How many NESA Chemistry questions cover Quantitative Analysis?

AusGrader has 102 NESA Chemistry questions on Quantitative Analysis, all with instant AI grading and detailed marking feedback.

HSC Chemistry — Quantitative Analysis Questions & Answers

Q23

2020

VCAA

1 mark

Q23

1 mark

Use the following information to answer the question.

A solution of citric acid, $\text{C}_3\text{H}_5\text{O(COOH)}_3$ , was analysed by titration.
25.0 mL aliquots of the $\text{C}_3\text{H}_5\text{O(COOH)}_3$ solution were titrated against a standardised solution of 0.0250 M sodium hydroxide, NaOH. Phenolphthalein indicator was used and the average titre was found to be 24.0 mL.

Based on the titration, the concentration of $\text{C}_3\text{H}_5\text{O(COOH)}_3$ in the solution was

A

$8.0 \times 10^{-3} \text{ M}$

B

$8.7 \times 10^{-3} \text{ M}$

C

$2.6 \times 10^{-2} \text{ M}$

D

$7.2 \times 10^{-2} \text{ M}$

Reveal Answer

A

$8.0 \times 10^{-3} \text{ M}$

Correct Answer

Citric acid is triprotic, reacting with NaOH in a 1:3 ratio. The moles of NaOH used is $0.0250 \times 0.0240 = 6.0 \times 10^{-4}$ mol, so the moles of acid is $2.0 \times 10^{-4}$ mol. Dividing by the aliquot volume (0.0250 L) gives $8.0 \times 10^{-3}$ M.

B

$8.7 \times 10^{-3} \text{ M}$

This incorrect answer is obtained by mistakenly swapping the volumes of the acid and base in the concentration calculation.

C

$2.6 \times 10^{-2} \text{ M}$

This error occurs if the volumes of the acid and base are swapped and the 1:3 stoichiometric ratio between citric acid and NaOH is ignored.

D

$7.2 \times 10^{-2} \text{ M}$

This incorrect result comes from multiplying the moles of NaOH by 3 instead of dividing by 3 to find the moles of the triprotic citric acid.

Q6

2024

SCSA

1 mark

Q6

1 mark

Use the following information to answer the question.

A group of students conducted a series of titrations to determine the concentration of acetic acid in vinegar using the following steps:

i. A sample of vinegar was pipetted into a volumetric flask that had been rinsed with the vinegar and then deionised water added up to the mark.
ii. The volumetric flask was stoppered, and the diluted solution mixed thoroughly.
iii. Aliquots of the diluted vinegar solution were pipetted into conical flasks that had been rinsed with deionised water and a few drops of indicator added to each flask.
iv. A standardised sodium hydroxide solution was added to a burette that had been rinsed with deionised water.
v. Two samples of diluted vinegar were titrated against the sodium hydroxide solution and both values were used to calculate the concentration of the vinegar.

Phenolphthalein was chosen as the indicator for the titration. Which of the following best explains why this was an appropriate indicator?

A

The titration was between a strong base and a weak acid, so the final solution would be slightly basic.

B

Phenolphthalein changes colour in the basic range, at a similar pH to the equivalence point of the titration.

C

Phenolphthalein is pink in the acidic range and colourless in the basic range, and so the end point is easy to identify.

D

At the equivalence point $[OH^-] > [H^+]$ ; therefore, phenolphthalein is an appropriate indicator as its colour change occurs at a basic pH.

Reveal Answer

A

The titration was between a strong base and a weak acid, so the final solution would be slightly basic.

While this statement is true, it fails to mention phenolphthalein's specific properties or its colour change range, which is necessary to explain why it was chosen as the indicator.

B

Phenolphthalein changes colour in the basic range, at a similar pH to the equivalence point of the titration.

Although this statement is generally correct, it is not the best explanation because it lacks the specific chemical justification for the basic equivalence point that is provided in Option D.

C

Phenolphthalein is pink in the acidic range and colourless in the basic range, and so the end point is easy to identify.

This option states the incorrect colour change. Phenolphthalein is colourless in acidic solutions and turns pink in basic solutions, not the other way around.

D

At the equivalence point $[OH^-] > [H^+]$ ; therefore, phenolphthalein is an appropriate indicator as its colour change occurs at a basic pH.

Correct Answer

The titration of a weak acid (acetic acid) with a strong base (sodium hydroxide) produces a basic salt, resulting in $[OH^-] > [H^+]$ at the equivalence point. Phenolphthalein is the ideal indicator because its colour change interval (pH 8.2-10.0) aligns perfectly with this basic equivalence point.

Q14

2021

QCAA

Paper 1

1 mark

Q14

1 mark

Analyse the data to determine the relative strengths of acids from strongest to weakest.

