WAEC Chemistry Practical Questions and Answers 2026

*WAEC CHEMISTRY PRACTICAL (PAPER 3 – ALT TO PRACTICAL)* *COMPLETE ANSWERS – QUESTIONS 1, 2 & 3* =========================================================== QUESTION 1: TITRATION (D vs E) D = H₂C₂O₄·2H₂O (oxalic acid dihydrate) – 1.58 g in 250 cm³ solution E = impure KMnO₄(aq) – 3.50 g/dm³ Pipette volume = 25.0 cm³ (assumed – use your actual) Indicator = none (KMnO₄ is self-indicating) Dilute H₂SO₄ added, mixture heated to about 50°C (a) Titration results table (SAMPLE – replace with your readings) Burette readings (cm³): Rough titre = 24.80 1st accurate titre = 24.60 2nd accurate titre = 24.70 Average volume of E used = (24.60 + 24.70) ÷ 2 = 49.30 ÷ 2 = 24.65 cm³ (b) Calculations (i) Concentration of D in g/dm³ Mass of H₂C₂O₄·2H₂O in 250 cm³ = 1.58 g Concentration = mass ÷ volume in dm³ = 1.58 ÷ 0.250 = 6.32 g/dm³ Answer: 6.32 g/dm³ (ii) Concentration of D in mol/dm³ Molar mass of H₂C₂O₄·2H₂O: H₂ = 2×1 = 2 C₂ = 2×12 = 24 O₄ = 4×16 = 64 2H₂O = 2×(18) = 36 Total = 2 + 24 + 64 + 36 = 126 g/mol Concentration in mol/dm³ = (concentration in g/dm³) ÷ molar mass = 6.32 ÷ 126 = 0.0502 mol/dm³ Answer: 0.0502 mol/dm³ (iii) Concentration of MnO₄⁻ in E in mol/dm³ From the equation: 5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂ Mole ratio: 5 mol C₂O₄²⁻ : 2 mol MnO₄⁻ Moles of D in pipetted volume: Volume of D = 25.0 cm³ = 0.025 dm³ Moles of D = concentration × volume = 0.0502 × 0.025 = 0.001255 mol Moles of MnO₄⁻ in average titre = (moles of D) × (2/5) = 0.001255 × 0.4 = 0.000502 mol Volume of E used = 24.65 cm³ = 0.02465 dm³ Concentration of MnO₄⁻ in E = moles ÷ volume = 0.000502 ÷ 0.02465 = 0.0204 mol/dm³ Answer: 0.0204 mol/dm³ (iv) Concentration of MnO₄⁻ in E in g/dm³ Molar mass of MnO₄⁻ = Mn(55) + O₄(64) = 119 g/mol Concentration in g/dm³ = concentration in mol/dm³ × molar mass = 0.0204 × 119 = 2.43 g/dm³ Answer: 2.43 g/dm³ (v) Percentage purity of KMnO₄ in E Given mass of impure KMnO₄ = 3.50 g/dm³ Molar mass of KMnO₄ = K(39) + Mn(55) + O₄(64) = 158 g/mol In pure KMnO₄, 1 mol KMnO₄ contains 1 mol MnO₄⁻ So concentration of KMnO₄ in g/dm³ = concentration of MnO₄⁻ in mol/dm³ × molar mass of KMnO₄ = 0.0204 × 158 = 3.22 g/dm³ Percentage purity = (mass of pure KMnO₄ ÷ mass of impure) × 100 = (3.22 ÷ 3.50) × 100 = 92.0% Answer: 92.0% =========================================================== QUESTION 2: QUALITATIVE ANALYSIS (F) F = mixture of inorganic salt + carbohydrate From specimen context, likely a metal salt (e.g., ZnSO₄, CuSO₄, or Pb(NO₃)₂) with starch or glucose. All tests, observations and inferences in tabular form (draw table in booklet) (a) F + distilled water → stir → filter. Test filtrate with litmus. +-------------------+---------------------+-----------------------------+ | Test / Observation| Inference | Conclusion | +-------------------+---------------------+-----------------------------+ | F partially | Inorganic salt is | Salt is soluble in water | | dissolves | soluble | | | Residue remains | Carbohydrate is | Carbohydrate is insoluble | | | starch (insoluble) | in cold water | | Filtrate turns | Solution is acidic | Hydrolysis of salt gives | | blue litmus red | | acidic solution (e.g., | | | | Zn²⁺ or Cu²⁺ salt) | +-------------------+---------------------+-----------------------------+ (b)(i) Filtrate + Zinc dust +-------------------+---------------------+-----------------------------+ | Test / Observation| Inference | Conclusion | +-------------------+---------------------+-----------------------------+ | Bubbles of gas | Gas evolved is | Presence of H⁺ ions (acidic | | evolved | hydrogen (H₂) | solution); Zn displaces H⁺ | | Pop sound with | | | | burning splint | | | +-------------------+---------------------+-----------------------------+ (b)(ii) Filtrate + NH₃(aq) in drops, then excess +-------------------+---------------------+-----------------------------+ | Test / Observation| Inference | Conclusion | +-------------------+---------------------+-----------------------------+ | Drops: coloured | Cu²⁺ or Fe²⁺ or | Transition metal ion | | precipitate forms | Ni²⁺ present | present | | Excess: | Cu²⁺ forms deep | If blue precipitate: Cu²⁺ | | precipitate | blue solution | If dirty green: Fe²⁺ | | dissolves (or not)| Zn²⁺ gives white, | If white insoluble: Pb²⁺ | | | then dissolves | (varies by actual salt) | +-------------------+---------------------+-----------------------------+ (b)(iii) Filtrate + BaCl₂(aq) then dilute HCl +-------------------+---------------------+-----------------------------+ | Test / Observation| Inference | Conclusion | +-------------------+---------------------+-----------------------------+ | White precipitate | SO₄²⁻ present | Salt is a sulphate | | forms with BaCl₂ | (barium sulphate) | e.g., CuSO₄, ZnSO₄, | | Precipitate does | insoluble in dilute | FeSO₄, or (NH₄)₂SO₄ | | NOT dissolve in | HCl | | | dilute HCl | | | +-------------------+---------------------+-----------------------------+ (c)(i) Residue + distilled water → boil +-------------------+---------------------+-----------------------------+ | Test / Observation| Inference | Conclusion | +-------------------+---------------------+-----------------------------+ | Mixture becomes | Starch granules | Carbohydrate is starch | | cloudy/paste-like | swell and thicken | (insoluble in cold water | | and thickens | in hot water | but forms paste when boiled)| +-------------------+---------------------+-----------------------------+ (c)(ii) Add iodine solution to boiled mixture +-------------------+---------------------+-----------------------------+ | Test / Observation| Inference | Conclusion | +-------------------+---------------------+-----------------------------+ | Deep blue-black | Starch present | Carbohydrate is confirmed | | colour appears | (Iodine-starch | as starch (amylose complex) | | | complex formed) | | +-------------------+---------------------+-----------------------------+ Overall conclusion: F contains a soluble metallic sulphate (e.g., CuSO₄, ZnSO₄, or FeSO₄) and insoluble starch. =========================================================== QUESTION 3: THEORY (TITRATION AND APPARATUS) (a)(i) Why are burette readings taken at eye level? To avoid parallax error. If the eye is above or below the meniscus, the reading will be either higher or lower than the true value. The meniscus should be at eye level for accurate reading. (a)(ii) Why is the funnel removed before taking the initial burette reading? To prevent drops of liquid from the funnel falling into the burette, which would increase the volume and cause an inaccurate initial reading. (a)(iii) Why is the burette rinsed with the acid solution? To remove any water or other liquid that may dilute the acid solution. Rinsing ensures that the solution in the burette is at the correct concentration. (b) List three laboratory apparatus required for the preparation of a standard solution. 1. Analytical balance (weighing balance) 2. Volumetric flask (e.g., 250 cm³ or 1000 cm³) 3. Weighing bottle or watch glass (or spatula and beaker) =========================================================== END OF ANSWERS =========================================================== IMPORTANT NOTES FOR THE EXAM 1. Replace SAMPLE burette readings (24.60, 24.70) with your actual readings.

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ANSWERS

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