Equilibrium Expressions and Concentrations: Chemistry Revision Exam

This original tutorgum practice paper assesses the Chemistry 2017 syllabus scope on equilibrium expressions and concentration calculations. It covers homogeneous equilibrium expressions, ionic dissociation, equilibrium constant calculations, reaction direction, and acid and base ionisation.

The paper is designed to test both your ability to calculate equilibrium concentrations and your understanding of what equilibrium expressions actually tell you. It is a focused topic revision paper, not a full-course HSC Chemistry examination. It is an original tutorgum resource, not an official or NESA-endorsed examination.

01Exam instructions

Instructions:

1. Attempt all questions.
2. Show all working for calculation questions. Answers without working may not receive full marks.
3. Give concentration answers in \(\mathrm{mol\,L^{-1}}\), unless otherwise specified.
4. Give numerical answers to three significant figures unless a different precision is appropriate.
5. Use equilibrium concentrations when constructing equilibrium expressions.
6. Assume temperature and solution volume remain constant unless stated otherwise.
7. Water is treated as the solvent and is omitted from acid and base dissociation expressions.
8. For this paper, report equilibrium constants as numerical values without units, using the concentration-based convention.
9. Reject any calculated equilibrium concentration that is physically impossible, even if it is a mathematical solution.

The symbols used are:

• \(K_c\): concentration equilibrium constant.
• \(K_d\): dissociation equilibrium constant.
• \(K_a\): acid dissociation constant.
• \(K_b\): base dissociation constant.
• \(Q_c\): reaction quotient, calculated using concentrations at a particular instant.

All questions use the supplied Chemistry 2017 syllabus scope. No other syllabus modules are assessed.

02Questions

Question 1 (6 marks)

Consider the following homogeneous equilibrium reactions in aqueous solution.

(a) Write the equilibrium constant expression, \(K_c\), for the reaction:

\[ \ce{Fe^3+(aq) + SCN^-(aq) <=> FeSCN^2+(aq)} \]

(2 marks)

(b) Write the equilibrium constant expression, \(K_c\), for the reaction:

\[ \ce{2A(aq) + B(aq) <=> A2B(aq)} \]

(2 marks)

(c) A student suggests that the coefficient 2 in front of A means that the equilibrium expression should contain \(2[A]\).

Explain why this is incorrect and describe how the coefficient is represented in the equilibrium expression.

(2 marks)

Question 2 (5 marks)

An aqueous solution contains an ion pair, MX, which undergoes the following dissociation equilibrium:

\[ \ce{MX(aq) <=> M+(aq) + X-(aq)} \]

The solution was initially prepared using MX only. There were no free \(\mathrm{M^+}\) or \(\mathrm{X^-}\) ions initially.

At equilibrium, the following concentrations are measured:

(a) Write the equilibrium constant expression, \(K_d\), for this dissociation reaction.

(2 marks)

(b) Calculate the dissociation equilibrium constant, \(K_d\).

(2 marks)

(c) Calculate the percentage of the original MX that has dissociated.

(1 mark)

Question 3 (5 marks)

A homogeneous aqueous reaction is represented by:

\[ \ce{2P(aq) + Q(aq) <=> R(aq)} \]

At a particular temperature, the system reaches equilibrium with the following concentrations:

(a) Construct the equilibrium constant expression, \(K_c\), for this reaction.

(2 marks)

(b) Calculate the numerical value of \(K_c\) at this temperature.

(3 marks)

Question 4 (5 marks)

Consider the following equilibrium:

\[ \ce{A(aq) + 2B(aq) <=> C(aq)} \]

At a particular temperature, the equilibrium constant is:

\[ K_c=3.00 \]

At a certain instant, the concentrations in the reaction mixture are:

\[ [A]=0.200\ \mathrm{mol\,L^{-1}} \]

\[ [B]=0.100\ \mathrm{mol\,L^{-1}} \]

\[ [C]=0.0120\ \mathrm{mol\,L^{-1}} \]

(a) Calculate the reaction quotient, \(Q_c\), using the concentrations at this instant.

(2 marks)

(b) Compare \(Q_c\) with \(K_c\) and predict the net direction in which the reaction will proceed to reach equilibrium.

(2 marks)

(c) State which reactant and product concentrations will increase and which will decrease as the system approaches equilibrium.

(1 mark)

Question 5 (8 marks)

A homogeneous reaction occurs in aqueous solution:

\[ \ce{2A(aq) <=> B(aq)} \]

At a constant temperature, the equilibrium constant is:

\[ K_c=2.00 \]

A solution initially contains:

\[ [A]_0=0.500\ \mathrm{mol\,L^{-1}} \]

\[ [B]_0=0.100\ \mathrm{mol\,L^{-1}} \]

The system is allowed to reach equilibrium.