Acid	$K_a$ value
Nitrous acid	$K_a = 4.00 \times 10^{-4}$
Ethanoic acid	$K_a = 1.76 \times 10^{-5}$
Hydrofluoric acid	$K_a = 7.20 \times 10^{-4}$
Chloroethanoic acid	$K_a = 1.40 \times 10^{-3}$

A

chloroethanoic, ethanoic, nitrous, hydrofluoric

B

chloroethanoic, hydrofluoric, nitrous, ethanoic

C

ethanoic, nitrous, hydrofluoric, chloroethanoic

D

ethanoic, hydrofluoric, nitrous, chloroethanoic

Reveal Answer

A

chloroethanoic, ethanoic, nitrous, hydrofluoric

This order is incorrect because ethanoic acid has the smallest $K_a$ value ( $1.76 \times 10^{-5}$ ), making it the weakest acid, so it should be listed last rather than second.

B

chloroethanoic, hydrofluoric, nitrous, ethanoic

Correct Answer

Acid strength corresponds to the magnitude of the $K_a$ value. The correct order from largest to smallest $K_a$ is chloroethanoic ( $1.40 \times 10^{-3}$ ) > hydrofluoric ( $7.20 \times 10^{-4}$ ) > nitrous ( $4.00 \times 10^{-4}$ ) > ethanoic ( $1.76 \times 10^{-5}$ ).

C

ethanoic, nitrous, hydrofluoric, chloroethanoic

This option ranks the acids from weakest to strongest (increasing $K_a$ ), but the question asks for the order from strongest to weakest.

D

ethanoic, hydrofluoric, nitrous, chloroethanoic

This option incorrectly lists ethanoic acid as the strongest; however, it has the lowest $K_a$ value ( $1.76 \times 10^{-5}$ ), which indicates it is actually the weakest acid in the group.

Q19

2025

NESA

1 mark

Q19

1 mark

0.1 mol of solid sodium acetate is dissolved in 500 mL of 0.1 mol L $^{-1}$ HCl in a beaker. This solution has a pH of 4.8.

500 mL of distilled water is then added to the beaker.

What is the pH of the final solution?

A

2.4

B

4.5

C

4.8

D

5.1

Reveal Answer

A

2.4

This incorrectly assumes the pH halves upon dilution. The solution is a buffer, so its pH depends on the ratio of acid to conjugate base, not just the total volume.

B

4.5

Diluting a buffer does not decrease its pH, as the ratio of the weak acid to its conjugate base remains unchanged.

C

4.8

Correct Answer

The initial mixture forms a buffer solution. Adding water dilutes the weak acid and its conjugate base equally, keeping their ratio constant in the Henderson-Hasselbalch equation, so the pH remains 4.8.

D

5.1

This incorrectly assumes the pH increases upon dilution (like an unbuffered acid). Because this is a buffer solution, it resists changes to its pH when diluted.

Q12

2022

VCAA

1 mark

Q12

1 mark

Enzymes are commonly not effective in acidic conditions because acids

A

change the charges on the enzymes.

B

react with the enzymes to form zwitterions.

C

esterify the enzymes into smaller molecules.

D

react with the carboxyl groups on the enzymes' amino acid residues.

Reveal Answer

A

change the charges on the enzymes.

Correct Answer

Acidic conditions increase the concentration of $H^+$ ions, which protonate amino acid side chains. This alters the enzyme's overall charge distribution, disrupting the ionic bonds that maintain its functional 3D structure.

B

react with the enzymes to form zwitterions.

Zwitterions are molecules with both positive and negative charges that net to zero, typically occurring at an amino acid's isoelectric point. In highly acidic conditions, amino acids become positively charged cations, not zwitterions.

C

esterify the enzymes into smaller molecules.

Acids do not esterify enzymes into smaller molecules. Breaking down an enzyme's protein chain into smaller molecules would involve the hydrolysis of peptide bonds, not esterification.

D

react with the carboxyl groups on the enzymes' amino acid residues.

While acidic conditions do protonate carboxylate groups into neutral carboxyl groups, this is only a partial explanation. The loss of enzyme effectiveness is due to the overall change in charges across all ionizable groups, which disrupts the active site.

Q9

2022

SCSA

1 mark

Q9

1 mark

Consider an acid-base titration between hydrochloric acid solution and ammonia solution. Which of the following actions is least likely to cause an error when calculating the concentration of hydrochloric acid?

A

cleaning the pipette with distilled water before each titration

B

rinsing the sides of the conical flask with distilled water during the titration

C

measuring the ammonia solution in a 20 mL measuring cylinder

D

leaving the funnel in the burette for each titration

Reveal Answer

A

cleaning the pipette with distilled water before each titration

Rinsing the pipette with distilled water dilutes the solution being transferred, which alters the number of moles delivered and causes a calculation error.

B

rinsing the sides of the conical flask with distilled water during the titration

Correct Answer

Adding distilled water to the conical flask does not change the total number of moles of acid or base present, so it will not affect the titration results or the calculated concentration.

C

measuring the ammonia solution in a 20 mL measuring cylinder

A measuring cylinder is not precise enough for analytical titrations; a volumetric pipette must be used to avoid significant volume measurement errors.

