Conjugate Pairs and Acid-Base Behaviour: Chemistry Mock Exam

This original tutorgum practice paper assesses conjugate acid-base pairs, proton-transfer reactions, acid and base dissociation, amphiprotic salts, and the difference between acid/base strength and concentration. It is a focused revision assessment based on the supplied Chemistry 2017 syllabus scope, not a full-course examination or an official, endorsed, past, or predicted NESA paper.

01Exam instructions

Assessed content

This paper covers:

• Identifying conjugate acid-base pairs and explaining proton transfer.
• Writing ionic equations for acid and base dissociation in water.
• Explaining amphiprotic behaviour and writing equations for sodium hydrogen carbonate and potassium dihydrogen phosphate.
• Distinguishing strong and weak acids and bases.
• Distinguishing concentrated and dilute acids and bases using solution data and particle behaviour.

The paper does not assess pH calculations, titrations, buffers, equilibrium constant calculations, or material from other syllabus editions.

Working conventions

1. Include state symbols in all chemical equations.
2. When showing an acid transferring a proton to water, represent the product as \(\ce{H3O+}\), rather than using \(\mathrm{H}^{+}\) as shorthand.
3. Use a forward arrow (\(\rightarrow\)) for reactions modelled as essentially complete, and an equilibrium arrow (\(\rightleftharpoons\)) for weak acid and base reactions.
4. Assume aqueous solutions are at 25°C.
5. Show your working and explain your reasoning where requested.

These are the suggested conditions for this tutorgum practice paper, not statements about current official examination rules.

02Questions

Question 1 (6 marks)

Consider the following proton-transfer reactions.

(a) Ammonia reacts with water according to the equation:

\[ \ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)} \]

Identify:

• The Brønsted-Lowry acid and base.
• The two conjugate acid-base pairs.

(3 marks)

(b) Hydrogen carbonate ions react with hydronium ions according to:

\[ \ce{HCO3^-(aq) + H3O+(aq) -> H2CO3(aq) + H2O(l)} \]

Identify:

• The Brønsted-Lowry acid and base.
• The two conjugate acid-base pairs.

(3 marks)

Question 2 (8 marks)

Write a balanced ionic equation showing how each of the following substances behaves in water.

Include state symbols and use an appropriate reaction arrow.

(a) Hydrobromic acid, \(\ce{HBr}\), a strong acid. (2 marks)

(b) Methanoic acid, \(\ce{HCOOH}\), a weak acid. (2 marks)

(c) Potassium hydroxide, \(\ce{KOH}\), a strong base. (2 marks)

(d) Methylamine, \(\ce{CH3NH2}\), a weak base. (2 marks)

Question 3 (8 marks)

A chemistry student prepares four aqueous solutions. The following data are recorded.

The analytical concentration is the total concentration of solute added per litre of solution, before considering ionisation or dissociation.

(a) Classify each solute as a strong acid, weak acid, strong base, or weak base. (4 marks)

(b) Identify:

• The most concentrated solution.
• The most dilute solution.

(2 marks)

(c) A student argues that solution D must contain a strong base because it has the highest concentration.

Explain why this conclusion is incorrect. Then explain why solutions A and B have different acid strengths despite having the same analytical concentration. Use the supplied data in your response. (2 marks)

Question 4 (10 marks)

Sodium hydrogen carbonate, \(\ce{NaHCO3}\), and potassium dihydrogen phosphate, \(\ce{KH2PO4}\), are ionic compounds that dissolve in water.

(a) Define the term amphiprotic and identify the amphiprotic ion present in each of these salts. (2 marks)

(b) Write a balanced ionic dissociation equation for each salt dissolving completely in water.

Include state symbols. (2 marks)

(c) Consider the hydrogen carbonate ion, \(\mathrm{HCO_{3}}^{-}\).

Write two balanced equations showing its reaction with water:

• One in which hydrogen carbonate acts as an acid.
• One in which hydrogen carbonate acts as a base.

Label the role of hydrogen carbonate in each equation. (3 marks)

(d) Consider the dihydrogen phosphate ion, \(\mathrm{H_{2}PO_{4}}^{-}\).

Write two balanced equations showing its reaction with water:

• One in which dihydrogen phosphate acts as an acid.
• One in which dihydrogen phosphate acts as a base.

Label the role of dihydrogen phosphate in each equation. (3 marks)

Question 5 (8 marks)

A student is asked to write explicit equations showing how different substances behave in water.