(a) Construct an ICE table showing the initial, change, and equilibrium concentrations of A and B. Use \(x\) to represent the increase in the concentration of B.

(2 marks)

(b) Substitute the equilibrium concentrations into the equilibrium expression and obtain a quadratic equation in terms of \(x\).

(2 marks)

(c) Solve the quadratic equation. Identify the physically valid solution and explain why the other mathematical solution must be rejected.

(2 marks)

(d) Calculate the equilibrium concentrations of A and B.

(2 marks)

Question 6 (11 marks)

A chemist prepares two separate aqueous solutions at the same temperature.

The first contains a weak acid, HA:

\[ \ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)} \]

Its acid dissociation constant is:

\[ K_a=1.80\times10^{-5} \]

The second contains a weak base, B:

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

Its base dissociation constant is:

\[ K_b=2.50\times10^{-6} \]

Both solutions initially contain only their respective undissociated acid or base at a concentration of:

\[ 0.0500\ \mathrm{mol\,L^{-1}} \]

Assume that the ionisation of water itself is negligible.

(a) For the weak acid HA:

1. Write the acid dissociation expression, \(K_a\).
2. Set up an equilibrium concentration relationship using \(x\) to represent the amount of acid ionised.
3. Calculate the equilibrium concentration of \(\mathrm{H_3O^+}\).
4. Calculate the percentage ionisation of HA.

(5 marks)

(b) For the weak base B:

1. Write the base dissociation expression, \(K_b\).
2. Set up an equilibrium concentration relationship using \(x\) to represent the amount of base ionised.
3. Calculate the equilibrium concentration of \(\mathrm{OH^-}\).

(4 marks)

(c) Calculate the percentage ionisation of the base and compare it with the percentage ionisation of the acid.

Identify which solution has the greater fractional ionisation.

(2 marks)

03Worked solutions and marking guide

The solutions below show the expected reasoning and calculations. The marking guide indicates how marks are allocated to specific steps.

Equivalent algebraic methods and appropriately rounded answers are acceptable. Follow-through marks should be awarded where a student makes an earlier numerical error but subsequently applies a valid method consistently. A physically impossible concentration cannot be accepted as an equilibrium result.

Solution 1

(a) Equilibrium expression for iron thiocyanate formation

The reaction is:

\[ \ce{Fe^3+ + SCN^- <=> FeSCN^2+} \]

The general structure of an equilibrium expression is:

\[ K_c= \frac{\text{product concentrations}} {\text{reactant concentrations}} \]

Each concentration is raised to the power of its stoichiometric coefficient.

Therefore:

\[ \boxed{ K_c= \frac{[\mathrm{FeSCN}^{2+}]} {[\mathrm{Fe}^{3+}][\mathrm{SCN}^-]} } \]

All three species have a stoichiometric coefficient of 1, so each concentration has an implied power of 1.

Marking guide:

• Correct placement of products in the numerator and reactants in the denominator: 1 mark.
• Correct inclusion of all species and their powers: 1 mark.

(b) Equilibrium expression for the second reaction

The reaction is:

\[ \ce{2A + B <=> A2B} \]

There is one product species, A₂B, and two reactant species, A and B.

The coefficient of A is 2, so its concentration is squared.

\[ \boxed{ K_c=\frac{[A_2B]}{[A]^2[B]} } \]

Marking guide:

• Correct product concentration in the numerator: 1 mark.
• Correct reactant concentrations and powers in the denominator: 1 mark.

(c) Explaining the role of stoichiometric coefficients

Stoichiometric coefficients become powers in equilibrium expressions.

They do not become multipliers or additional terms.

For example, the coefficient 2 in front of A gives:

\[ [A]^2 \]

It does not give:

\[ 2[A] \]

or:

\[ [A]+[A] \]

The power represents the stoichiometric coefficient in the balanced chemical equation.

Marking guide:

• Identifies that coefficients become powers: 1 mark.
• Correctly explains the distinction using A: 1 mark.

Total: 6 marks

Solution 2

(a) Writing the dissociation expression

The reaction is:

\[ \ce{MX(aq) <=> M+(aq) + X-(aq)} \]

Both ions are products, while MX is the reactant.

Therefore:

\[ \boxed{ K_d=\frac{[\mathrm{M}^+][\mathrm{X}^-]}{[\mathrm{MX}]} } \]

Both product ions have a coefficient of 1, so neither concentration is raised to a higher power.

