2026 WAEC GCE CHEMISTRY PRACTICAL ANSWER


(1)


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(2)


(2bi)
Chemical formula of C: CuSO₃

(2bii)
IUPAC name: Copper(II) trioxosulphate(IV)

(2c)
(i) CuSO₃(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l) + SO₂(g)
(ii) CuCl₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaCl(aq)
Ionic form: Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s)
(iii) CuSO₃(s) → CuO(s) + SO₂(g) (heat)

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(3ai)
Burette. A pipette can't measure 10.50 cm³.

(3aii)
Separating funnel. Tetrachloromethane and water are immiscible.

(3b)
(i) Phenolphthalein: colourless to pink.
(ii) Methyl orange: red to orange/yellow.
If the acid is in the burette instead, the changes are reversed: pink to colourless, and yellow to orange/red.

(3c)
The solubility of NaCl changes very little with temperature. Cooling its saturated solution therefore deposits almost no crystals. The solubility of CuSO₄ falls sharply on cooling, so its crystals (CuSO₄·5H₂O) separate out.

(3d)
(i) Pipette (or burette), to measure the stock solution
(ii) Volumetric flask
(iii) Wash bottle (distilled water)

(3e)
36% HCl is more concentrated. It contains 36 g of HCl per 100 g of solution, which is more solute per unit mass than 33%.

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2026 WAEC GCE CHEMISTRY THEORY ANSWER


(1a)
Incomplete combustion of butane gives carbon(II) oxide (CO) and water, and sometimes carbon (soot).
2C4H10 + 9O2 → 8CO + 10H2O

(1b)
(i) Sodium trioxocarbonate(IV) and calcium trioxocarbonate(IV): add water and filter. Na2CO3 dissolves, CaCO3 stays as residue.
(ii) KCl and NH4Cl: sublimation. NH4Cl sublimes and KCl does not.

(1c)
(i) Q is chloroethene (vinyl chloride): CH2=CHCl
H H
\ /
C = C
/ \
H Cl
(ii) Addition polymerization.

(1d)
(i) Structural isomers are compounds with the same molecular formula but different structural arrangements of atoms.
(ii) CH3CH2CH2CH3 (butane)
CH3CH(CH3)CH3 (2-methylpropane)

(1e)
15P: 1s2 2s2 2p6 3s2 3p3. Valence electrons = 5.
7N3-: 1s2 2s2 2p6. Valence electrons = 8.

(1f)
(i) s and p orbitals (2s and 2p).
(ii) 8 electrons.

(1g)
(i) Impurities raise the boiling point.
(ii) Impurities lower the melting point (and the substance melts over a range).

(1h)
(i) O2
(ii) O2-

(1i)
(i) Diesel and water: separating funnel.
(ii) Marble chips and sulphur powder: sieve (the chips are coarse, the powder is fine).

(1j)
(i)It is dynamic: the forward and backward reaction rates are equal.
(ii)The concentrations of reactants and products remain constant (it occurs in a closed system).

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(2)



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(3ai)
Zinc hydroxide chloride (zinc hydroxychloride).

(3aii)
Double salt.

(3aiii)
(I) Basic salt: Zn(OH)Cl (or Mg(OH)Cl)
(II) Amphoteric oxide: ZnO (or Al2O3)
(III) Complex salt: K4[Fe(CN)6] (or [Cu(NH3)4]SO4)

(3bi)
(I) CH3CHOHCH2CH3: secondary alkanol (butan-2-ol)
(II) CH3CH2CH2OH: primary alkanol (propan-1-ol)
(III) (CH3)2COHCH2CH3: tertiary alkanol (2-methylbutan-2-ol)

(3bii)
Warm each alkanol separately with acidified potassium dichromate(VI) (or acidified KMnO4).
Propan-1-ol (primary): the orange solution turns green (it is oxidized).
(CH3)2COHCH2CH3 (tertiary): no colour change; the solution stays orange (no oxidation).