They submit the following equations:

Equation (i)

\[ \ce{HCl(aq) -> H+(aq) + Cl-(aq)} \]

Equation (ii)

\[ \ce{NH3(aq) -> NH4+(aq) + OH-(aq)} \]

The student claims that both substances simply split into ions when dissolved in water.

(a) Explain what equation (i) leaves out when showing the explicit proton-transfer reaction of hydrochloric acid with water. Rewrite the equation correctly, including all relevant species and state symbols. (3 marks)

(b) Explain why equation (ii) does not correctly represent the behaviour of ammonia in water. Rewrite the equation to show its actual behaviour. (3 marks)

(c) Identify the two conjugate acid-base pairs in your corrected equation for ammonia. (2 marks)

03Worked solutions and marking guide

Solution 1

(a) Ammonia and water

The balanced reaction is:

\[ \ce{NH3 + H2O <=> NH4+ + OH-} \]

Ammonia, \(\ce{NH3}\), accepts a proton from water. Therefore, ammonia acts as the Brønsted-Lowry base.

Water donates a proton to ammonia. Therefore, water acts as the Brønsted-Lowry acid.

The conjugate acid-base pairs are:

1. \(\ce{NH4+ / NH3}\): ammonia accepts a proton to become its conjugate acid, ammonium.
2. \(\ce{H2O / OH-}\): water donates a proton to become its conjugate base, hydroxide.

Remember that conjugate acid-base pairs differ by exactly one proton, \(\mathrm{H}^{+}\).

Marking guide (3 marks)

(b) Hydrogen carbonate and hydronium

The reaction is:

\[ \ce{HCO3^- + H3O+ -> H2CO3 + H2O} \]

Hydronium, \(\ce{H3O+}\), donates a proton to hydrogen carbonate. It therefore acts as the Brønsted-Lowry acid.

Hydrogen carbonate, \(\mathrm{HCO_{3}}^{-}\), accepts that proton and becomes carbonic acid, \(\ce{H2CO3}\). It therefore acts as the Brønsted-Lowry base.

The conjugate acid-base pairs are:

1. \(\ce{H2CO3 / HCO3-}\): hydrogen carbonate accepts a proton to form carbonic acid.
2. \(\ce{H3O+ / H2O}\): hydronium donates a proton to form water.

Notice that hydrogen carbonate is acting as a base here. Its behaviour can change in a different reaction.

Marking guide (3 marks)

Solution 2

(a) Hydrobromic acid, \(\ce{HBr}\)

Hydrobromic acid is a strong acid. It transfers its proton to water, and the reaction is modelled as essentially complete.

\[ \boxed{\ce{HBr(aq) + H2O(l) -> H3O+(aq) + Br-(aq)}} \]

Water accepts a proton from HBr, producing hydronium and bromide ions.

Marking guide (2 marks)

• 1 mark for the correctly balanced equation, including the relevant species and state symbols.
• 1 mark for using the forward arrow to represent essentially complete ionisation.

(b) Methanoic acid, \(\ce{HCOOH}\)

Methanoic acid is a weak acid. Only a small proportion of its molecules ionise in water, so an equilibrium arrow is appropriate.

\[ \boxed{\ce{HCOOH(aq) + H2O(l) <=> H3O+(aq) + HCOO-(aq)}} \]

Methanoic acid donates a proton to water, forming hydronium and methanoate ions.

Marking guide (2 marks)

• 1 mark for the correctly balanced equation, including state symbols.
• 1 mark for using the equilibrium arrow to represent partial ionisation.

(c) Potassium hydroxide, \(\ce{KOH}\)

Potassium hydroxide is a strong base. As an ionic compound, it dissociates completely into its constituent ions when dissolved in water.

\[ \boxed{\ce{KOH(aq) -> K+(aq) + OH-(aq)}} \]

The hydroxide ions produced are responsible for the basic behaviour of the solution.

Marking guide (2 marks)

• 1 mark for the correctly balanced dissociation equation, including state symbols.
• 1 mark for using the forward arrow to represent essentially complete dissociation.

(d) Methylamine, \(\ce{CH3NH2}\)

Methylamine is a weak base. It accepts a proton from water, producing methylammonium and hydroxide ions.

\[ \boxed{\ce{CH3NH2(aq) + H2O(l) <=> CH3NH3+(aq) + OH-(aq)}} \]

Only a fraction of the methylamine molecules react with water, so an equilibrium arrow is required.

Marking guide (2 marks)

• 1 mark for the correctly balanced proton-transfer equation, including state symbols.
• 1 mark for using the equilibrium arrow to represent partial reaction.