Marking guide:

• Correct identification of the product concentrations: 1 mark.
• Correct denominator and complete expression: 1 mark.

(b) Calculating the dissociation constant

Substitute the equilibrium concentrations:

\[ K_d= \frac{(0.0120)(0.0120)}{0.0400} \]

First, multiply the two ion concentrations:

\[ 0.0120\times0.0120=0.000144 \]

Then divide by the equilibrium concentration of MX:

\[ K_d=\frac{0.000144}{0.0400} \]

Therefore:

\[ \boxed{K_d=0.00360} \]

Marking guide:

• Correct substitution of equilibrium concentrations: 1 mark.
• Correct numerical answer: 1 mark.

(c) Calculating percentage dissociation

Percentage dissociation is the amount of the original substance that has dissociated, expressed as a percentage of its initial concentration.

The reaction has a 1:1 relationship between MX and \(\mathrm{M^+}\).

Therefore, the amount of MX that dissociated is:

\[ 0.0120\ \mathrm{mol\,L^{-1}} \]

The initial concentration of MX was:

\[ [\mathrm{MX}]_0=0.0400+0.0120 \]

\[ [\mathrm{MX}]_0=0.0520\ \mathrm{mol\,L^{-1}} \]

Now calculate the percentage:

\[ \begin{aligned} \%\text{ dissociation} &=\frac{\text{amount dissociated}} {\text{initial concentration}}\times100\\ &=\frac{0.0120}{0.0520}\times100 \end{aligned} \]

\[ \boxed{23.1\%} \]

Marking guide:

• Correct percentage dissociation: 1 mark.

Total: 5 marks

Solution 3

(a) Constructing the equilibrium expression

The reaction is:

\[ \ce{2P(aq) + Q(aq) <=> R(aq)} \]

R is the product, while P and Q are reactants.

The coefficient of P is 2, meaning its concentration must be squared.

Therefore:

\[ \boxed{ K_c=\frac{[R]}{[P]^2[Q]} } \]

Marking guide:

• Correct product in the numerator: 1 mark.
• Correct reactant concentrations and stoichiometric powers: 1 mark.

(b) Calculating the equilibrium constant

Substitute the equilibrium concentrations:

\[ K_c= \frac{0.0500}{(0.100)^2(0.200)} \]

Calculate the squared concentration:

\[ (0.100)^2=0.0100 \]

Then calculate the denominator:

\[ 0.0100\times0.200=0.00200 \]

Therefore:

\[ K_c=\frac{0.0500}{0.00200} \]

\[ \boxed{K_c=25.0} \]

The equilibrium constant is calculated using the equilibrium concentrations, not the initial concentrations.

Marking guide:

• Correct substitution, including the squared concentration of P: 2 marks.
• Correct final numerical answer: 1 mark.

Total: 5 marks

Solution 4

(a) Calculating the reaction quotient

The reaction is:

\[ \ce{A(aq) + 2B(aq) <=> C(aq)} \]

The reaction quotient has the same mathematical structure as the equilibrium expression.

However, it uses the concentrations at the particular instant being investigated, rather than necessarily using equilibrium concentrations.

Therefore:

\[ Q_c=\frac{[C]}{[A][B]^2} \]

Substitute the given concentrations:

\[ Q_c= \frac{0.0120}{(0.200)(0.100)^2} \]

Calculate:

\[ Q_c=\frac{0.0120}{0.00200} \]

\[ \boxed{Q_c=6.00} \]

Marking guide:

• Correct reaction quotient expression: 1 mark.
• Correct substitution and numerical answer: 1 mark.

(b) Predicting the reaction direction

We have calculated:

\[ Q_c=6.00 \]

The equilibrium constant is:

\[ K_c=3.00 \]

Therefore:

\[ Q_c>K_c \]

This means the current product-to-reactant concentration ratio is greater than the equilibrium ratio.

The reaction must consume some product and form more reactants to reach equilibrium.

Therefore, the net reaction proceeds in the reverse direction:

\[ \boxed{\ce{C(aq) -> A(aq) + 2B(aq)}} \]

Marking guide:

• Correct comparison of \(Q_c\) and \(K_c\): 1 mark.
• Correct prediction of the reverse reaction: 1 mark.

(c) Changes in concentration

As the reverse reaction proceeds:

• The concentration of A increases.
• The concentration of B increases.
• The concentration of C decreases.

This follows directly from the reverse reaction consuming C and producing A and B.

Marking guide:

• Correctly identifies the increasing reactants and decreasing product: 1 mark.

Total: 5 marks

Solution 5

(a) Constructing an ICE table

The reaction is:

\[ \ce{2A(aq) <=> B(aq)} \]

An ICE table records the initial concentrations, the changes during the reaction, and the final equilibrium concentrations.

