Let's solve each part step by step: --- # **Question 2 Solution** --- ## **2.1 Leaching Reaction and Copper Specie** **Leaching reaction equation:** Malachite: \( \ce{Cu2CO3(OH)2} \) Dilute sulfuric acid: \( \ce{H2SO4} \) The leaching reaction is: \[ \ce{Cu2CO3(OH)2 (s) + 2 H2SO4 (aq) -> 2 Cu^{2+} (aq) + 2 SO4^{2-} (aq) + CO2 (g) + 3 H2O (l)} \] **Copper specie in solution:** Copper exists as the **cupric ion (\(\ce{Cu^{2+}}\))** in the leach solution. --- ## **2.2 Why Concentrated Sulphuric Acid Is Not Used** **Reason:** Concentrated sulfuric acid is **not used** because it is highly corrosive, poses safety hazards, and can cause excessive dissolution of unwanted gangue minerals, leading to impurities in solution and higher acid consumption. **Dilute acid** is safer, cheaper, and effective for selective copper extraction. --- ## **2.3 Solvent Extraction Reaction with Quinoline** **General reaction:** Assume the extractant quinoline (represented as QH) reacts with \(\ce{Cu^{2+}}\): \[ \ce{Cu^{2+} (aq) + 2 QH (org) <=> CuQ2 (org) + 2 H^+ (aq)} \] - Copper(II) ions in aqueous phase react with quinoline in organic phase to form a copper-quinolate complex, which is soluble in organic phase. --- ## **2.4 Benchmark Solvent Extraction** **Given:** - Volume of aqueous phase (\(V_a\)) = 300 mL - Volume of organic phase (\(V_o\)) = 240 mL - Initial copper concentration (\(C_\)) = 6.5 g/dm³ - Distribution coefficient (\(D\)) = 2.5 --- ### **2.4.1 Exponential Equation for Raffinate Concentration** Let: - \(C_n\): Copper concentration in aqueous phase after \(n\) stages - \(C_\): Initial copper concentration - \(V_a\): Volume of aqueous phase - \(V_o\): Volume of organic phase - \(D\): Distribution coefficient (\(D = \frac{\text{[Cu]}_{org}}{\text{[Cu]}_{aq}}\)) **In a single stage:** Fraction remaining in aqueous phase after one extraction: \[ \frac{C_1}{C_} = \frac{1}{1 + D \frac{V_o}{V_a}} \] **For \(n\) stages:** \[ \frac{C_n}{C_} = \left( \frac{1}{1 + D \frac{V_o}{V_a}} \right)^n \] So, \[ C_n = C_ \left( \frac{1}{1 + D \frac{V_o}{V_a}} \right)^n \] --- ### **2.4.2 Calculate Final Raffinate Concentration** Given: - \(D = 2.5\) - \(V_o = 240~\text{ml}\) - \(V_a = 300~\text{ml}\) - \(C_ = 6.5~\text{g/dm}^3\) - \(D \frac{V_o}{V_a} = 2.5 \times \frac{240}{300} = 2.5 \times .8 = 2.\) So, \[ \frac{C_n}{C_} = \left( \frac{1}{1 + 2.} \right)^n = \left( \frac{1}{3} \right)^n \] - **(i) 1 step (\(n=1\))** \[ C_1 = 6.5 \left( \frac{1}{3} \right)^1 = 6.5 \times .333 = \boxed{2.17~\text{g/dm}^3} \] - **(ii) 6 steps (\(n=6\))** \[ C_6 = 6.5 \left( \frac{1}{3} \right)^6 = 6.5 \times (.333)^6 \approx 6.5 \times .00137 = \boxed{.00891~\text{g/dm}^3} \] - **(iii) 8 steps (\(n=8\))** \[ C_8 = 6.5 \left( \frac{1}{3} \right)^8 = 6.5 \times (.333)^8 \approx 6.5 \times .000152 = \boxed{.00099~\text{g/dm}^3} \] --- ## **2.5 Effect of Large Number of Extraction Steps** **General effect:** Using a larger number of extraction steps **significantly reduces the final raffinate concentration** (amount of copper left in aqueous phase). Each step removes a fraction of the remaining copper, so more steps mean more complete extraction and lower residual copper in the raffinate. --- # **Summary Table** | Steps (\(n\)) | Final Raffinate Conc. (\(C_n\)) (g/dm³) | |:-------------:|:---------------------------------------:| | 1 | 2.17 | | 6 | .00891 | | 8 | .00099 | --- **End of Solution.**