Solution 3

(a) Classifying the four solutions

The strength of an acid or base describes the extent to which it ionises or dissociates in water.

Using the percentage data:

A high percentage of ionisation or dissociation indicates strong acid or base behaviour.

A low percentage indicates weak acid or base behaviour.

Marking guide (4 marks)

• 1 mark for correctly identifying A as a strong acid.
• 1 mark for correctly identifying B as a weak acid.
• 1 mark for correctly identifying C as a strong base.
• 1 mark for correctly identifying D as a weak base.

(b) Identifying the most concentrated and most dilute solutions

Concentration describes the amount of solute dissolved per unit volume of solution.

The four concentrations are:

• A: \(0.100\ \mathrm{mol\,L^{-1}}\)
• B: \(0.100\ \mathrm{mol\,L^{-1}}\)
• C: \(0.0100\ \mathrm{mol\,L^{-1}}\)
• D: \(0.500\ \mathrm{mol\,L^{-1}}\)

Therefore:

• Most concentrated: Solution D, at \(0.500\ \mathrm{mol\,L^{-1}}\).
• Most dilute: Solution C, at \(0.0100\ \mathrm{mol\,L^{-1}}\).

Marking guide (2 marks)

• 1 mark for identifying D as the most concentrated.
• 1 mark for identifying C as the most dilute.

(c) Explaining the difference between strength and concentration

The student’s conclusion about solution D is incorrect because concentration and strength describe different properties.

Solution D contains the highest concentration of solute, at \(0.500\ \mathrm{mol\,L^{-1}}\). However, only 1.0% of its ammonia molecules ionise in water. This makes ammonia a weak base, despite the solution being concentrated.

Solutions A and B have identical analytical concentrations of \(0.100\ \mathrm{mol\,L^{-1}}\), but their acid strengths differ.

• HCl in solution A is approximately 100% ionised.
• Ethanoic acid in solution B is only 1.3% ionised.

HCl is therefore a strong acid, while ethanoic acid is a weak acid.

The key distinction is that concentration measures how much solute is present, while strength describes how extensively that solute ionises or dissociates in water.

Marking guide (2 marks)

• 1 mark for explaining that D is concentrated because of its high solute concentration but weak because only 1.0% ionises.
• 1 mark for comparing A and B using their equal concentrations and substantially different percentages of ionisation.

Solution 4

(a) Defining amphiprotic behaviour

An amphiprotic substance can both donate and accept a proton, \(\mathrm{H}^{+}\), depending on the reaction partner.

The amphiprotic ions in the two salts are:

• Sodium hydrogen carbonate, \(\ce{NaHCO3}\): hydrogen carbonate ion, \(\mathrm{HCO_{3}}^{-}\).
• Potassium dihydrogen phosphate, \(\ce{KH2PO4}\): dihydrogen phosphate ion, \(\mathrm{H_{2}PO_{4}}^{-}\).

Both ions can act as either Brønsted-Lowry acids or Brønsted-Lowry bases.

Marking guide (2 marks)

• 1 mark for explaining that an amphiprotic substance can donate and accept a proton.
• 1 mark for correctly identifying both amphiprotic ions.

(b) Ionic dissociation equations

When sodium hydrogen carbonate dissolves in water, it separates into sodium and hydrogen carbonate ions.

\[ \boxed{\ce{NaHCO3(s) -> Na+(aq) + HCO3-(aq)}} \]

When potassium dihydrogen phosphate dissolves in water, it separates into potassium and dihydrogen phosphate ions.

\[ \boxed{\ce{KH2PO4(s) -> K+(aq) + H2PO4-(aq)}} \]

The sodium and potassium ions are spectator ions in the subsequent acid-base reactions. The hydrogen-containing anions are the species that demonstrate amphiprotic behaviour.

Marking guide (2 marks)

• 1 mark for the correctly balanced sodium hydrogen carbonate dissociation equation.
• 1 mark for the correctly balanced potassium dihydrogen phosphate dissociation equation.

(c) Amphiprotic behaviour of hydrogen carbonate

Hydrogen carbonate, \(\mathrm{HCO_{3}}^{-}\), can act as both an acid and a base.

Acting as an acid

An acid donates a proton.

Hydrogen carbonate donates a proton to water, producing carbonate and hydronium ions.

\[ \boxed{\ce{HCO3^-(aq) + H2O(l) <=> CO3^2-(aq) + H3O+(aq)}} \]

Here, \(\mathrm{HCO_{3}}^{-}\) acts as the acid, while water acts as the base.

