WAEC Chemistry 2021
Questions & Answers

Distractor check: A student might select glucose, sucrose, or fatty acids by confusing protein building blocks with carbohydrates or lipids. Reasoning to the answer: Proteins are large polymers built from amino acid monomers linked together by peptide bonds. When proteins undergo hydrolysis in the presence of dilute mineral acids, these peptide bonds are cleaved to release free amino acids. Common mistake: Confusing the monomer units of proteins with those of carbohydrates or fats.

Distractor check: A student might pick carbon monoxide or hydrogen peroxide by mistakenly associating common pollutants or household chemicals with precipitation acidity. Reasoning to the answer: Nitrogen(IV) oxide is a major contributor to acid rain because it reacts directly with atmospheric water and oxygen to produce corrosive nitric acid according to the equation 4NO₂ + 2H₂O + O₂ → 4HNO₃. Common mistake: Confusing nitrogen dioxide's acid-rain-forming properties with other gaseous pollutants.

Distractor check: A student might pick C₂H₆, C₂H₄, or CH₂ by miscalculating the molecular mass scaling factor or confusing empirical and molecular formulas. Reasoning to the answer: A carbon-to-hydrogen ratio of 1:2 gives an empirical formula of CH₂ with a mass unit of 14. Dividing the given molecular mass of 56 by this empirical mass yields a multiplier of 4, resulting in the molecular formula C₄H₈. Common mistake: Stopping at the empirical formula instead of scaling it up using the molecular mass.

Distractor check: A student might pick 30, 44, or 74 by miscalculating individual atomic contributions or miscounting the constituent atoms of ethanoic acid. Reasoning to the answer: The relative molecular mass of ethanoic acid, CH₃COOH, is determined by summing the atomic masses of its component atoms: two carbon atoms at 12 each, four hydrogen atoms at 1 each, and two oxygen atoms at 16 each, giving 2(12) + 4(1) + 2(16) = 60. Common mistake: Incorrectly summing or omitting individual atomic weights during molecular mass calculation.

Distractor check: A student might select electrical conductivity or the nature of oxides formed by assuming corrosion prevention relies on surface coatings rather than electrochemical principles. Reasoning to the answer: Cathodic protection works by attaching the metal you want to protect to a more reactive sacrificial metal that acts as an anode. This strategy relies on the relative oxidation tendencies and standard electrode potentials of the metals involved. Common mistake: Confusing sacrificial cathodic protection with physical barrier coatings.

Distractor check: A student might choose solid NO₂, saturated NaCl, or acidified KMnO₄ by guessing based on general chemical reagents without matching the specific reduction reaction needed for chlorine gas. Reasoning to the answer: Discharged chlorine gas in humid air is a strong oxidizing agent that can be safely neutralized and removed by reduction when treated with acidified iron(II) sulfate, producing chloride ions and iron(III) ions. Common mistake: Choosing generic oxidizing reagents instead of a reducing agent like Fe²⁺ to neutralize chlorine gas.

Students might mistakenly choose B because they see multiple methyl groups, assuming any branching automatically makes an alcohol tertiary. To determine the correct classification, we must look at the specific carbon that bears the hydroxyl (-OH) group; in (CH₃)₃COH, this central carbon is directly linked to three distinct methyl groups rather than just two or one. Because tertiary alcohols are defined by having their -OH group bonded to a carbon that is itself connected to three other carbon atoms, option C is the right choice. Common mistake: Confusing a secondary alcohol (which has the carbon bonded to two carbons) with a tertiary alcohol due to the presence of branching.

A student might be tempted by option A or B, thinking that mixing random salt solutions will always yield a visible reaction. When we combine potassium sulfate and barium chloride, however, a double displacement reaction takes place where barium ions react with sulfate ions to produce barium sulfate, represented by the balanced equation K₂SO₄ + BaCl₂ → BaSO₄ + 2KCl. This specific compound, barium sulfate, forms an insoluble white precipitate in water, whereas the other pairs exclusively produce fully soluble products that remain dissolved. Common mistake: Assuming all ionic combinations will produce a precipitate without checking the solubility rules for the resulting products.

A test-taker might select option D, knowing that higher temperatures generally speed up chemical processes, but the question specifically asks what increases the rate through direct particle frequency in this reaction setup. According to collision theory, raising the concentration of hydrochloric acid increases the total count of reactive H⁺ ions present in the mixture, which directly leads to a greater frequency of effective collisions with the magnesium surface. Volume alone will not alter this rate if the overall concentration remains unchanged. Common mistake: Confusing the effects of concentration and total solution volume on the frequency of reactant collisions.

One might incorrectly select option A, assuming reactions are most vigorous and therefore shift pH fastest right at the start. During the titration of a strong acid with a strong base, the steepest pH change actually happens at the end point—also known as the equivalence point—because a sudden stoichiometric shift occurs as you transition from a slight excess of acid to a slight excess of base with minimal buffering capacity in the mixture, producing a nearly vertical jump on the titration curve. Common mistake: Assuming the rate of pH change is uniform throughout the entire titration curve or highest at the very beginning.