(3ci)
(I) Carbon dating: 14C
(II) Reference scale for relative atomic mass: 12C

(3cii)
Relative atomic mass = (98.90 × 12 + 1.10 × 13) / 100
= (1186.8 + 14.3) / 100
= 12.011

(3ciii)
Relative atomic mass is a ratio: the average mass of an atom of the element divided by 1/12 of the mass of a 12C atom. The units cancel, so it has no unit.

(3di)
(α) With hydrogen bromide: bromoethane, CH3CH2Br
(β) With acidified KMnO4: ethane-1,2-diol (the purple colour is decolourized)
(γ) With acidified water (hydration): ethanol, CH3CH2OH

(3dii)
Polyethene (polythene).

(3diii)
(i)Making plastic bags and packaging films.
(ii)Making bowls, buckets and bottles and insulating electric wires.

(3e)
(i)Observation / identifying the problem.
(ii)Formulating a hypothesis.
(iii)Experimentation (testing the hypothesis).
(iv)Recording and analysing the data, then drawing a conclusion (theory or law)

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(4ai)
(i) Carbon monoxide – CO
(ii) Carbon dioxide – CO₂

(4aii)
I - CO
II - CO₂
III - CO

(4bi)
(i) Bituminous coal
(ii) Anthracite

(4bii)
(i) Coal tar
(ii) Ammoniacal liquor

(4ci)
Ammonium nitrite — NH₄NO₂

(4cii)
NH₄NO₂ → N₂ + 2H₂O

(4ciii)
(i) Manufacture of ammonia
(ii) Providing an inert atmosphere

(4di)
Haematite — Fe₂O₃

(4dii)
(i) Coke
(ii) Limestone

(4diii)
Hot air supports the combustion of coke and produces heat and carbon monoxide.

(4div)
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Molar mass of Fe₃O₄ = (3 × 56) + (4 × 16) = 232 g/mol
Moles of Fe₃O₄ = 5.80 ÷ 232 = 0.025 mol
Moles of Fe = 0.025 × 3 = 0.075 mol
Mass of Fe = 0.075 × 56
= 4.20 g

(4ei)
(i) Diamond has a giant covalent structure.
(ii) It contains strong covalent bonds requiring much energy to break.

(4eii)
(i) Lubricant: Its layers can slide over one another easily.
(ii) Electrode: It conducts electricity.

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(5ai)
(i) A transition element is a d-block element that has a partially filled d-subshell in its atom or in one or more of its ions.

(5aii)
(a) They show variable oxidation states.
(b) They form coloured compounds and complex ions.
(c) They also act as catalysts.

(5aiii)
Titanium, as Ti3+ (3d1), is surrounded by water ligands in solution, forming [Ti(H2O)6]3+. The ligands split the d-orbitals into two energy levels. The d-electron absorbs a certain wavelength of visible light and is promoted to the higher level (d-d transition). The remaining wavelengths are transmitted, so the ion appears coloured (purple).

(5bi)
(i) An equilibrium reaction is a reversible reaction in which the forward and backward rates are equal, so the concentrations of reactants and products stay constant.

(5bii)
(a) concentration,
(b) pressure (for gases), and
(c) temperature.

(5biii)
Kp = Kc(RT)^Δn
Δn = 2 − 3 = −1
Kc = Kp × RT
= 1.9 × 10^3 × 8.314 × 298
= 4.71 × 10^6

(5ci)
(i) Hydrolysis is the reaction of the ions of a salt with water to produce an acidic or basic solution.

(5cii)
(a) NH4Cl: turns blue litmus red (acidic).
(b) CH3COONa: turns red litmus blue (alkaline).
(c) KCl: no effect on litmus (neutral).

(5ciii)
(a) [H+] = 10^-pH = 10^-5.50 = 3.16 × 10^-6 mol dm^-3
(b) Moles of H+ in 250 cm3 = 3.16 × 10^-6 × 0.250 = 7.91 × 10^-7 mol
Number of H+ ions = 7.91 × 10^-7 × 6.02 × 10^23 = 4.76 × 10^17 ions