Acting as a base

A base accepts a proton.

Hydrogen carbonate accepts a proton from water, producing carbonic acid and hydroxide ions.

\[ \boxed{\ce{HCO3^-(aq) + H2O(l) <=> H2CO3(aq) + OH-(aq)}} \]

Here, \(\mathrm{HCO_{3}}^{-}\) acts as the base, while water acts as the acid.

The same hydrogen carbonate ion can therefore participate in two different proton-transfer reactions.

Marking guide (3 marks)

• 1 mark for the correctly balanced equation showing hydrogen carbonate acting as an acid.
• 1 mark for the correctly balanced equation showing hydrogen carbonate acting as a base.
• 1 mark for correctly labelling both roles.

(d) Amphiprotic behaviour of dihydrogen phosphate

Dihydrogen phosphate, \(\mathrm{H_{2}PO_{4}}^{-}\), also acts as both an acid and a base.

Acting as an acid

Dihydrogen phosphate donates a proton to water.

\[ \boxed{\ce{H2PO4^-(aq) + H2O(l) <=> HPO4^2-(aq) + H3O+(aq)}} \]

The dihydrogen phosphate ion acts as the acid because it donates a proton.

Water acts as the base because it accepts that proton.

Acting as a base

Dihydrogen phosphate accepts a proton from water.

\[ \boxed{\ce{H2PO4^-(aq) + H2O(l) <=> H3PO4(aq) + OH-(aq)}} \]

The dihydrogen phosphate ion acts as the base because it accepts a proton.

Water acts as the acid because it donates that proton.

These two reactions demonstrate why \(\mathrm{H_{2}PO_{4}}^{-}\) is classified as amphiprotic.

Marking guide (3 marks)

• 1 mark for the correctly balanced equation showing dihydrogen phosphate acting as an acid.
• 1 mark for the correctly balanced equation showing dihydrogen phosphate acting as a base.
• 1 mark for correctly labelling both roles.

Solution 5

(a) Correcting the hydrochloric acid equation

The student’s equation is:

\[ \ce{HCl(aq) -> H+(aq) + Cl-(aq)} \]

This is a shorthand representation of hydrochloric acid ionisation. However, it does not explicitly show water accepting the proton from hydrochloric acid.

When writing the complete proton-transfer reaction, water must be included as a reactant.

Hydrochloric acid donates a proton to water, forming hydronium and chloride ions.

The correct equation is:

\[ \boxed{\ce{HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)}} \]

The forward arrow is appropriate because hydrochloric acid is a strong acid and ionises essentially completely in water.

Marking guide (3 marks)

• 1 mark for explaining that water accepts the proton and is missing from the original equation.
• 1 mark for writing the correctly balanced equation with hydronium and chloride ions.
• 1 mark for using the forward arrow to represent essentially complete ionisation.

(b) Correcting the ammonia equation

The student’s equation is:

\[ \ce{NH3(aq) -> NH4+(aq) + OH-(aq)} \]

This equation is incorrect because ammonia does not simply split into ammonium and hydroxide ions.

Ammonia is a weak base. It accepts a proton from water to form ammonium ions, while water produces hydroxide ions.

The correct equation is:

\[ \boxed{\ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)}} \]

The equilibrium arrow is important.

Ammonia is a weak base, meaning only a fraction of its molecules react with water. The forward and reverse reactions can therefore occur.

Marking guide (3 marks)

• 1 mark for explaining that ammonia accepts a proton from water rather than simply dissociating.
• 1 mark for writing the correctly balanced proton-transfer equation.
• 1 mark for using the equilibrium arrow to represent partial reaction.

(c) Identifying the conjugate acid-base pairs

The corrected equation is:

\[ \ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)} \]

The two conjugate acid-base pairs are:

1. \(\ce{NH4+ / NH3}\): ammonia accepts a proton to become ammonium.
2. \(\ce{H2O / OH-}\): water donates a proton to become hydroxide.

Each pair differs by one proton.

Marking guide (2 marks)

• 1 mark for correctly identifying \(\ce{NH4+ / NH3}\).
• 1 mark for correctly identifying \(\ce{H2O / OH-}\).

04Final mark allocation

05Key concepts to review after the paper

Use your results to identify which concepts need more revision.

One particularly important distinction to remember is that strength and concentration are independent properties. A concentrated solution can contain a weak acid or base, while a dilute solution can contain a strong acid or base.

Being able to explain that distinction using both chemical equations and solution data is an important part of mastering this topic.