1. **Problem Statement:** A student used excess NaOH solution C to expel NH3 gas from (NH4)2SO4. A 25.0cm3 portion of solution C was titrated with 0.195 mol dm⁻³ H2SO4. Given titration data and reaction equation, we need to answer several questions including apparatus used, indicator suitability, titration repetition reasons, errors, and calculations of NaOH concentration and mass of (NH4)2SO4. 2. **(a)(i) Apparatus to measure 25.0cm³ portion:** A pipette is used to measure 25.0cm³ because it delivers a fixed, accurate volume of liquid, ensuring precise titration results. 3. **(a)(ii) Suitability of phenolphthalein indicator:** Phenolphthalein is suitable because it changes color in the pH range 8.2 to 10, which matches the endpoint of titrating a strong base (NaOH) with a strong acid (H2SO4). 4. **(a)(iii) Why few drops of phenolphthalein used:** Few drops are used to avoid excess indicator which can cause a faint color change and affect the accuracy of detecting the endpoint. 5. **(a)(iv) Reason for repeating titration:** Repeating titrations ensures reliability and accuracy by allowing calculation of an average titre and identification of anomalous results. 6. **(a)(v) Three possible experimental errors:** - Parallax error reading burette. - Not mixing solution thoroughly during titration. - Overshooting the endpoint by adding titrant too quickly. 7. **(b)(i) Calculate concentration of excess NaOH:** Average volume of H2SO4 used = $$\frac{20.55 + 20.85 + 20.25}{3} = 20.55\,cm^3$$ Convert to dm³: $$20.55\,cm^3 = 0.02055\,dm^3$$ Moles of H2SO4 used in titration: $$0.195 \times 0.02055 = 0.004005\,mol$$ From reaction: $$2\,NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$$ Mole ratio NaOH:H2SO4 = 2:1 Moles NaOH in 25.0cm³ portion: $$2 \times 0.004005 = 0.00801\,mol$$ Concentration of NaOH: $$\frac{0.00801}{0.025} = 0.3204\,mol\,dm^{-3}$$ 8. **(b)(ii) Calculate value of x in (NH4)2SO4:** Total moles of NaOH initially: $$2.05 \times 1.00 = 2.05\,mol$$ Moles NaOH reacted with H2SO4 in titration: $$0.00801 \times \frac{1000}{25} = 0.3204\,mol$$ Excess NaOH moles after reaction with (NH4)2SO4: $$2.05 - 0.3204 = 1.7296\,mol$$ Moles NaOH reacted with (NH4)2SO4: $$0.3204\,mol$$ Reaction between (NH4)2SO4 and NaOH to release NH3: $$(NH_4)_2SO_4 + 2NaOH \to Na_2SO_4 + 2NH_3 + 2H_2O$$ Moles (NH4)2SO4 reacted = $$\frac{0.3204}{2} = 0.1602\,mol$$ Molar mass of (NH4)2SO4: $$2(14 + 4 \times 1) + 32 + 4 \times 16 = 132\,g/mol$$ Mass of (NH4)2SO4: $$0.1602 \times 132 = 21.15\,g$$ Therefore, $$x = 21.15$$ grams. 9. **Summary:** - Pipette used for accurate volume measurement. - Phenolphthalein suitable for strong base-strong acid titration. - Few drops prevent excess indicator effect. - Repetition ensures accuracy. - Errors include parallax, mixing, and endpoint overshoot. - NaOH concentration calculated as $$0.3204\,mol\,dm^{-3}$$. - Mass of (NH4)2SO4 calculated as $$21.15\,g$$.