1. **Problem 1a(i):** Mention the laboratory apparatus used to measure the 25.0cm³ portion of solution C and state why it should be used. - The apparatus used is a **pipette**. - A pipette is used because it measures a fixed volume of liquid accurately and precisely, which is essential for titration. 2. **Problem 1a(ii):** State whether phenolphthalein indicator is suitable or unsuitable. Give reason. - Phenolphthalein is **suitable** because it changes color in the pH range of about 8.2 to 10, which corresponds to the endpoint of titration between a strong base (NaOH) and a strong acid (H2SO4). 3. **Problem 1a(iii):** State why few drops of phenolphthalein indicator were used. - Few drops are used to avoid excess indicator which can cause a false color change and affect the accuracy of the titration. 4. **Problem 1a(iv):** Explain why the student needed to repeat another titration exercise. - Repeating titrations ensures accuracy and reliability of results by confirming consistent volume measurements and endpoint detection. 5. **Problem 1a(v):** State three experimental errors that might be committed. - Reading the burette incorrectly (parallax error). - Not mixing the solution thoroughly before titration. - Adding indicator in excess or too little. 6. **Problem 1b(i):** Calculate the concentration of the excess NaOH solution. - Average volume of H2SO4 used = $\frac{20.55 + 20.85 + 20.25}{3} = 20.55$ cm³ (approximate) - Convert to dm³: $20.55 \text{ cm}^3 = 0.02055 \text{ dm}^3$ - Moles of H2SO4 used in 25.0 cm³ portion: $$n = c \times V = 0.195 \times 0.02055 = 0.00400575 \text{ mol}$$ - From the reaction: $2 \text{NaOH} + \text{H}_2\text{SO}_4 \to \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}$ - Moles of NaOH in 25.0 cm³ portion = $2 \times 0.00400575 = 0.0080115$ mol - Concentration of NaOH in solution C: $$c = \frac{n}{V} = \frac{0.0080115}{0.025} = 0.32046 \text{ mol dm}^{-3}$$ 7. **Problem 1b(ii):** Calculate the value of $x$ in $(\text{NH}_4)_2\text{SO}_4$. - Total moles of NaOH initially = $2.05 \times 1.00 = 2.05$ mol - Moles of NaOH reacted with H2SO4 (titrated) = concentration $\times$ volume of excess solution titrated - Volume of excess solution C = 1.00 dm³ - Moles of NaOH neutralized by H2SO4 in entire 1.00 dm³: $$0.32046 \times 1.00 = 0.32046 \text{ mol}$$ - Moles of NaOH reacted with $(\text{NH}_4)_2\text{SO}_4$: $$2.05 - 0.32046 = 1.72954 \text{ mol}$$ - Reaction between $(\text{NH}_4)_2\text{SO}_4$ and NaOH: $$(\text{NH}_4)_2\text{SO}_4 + 4 \text{NaOH} \to 2 \text{NH}_3 + 4 \text{H}_2\text{O} + \text{Na}_2\text{SO}_4$$ - Moles of $(\text{NH}_4)_2\text{SO}_4$: $$\frac{1.72954}{4} = 0.432385 \text{ mol}$$ - Molar mass of $(\text{NH}_4)_2\text{SO}_4$: $$2 \times (14 + 4 \times 1) + 32 + 4 \times 16 = 132 \text{ g mol}^{-1}$$ - Mass $x$ g: $$x = 0.432385 \times 132 = 57.04 \text{ g}$$ --- 8. **Problem 2:** Complete the table based on tests and observations. | TEST | OBSERVATION | INFERENCE | |---|---|---| | a.i | G(s) + distilled water + stir + filter | Black residue and colourless filtrate | G is partially soluble | | a.ii | Filtrate + litmus | Litmus turns red | Filtrate is acidic | | a.iii | Filtrate + HCl(aq) + warm | SO2 gas evolved (smell and test) | SO3²⁻ ion present | | a.iv | Filtrate + NH3(aq) drops | White gelatinous precipitate formed | Presence of Al³⁺ or similar ions | | a.iv (excess NH3) | Precipitate dissolved | Complex ion formation | | b.i | Residue + dilute HCl | Effervescence, colourless odourless gas turns lime water milky; pale blue solution remains | Residue contains carbonate and Cu²⁺ ions | | b.ii | Residue + NaOH drops | Blue gelatinous precipitate | Cu²⁺ ions present | | b.ii (excess NaOH) | Precipitate dissolved | Formation of soluble complex | | b.iii | Residue + NH3 drops | Blue gelatinous precipitate | Cu²⁺ ions present | | b.iii (excess NH3) | Precipitate dissolved | Complex ion formation | | c.i | H(aq) + Million's reagent | Brick-red precipitate | Presence of Cu²⁺ ions | | c.ii | H(aq) + Conc. HNO3 | No reaction or specific observation | Confirm organic compound | --- 9. **Problem 3a:** Identify D, E, F, G and state method of preparation. - i. D = Zn metal; Method: Reaction of zinc metal with sulfuric acid to form ZnSO4. - ii. E = Cl⁻ ion (from NaCl or similar); Method: Precipitation reaction with AgNO3 to form AgCl(s). - iii. F = KOH or K-containing compound; Method: Neutralization of KOH with HNO3 to form KNO3. - iv. G = Fe metal; Method: Direct combination of Fe with chlorine gas to form FeCl3. --- 10. **Problem 3b(i):** Reason why hydrogen peroxide is stored in amber-colored reagent bottle. - To protect it from light which decomposes hydrogen peroxide, amber bottles block light. 11. **Problem 3b(ii):** Reason why ethyne gas is stored in propanone in steel cylinder. - Ethyne is dissolved in propanone to reduce explosion risk by lowering gas pressure and stabilizing it during storage.