1. **Problem 1: Titration and concentration calculations involving (NH4)2SO4 and NaOH** (i) The laboratory apparatus used to measure the 25.0 cm3 portion of solution C is a **pipette**. It should be used because it delivers an accurate and precise volume of liquid, which is essential for titration. (ii) Phenolphthalein indicator is **suitable** because it changes color in the pH range of about 8.2 to 10, which corresponds to the endpoint of the titration between a strong base (NaOH) and a strong acid (H2SO4). (iii) Few drops of phenolphthalein indicator are used to avoid excess indicator which could cause a false color change and affect the accuracy of the titration. (iv) The student needed to repeat the titration to ensure accuracy and reproducibility of results, as titrations can have slight variations. (v) Three experimental errors might include: - Parallax error when reading burette volumes. - Not mixing the solution thoroughly before taking readings. - Overshooting the endpoint by adding titrant too quickly. (b)(i) Calculate concentration of excess NaOH: - Average volume of H2SO4 used = $\frac{20.55 + 20.85 + 20.25}{3} = 20.55$ cm3 (approximate) - Convert to dm3: $20.55 \text{ cm}^3 = 0.02055$ dm3 - Moles of H2SO4 used in 25.0 cm3 portion: $$n = c \times V = 0.195 \times 0.02055 = 0.00400575 \text{ mol}$$ - From 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: $$c = \frac{n}{V} = \frac{0.0080115}{0.025} = 0.32046 \text{ mol dm}^{-3}$$ (ii) Calculate value of $x$ in $(NH_4)_2SO_4$: - Total moles of NaOH initially = $2.05 \times 1.00 = 2.05$ mol - Moles of NaOH remaining after reaction (from titration): $$\text{Volume of solution C} = 1.00 \text{ dm}^3$$ $$\text{Concentration} = 0.32046 \text{ mol dm}^{-3}$$ $$\text{Moles remaining} = 0.32046 \times 1.00 = 0.32046 \text{ mol}$$ - Moles of NaOH reacted with $(NH_4)_2SO_4$: $$2.05 - 0.32046 = 1.72954 \text{ mol}$$ - Reaction between $(NH_4)_2SO_4$ and NaOH releases NH3: $$(NH_4)_2SO_4 + 2NaOH \to 2NH_3 + Na_2SO_4 + 2H_2O$$ - Moles of $(NH_4)_2SO_4$ reacted = half moles of NaOH reacted: $$\frac{1.72954}{2} = 0.86477 \text{ mol}$$ - Molar mass of $(NH_4)_2SO_4$: $$2(14 + 4 \times 1) + 32 + 4 \times 16 = 2(18) + 32 + 64 = 36 + 32 + 64 = 132 \text{ g mol}^{-1}$$ - Mass $x$ g of $(NH_4)_2SO_4$: $$x = 0.86477 \times 132 = 114.07 \text{ g}$$ 2. **Problem 2: Completing the table for tests on mixture G and compound H** | TEST | OBSERVATION | INFERENCE | |---|---|---| | a. i | G(s) + distilled water + stir + filter: Black residue and colourless filtrate | G partially soluble; residue is insoluble solid | | ii | Filtrate + litmus: Filtrate is acidic | Filtrate contains acidic ions | | iii | Filtrate + HCl(aq) + warm: SO2 gas evolved | SO3^2- ions present in filtrate | | iv | Filtrate + NH3(aq) in drops: White gelatinous precipitate | Presence of Al^3+ or similar ions | | iv | Filtrate + NH3(aq) in excess: Precipitate dissolved | Amphoteric hydroxide present | | b. i | Residue + dilute HCl(aq): Effervescence, colourless odourless gas turns lime water milky; pale blue solution left | Residue contains carbonate (CO3^2-) and copper compounds | | b. ii | Solution from b(i) + NaOH drops in excess: Blue gelatinous precipitate, then dissolved | Cu^2+ ions present | | b. iii | Solution from b(i) + NH3 drops in excess: Cu^2+ confirmed | Cu^2+ ions confirmed | | c. i | H(aq) + million’s reagent: Brick-red precipitate | Presence of Cu^2+ ions | | c. ii | H(aq) + Conc. HNO3: No visible reaction | No reaction or inert to nitric acid | 3. **Problem 3: Identification of reagents and methods of salt preparation** (i) D: Zn metal; Method: **Reaction of metal with acid** (Zn + H2SO4 → ZnSO4 + H2) (ii) E: Cl^- ion (from NaCl or HCl); Method: **Precipitation reaction** (AgNO3 + Cl^- → AgCl(s)) (iii) F: KOH or K2CO3; Method: **Neutralization reaction** (KOH + HNO3 → KNO3 + H2O) (iv) G: Cl2 gas; Method: **Direct combination or displacement reaction** (Fe + Cl2 → FeCl3) 4. **Problem 4: Reasons for storage practices** (i) Hydrogen peroxide is stored in amber-colored reagent bottles to protect it from light which decomposes H2O2 into water and oxygen, reducing its effectiveness. (ii) Ethyne gas is stored in propanone in steel cylinders because ethyne is highly flammable and explosive; dissolving it in propanone stabilizes it and reduces explosion risk during storage and transport.