \( R \) and \( E \) are the endpoints of the tie line. --- ### **Final Answer Format** **You need to:** 1. Use the ternary diagram for the system at 20°C. 2. Plot the overall mixture point: (12.5% acetic acid, 37.5% water, 50% ether). 3. Draw a tie line through this point to the phase boundary. 4. Read off the compositions of raffinate and extract at the ends of the tie line. 5. Use the lever rule to determine the mass of each phase. 6. Calculate the mass of each component in each phase. --- #### **Summary Table (Example, assuming sample compositions):** | Phase | Mass (kg) | wt% Acetic Acid | wt% Water | wt% Ether | |------------|-----------|-----------------|-----------|-----------| | Raffinate | 16.67 | 7% | 90% | 3% | | Extract | 183.33 | 13% | 2% | 85% | > **Note:** For exact results, use the actual ternary diagram for acetic acid/water/isopropyl ether at 20°C. --- If you provide the ternary diagram or tie-line data, I can do the exact calculations for you!