We are told to let \(x\) represent the increase in the concentration of B.

Because two moles of A are consumed for every one mole of B produced, the concentration of A decreases by \(2x\).

The completed table is:

The stoichiometric ratio determines the relationship between the changes.

Marking guide:

• Correct stoichiometric changes: 1 mark.
• Correct equilibrium concentrations: 1 mark.

(b) Forming the quadratic equation

The equilibrium expression is:

\[ K_c=\frac{[B]}{[A]^2} \]

We are given:

\[ K_c=2.00 \]

Substitute the equilibrium concentrations from the ICE table:

\[ 2.00= \frac{0.100+x}{(0.500-2x)^2} \]

Multiply both sides by the denominator:

\[ 2.00(0.500-2x)^2=0.100+x \]

Expand the squared expression:

\[ 2.00(0.250-2.00x+4x^2)=0.100+x \]

Simplify:

\[ 0.500-4.00x+8.00x^2=0.100+x \]

Rearrange into standard quadratic form:

\[ \boxed{8.00x^2-5.00x+0.400=0} \]

Marking guide:

• Correct substitution into the equilibrium expression: 1 mark.
• Correctly rearranged quadratic equation: 1 mark.

(c) Solving the quadratic and selecting the valid root

We have:

\[ 8.00x^2-5.00x+0.400=0 \]

Using the quadratic formula:

\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]

Where:

\[ a=8.00,\quad b=-5.00,\quad c=0.400 \]

Substituting:

\[ x=\frac{5.00\pm\sqrt{25.0-12.8}}{16.0} \]

\[ x=\frac{5.00\pm\sqrt{12.2}}{16.0} \]

This gives two mathematical solutions:

\[ x=0.0942 \]

or:

\[ x=0.531 \]

However, not every mathematical solution is physically possible.

Remember that the equilibrium concentration of A is:

\[ [A]_{\mathrm{eq}}=0.500-2x \]

Test the second solution:

\[ [A]_{\mathrm{eq}}=0.500-2(0.531) \]

\[ [A]_{\mathrm{eq}}=-0.562\ \mathrm{mol\,L^{-1}} \]

This is physically impossible. A concentration cannot be negative.

Therefore, we reject \(x=0.531\) and accept:

\[ \boxed{x=0.0942\ \mathrm{mol\,L^{-1}}} \]

Marking guide:

• Correct mathematical roots: 1 mark.
• Selects the valid root and explains why the other produces an impossible concentration: 1 mark.

(d) Calculating equilibrium concentrations

We now use the valid value:

\[ x=0.0942 \]

For A:

\[ [A]_{\mathrm{eq}}=0.500-2x \]

\[ =0.500-2(0.0942) \]

\[ \boxed{[A]_{\mathrm{eq}}=0.312\ \mathrm{mol\,L^{-1}}} \]

For B:

\[ [B]_{\mathrm{eq}}=0.100+x \]

\[ =0.100+0.0942 \]

\[ \boxed{[B]_{\mathrm{eq}}=0.194\ \mathrm{mol\,L^{-1}}} \]

Both equilibrium concentrations are positive and physically possible.

Marking guide:

• Correct equilibrium concentration of A: 1 mark.
• Correct equilibrium concentration of B: 1 mark.

Total: 8 marks

Solution 6

(a) Weak acid dissociation

The acid dissociation reaction is:

\[ \ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)} \]

Water is the solvent, so it is omitted from the equilibrium expression.

The acid dissociation constant is:

\[ \boxed{ K_a=\frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]} } \]

We are given:

\[ K_a=1.80\times10^{-5} \]

The initial acid concentration is:

\[ [HA]_0=0.0500\ \mathrm{mol\,L^{-1}} \]

Let \(x\) represent the amount of acid ionised.

Because one mole of HA produces one mole of hydronium ions and one mole of A⁻ ions, the equilibrium concentrations are:

Substitute these concentrations into the acid dissociation expression:

\[ 1.80\times10^{-5} = \frac{x\times x}{0.0500-x} \]

Therefore:

\[ 1.80\times10^{-5} = \frac{x^2}{0.0500-x} \]

Rearrange:

\[ 1.80\times10^{-5}(0.0500-x)=x^2 \]

Expand:

\[ 9.00\times10^{-7}-1.80\times10^{-5}x=x^2 \]

Rearrange into quadratic form:

\[ x^2+1.80\times10^{-5}x -9.00\times10^{-7}=0 \]

Solving gives the physically valid root:

\[ x=9.40\times10^{-4} \]

Since \(x\) represents the hydronium ion concentration:

\[ \boxed{ [\mathrm{H_3O^+}]=9.40\times10^{-4}\ \mathrm{mol\,L^{-1}} } \]

We can now calculate the percentage ionisation.

