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A student uses a calibrated pH meter to measure an aqueous ethanoic acid solution. Figure 1 shows the meter reading.
State the pH of the solution shown in Figure 1.
Calculate the hydrogen ion concentration, [H⁺], in the solution. Give your answer in mol dm⁻³ to two significant figures.
At room temperature, of magnesium ribbon is added to excess dilute hydrochloric acid. Figure 1 shows the measured volume of gas collected over time.
Use Figure 1 to state the volume of gas collected after
Write a balanced equation, including state symbols, for the reaction of magnesium with dilute hydrochloric acid.
In a voltaic cell, electrode A undergoes the reaction
Identify electrode A as the anode or cathode and state its sign.
In an electrolytic cell, electrode B undergoes the reaction
Identify electrode B as the anode or cathode and state its sign.
Ethane reacts with bromine under ultraviolet light. Bromoethane is one product, and the reaction proceeds by a radical chain mechanism.
Write the two propagation equations for the reaction of ethane with bromine.
Write a termination equation involving two ethyl radicals.
State one reason why a mixture of organic products can form.
State two features of alkanes that contribute to their low reactivity.
The alkane shown reacts with under UV light. Which sequence shows initiation followed by the two propagation steps in which one H atom at C2 is replaced by Cl? A dot denotes one unpaired electron.
Answer: C. UV light causes homolytic fission of producing two chlorine radicals. A chlorine radical abstracts H from C2, forming and This radical reacts with to form and regenerate sustaining the chain. A: abstracts H at a terminal carbon, so the product is 1-chloropropane. B: uses radical–radical combination, a termination step that consumes the radicals rather than regenerating one. D: shows heterolytic fission, producing ions rather than the radicals needed to initiate the chain. Remember: a propagation step consumes one radical and produces another.
In a reaction mechanism for bromomethane, a curly arrow starts at the C–Br bond and points to Br. Which equation represents the products of this bond cleavage?
Answer: C. The curly arrow from the bond to Br shows the bonding electron pair moving to Br. Carbon loses its share of the pair, giving CH₃⁺; bromine gains both electrons, giving Br⁻.
A: splits the pair between the fragments, producing radicals by homolytic fission.
B: reverses the electron-pair movement, assigning both electrons to carbon.
D: represents cleavage of a C–H bond rather than the specified C–Br bond.
Remember: a curly arrow from a bond to an atom shows that the bonding pair moves to that atom.
Boron trifluoride reacts with ammonia to form a boron-containing adduct. A sample initially contains of Ammonia is added in measured amounts, and the amount of adduct formed is determined.
The following two stages occur when propanone reacts with cyanide ions in the presence of hydrogen cyanide:
In the product of the first stage, the carbon atom of is bonded to the carbon atom that was part of the carbonyl group.
Use Figure 1 to estimate the amount of at which the amount of adduct formed first becomes approximately constant. Hence, deduce the stoichiometric ratio of to in the adduct.
Write an equation for the formation of the adduct, showing the coordinate bond, and identify the Lewis acid and Lewis base.
State why the formation of the adduct in (b) is not a Brønsted–Lowry acid–base reaction.
A student claims, “Since each stage in the propanone reaction involves an electron-pair donor and an electron-pair acceptor, both stages must also be Brønsted–Lowry acid–base reactions.”
Annotate the two stages with curly arrows to show electron-pair movement. Identify the Lewis acid and Lewis base in each stage, and evaluate the student’s claim.
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