It is easy to mistakenly select option A, as students often generalize that heating a mixture always enhances solubility. When a salt dissolves via an exothermic process, heat is released as a product of the dissolution, meaning Le Chatelier's principle dictates that raising the ambient temperature will shift the equilibrium position back toward the reactants (undissolved salt). Consequently, this temperature increase suppresses the process, meaning the solubility of the salt actually decreases. Common mistake: Forgetting that Le Chatelier's principle applies in reverse to exothermic dissolution processes compared to endothermic ones.

A student might choose option A, mistakenly thinking that removing something implies the reverse of an addition process. The dehydration of ethanol proceeds via the chemical equation CH₃CH₂OH → C₂H₄ + H₂O, which involves the removal of a hydrogen atom and a hydroxyl group from adjacent carbon atoms in the molecular chain. Since atoms are taken away to form a double bond rather than added to one, this classifies the transformation as a classic elimination reaction. Common mistake: Misinterpreting the removal of elements as a substitution or addition step because water is a product.

One might select option B or C, forgetting which gases can actively participate in intermolecular bonding with water solvents. Ammonia (NH₃) dissolves readily at room temperature because it readily forms strong hydrogen bonds with water molecules and further reacts to generate ammonium and hydroxide ions (NH₄⁺ and OH⁻). The other listed gases lack this specific capability, resulting in significantly lower solubility levels. Common mistake: Overlooking the role of hydrogen bonding and assuming all gases share similar solubility properties in aqueous solutions.

A student might guess option A due to the familiar diatomic structure of standard elemental gases like nitrogen or oxygen. White phosphorus, however, adopts a unique structural arrangement consisting of four atoms bound together in a spatial configuration known as P₄ tetrahedral molecules, making it tetraatomic. This specific molecular geometry is exceptionally stable and directly accounts for its unique chemical properties. Common mistake: Assuming all non-metal elements naturally exist as diatomic molecules in their elemental state.

A test-taker could mistakenly choose option C, confusing the general preparation of soluble salts via acid-base neutralization with the specific synthesis of insoluble ones. Insoluble salts are most commonly produced through double decomposition—also known as precipitation or metathesis—where two soluble salt solutions are mixed so that their ions swap partners, yielding an insoluble solid precipitate alongside a soluble salt, as seen in AgNO₃ + NaCl → AgCl ↓ + NaNO₃. Common mistake: Confusing neutralization methods with double decomposition when dealing with insoluble reaction products.

One might select option B or C by making a calculation error or misplacing decimal points when handling the gas constant. To determine the correct value, we apply the ideal gas law equation rearranged for moles: n = PV / (RT). Substituting the given values gives n = (1 atm × 8.2 dm³) / (0.082 atm dm³ mol⁻¹ K⁻¹ × 320 K), which simplifies to 8.2 / 26.24, resulting in approximately 0.3125 mol, which rounds to 0.32 mol. Common mistake: Failing to correctly divide the product of pressure and volume by the product of the gas constant and temperature.

A student might mistakenly pick option A, confusing efflorescence with moisture removal when observing physical changes in salts exposed to air. Drying agents function properly because they are hygroscopic, meaning they possess the chemical property of actively pulling and absorbing moisture directly from the surrounding atmosphere to form stable hydrates, such as when calcium chloride (CaCl₂) dries gases. Common mistake: Mixing up hygroscopic absorption of water with efflorescent loss of water of crystallization.

A person might select option C by remembering that ammonia forms precipitates with various metal ions, but forgetting the distinct color associated with copper chemistry. When aqueous ammonia is introduced dropwise to copper(II) ions, a light blue precipitate of copper(II) hydroxide forms via Cu²⁺ + 2NH₃, but adding excess ammonia causes this precipitate to redissolve completely, yielding a deep blue complex solution containing [Cu(NH₃)₄]²⁺. Common mistake: Confusing the color of copper precipitates or forgetting that they can redissolve in excess reagent to form coordination complexes.

A student might choose option A, assuming that mixing two different salt solutions containing common metal and sulfate/nitrate ions will yield no observable reaction. Upon combining copper(II) sulfate and lead(II) nitrate, a precipitation reaction occurs according to the equation CuSO₄ + Pb(NO₃)₂ → PbSO₄ ↓ + Cu(NO₃)₂. Lead(II) sulfate is famously insoluble in water, meaning it immediately precipitates out as a distinct white solid. Common mistake: Assuming all double displacement reactions between clear salt solutions result in clear, unchanged mixtures.

One might select option A, overlooking the fact that it represents the exact same straight-chain molecule rather than a structural isomer. Butane has the molecular formula C₄H₁₀, and while option A is simply n-butane, option B displays 2-methylpropane (isobutane), which shares that identical atomic formula but features a branched carbon skeleton. Options C and D are incorrect because they represent propane and pentane, possessing different carbon chain lengths. Common mistake: Selecting the straight-chain version of a compound when asked to identify its structural isomer.