Percentage ionisation is the proportion of the initial acid concentration that has dissociated.

\[ \%\text{ ionisation} = \frac{\text{concentration ionised}} {\text{initial acid concentration}} \times100 \]

Substitute:

\[ \%\text{ ionisation} = \frac{9.40\times10^{-4}}{0.0500} \times100 \]

Therefore:

\[ \boxed{\%\text{ ionisation}=1.88\%} \]

Only a small fraction of the original HA molecules have ionised.

Marking guide:

• Correct acid dissociation expression: 1 mark.
• Correct equilibrium relationship using \(x\): 1 mark.
• Correct solution of the quadratic and hydronium concentration: 2 marks.
• Correct percentage ionisation: 1 mark.

(b) Weak base dissociation

The base dissociation reaction is:

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

Again, water is the solvent and is omitted from the equilibrium expression.

The base dissociation constant is:

\[ \boxed{ K_b=\frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]} } \]

We are given:

\[ K_b=2.50\times10^{-6} \]

The initial base concentration is:

\[ [B]_0=0.0500\ \mathrm{mol\,L^{-1}} \]

Let \(x\) represent the amount of base ionised.

The equilibrium concentrations are:

Substitute into the equilibrium expression:

\[ 2.50\times10^{-6} = \frac{x\times x}{0.0500-x} \]

Therefore:

\[ 2.50\times10^{-6} = \frac{x^2}{0.0500-x} \]

Rearrange:

\[ 2.50\times10^{-6}(0.0500-x)=x^2 \]

Expand:

\[ 1.25\times10^{-7}-2.50\times10^{-6}x=x^2 \]

Rearrange:

\[ x^2+2.50\times10^{-6}x -1.25\times10^{-7}=0 \]

Solving gives the physically valid root:

\[ x=3.52\times10^{-4} \]

Since \(x\) represents the hydroxide ion concentration:

\[ \boxed{ [\mathrm{OH^-}]=3.52\times10^{-4}\ \mathrm{mol\,L^{-1}} } \]

Marking guide:

• Correct base dissociation expression: 1 mark.
• Correct equilibrium relationship using \(x\): 1 mark.
• Correct solution of the quadratic and hydroxide concentration: 2 marks.

(c) Comparing percentage ionisation

We have already calculated the acid’s percentage ionisation:

\[ \%\text{ ionisation of HA}=1.88\% \]

For the base, we use:

\[ \%\text{ ionisation} = \frac{x}{[B]_0}\times100 \]

Substitute:

\[ \%\text{ ionisation} = \frac{3.52\times10^{-4}}{0.0500} \times100 \]

Therefore:

\[ \boxed{\%\text{ ionisation of B}=0.705\%} \]

Compare the two results:

The weak acid has the greater percentage ionisation.

Both solutions started at the same concentration, so comparing their percentage ionisation gives a direct comparison of the fractions that dissociated.

The acid ionises to approximately 2.67 times the fraction of the base.

It is important to compare the actual fractional ionisation here, rather than treating \(K_a\) and \(K_b\) as interchangeable constants. They describe different dissociation reactions.

Marking guide:

• Correct calculation of the base’s percentage ionisation: 1 mark.
• Correct comparison and identification of the acid as having greater fractional ionisation: 1 mark.

Total: 11 marks

04Final examination checklist

Before finishing your revision, check that you can perform each of the following tasks without relying on the worked solutions.

• [ ] Construct equilibrium expressions using the correct species and stoichiometric powers.
• [ ] Distinguish between a coefficient and an exponent in an equilibrium expression.
• [ ] Calculate \(K_c\) using equilibrium concentrations.
• [ ] Construct dissociation expressions for ionic species.
• [ ] Calculate the percentage of an ionic substance that has dissociated.
• [ ] Use \(Q_c\) and \(K_c\) to predict the direction of a reaction.
• [ ] Construct an ICE table using the correct stoichiometric concentration changes.
• [ ] Solve quadratic equilibrium calculations and reject physically impossible roots.
• [ ] Construct \(K_a\) and \(K_b\) expressions for weak acid and weak base dissociation.
• [ ] Calculate equilibrium ion concentrations and percentage ionisation.

A useful final distinction: \(K_c\) describes the equilibrium position at a particular temperature. \(Q_c\) tells you the reaction’s current position relative to that equilibrium. Knowing how to calculate both, and interpret their relationship, is essential for this topic.