D

leaving the funnel in the burette for each titration

Leaving the funnel in the burette can allow additional drops of solution to fall into the burette during the titration, leading to inaccurate volume readings.

Q6

2024

QCAA

Paper 1

1 mark

Q6

1 mark

The equivalence point of an acid–base titration occurs when the

A

pH equals the p $K_a$ .

B

pH stops changing.

C

indicator changes colour.

D

titrant completely neutralises the analyte.

Reveal Answer

A

pH equals the p $K_a$ .

The pH equals the p $K_a$ at the half-equivalence point, where exactly half of the weak acid has been converted to its conjugate base.

B

pH stops changing.

The pH changes most rapidly at the equivalence point rather than stopping; it only levels off significantly after excess titrant is added.

C

indicator changes colour.

The point where the indicator changes colour is called the end point, which is an experimental approximation that should ideally be close to the equivalence point.

D

titrant completely neutralises the analyte.

Correct Answer

The equivalence point is defined as the stoichiometric point where the moles of added titrant exactly equal the moles required to completely react with (neutralise) the analyte.

Q10

2024

QCAA

Paper 1

1 mark

Q10

1 mark

The acid dissociation constant ( $K_a$ ) represents the

A

pH of an acid solution.

B

strength of an acid solution.

C

concentration of an acid solution.

D

conjugate acid–base pairs of an acid solution.

Reveal Answer

A

pH of an acid solution.

The pH is a measure of the hydrogen ion concentration, $-\log[H^+]$ , which depends on both the acid's strength and its initial concentration, whereas $K_a$ is a constant specific to the acid itself.

B

strength of an acid solution.

Correct Answer

The acid dissociation constant ( $K_a$ ) quantifies the extent to which an acid dissociates in water; a larger $K_a$ value indicates a greater degree of ionization and thus a stronger acid.

C

concentration of an acid solution.

Concentration refers to the amount of solute in a given volume (molarity), while $K_a$ is an intrinsic property of the substance that remains constant regardless of concentration at a fixed temperature.

D

conjugate acid–base pairs of an acid solution.

Conjugate acid–base pairs are the chemical species related by the loss or gain of a proton, not the numerical constant ( $K_a$ ) that describes the equilibrium between them.

Q29

2025

SCSA

7 marks

Q29

A laboratory technician prepares a 1.0 L buffer solution containing 0.50 mol acetic acid, $CH_3COOH(aq)$ , and 0.50 mol sodium acetate, $CH_3COONa(aq)$ .

Q29a

2 marks

Write the equilibrium equation for the buffer solution.

Reveal Answer

Correct reactant and products
$CH_3COOH(aq) + H_2O(\ell) \rightleftharpoons CH_3COO^-(aq) + H_3O^+(aq)$
or
$CH_3COO^-(aq) + H_2O(\ell) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)$

Marking Criteria

Descriptor	Marks
Correct reactant and products	1
Equilibrium arrows used	1

Q29b

5 marks

If more than 0.67 L of 0.75 mol L $^{-1}$ sodium hydroxide solution is added to this buffer solution it will cause a significant change in pH of the buffer. Explain why lesser volumes of the base will cause only a minor change in pH. Include an appropriate balanced equation in your explanation.

Reveal Answer

Equation: $CH_3COOH(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + H_2O(\ell)$

Recognition that the [OH-] has increased.

As the [CH3COOH] is reduced by reaction with added OH-, the rate of the forward reaction will decrease/frequency of collisions between reactants will decrease (compared to the reverse reaction).
OR
The [CH3COO-] increases through the reaction of CH3COOH with OH-, the rate of the reverse reaction will increase/frequency of collisions between products will increase (compared to the forward reaction).

Recognition that when equilibrium is re-established, the [H3O+] has decreased/[OH-] increased (or most added OH- consumed).

Recognition that the pH is slightly higher.

Marking Criteria

Descriptor	Marks
Accept an appropriate equation which must show reaction with OH-	1
Recognition that the [OH-] has increased	1
Explanation of rate change: [CH3COOH] reduced (forward rate decreases) OR [CH3COO-] increases (reverse rate increases)	1
Recognition that when equilibrium is re-established, the [H3O+] has decreased/[OH-] increased OR most added OH- consumed	1
Recognition that the pH is slightly higher	1

Q15

2023

QCAA

Paper 1

1 mark

Q15

1 mark

Predict how a buffer solution, consisting of carbonic acid ( $\text{H}_2\text{CO}_3$ ) and hydrogen carbonate ions ( $\text{HCO}_3^-$ ), would react to resist a change in pH when a small amount of hydrochloric acid is added.

$\text{H}_2\text{CO}_3\text{(aq)} \rightleftharpoons \text{HCO}_3^-\text{(aq)} + \text{H}^+\text{(aq)} \rightleftharpoons \text{CO}_3^{2-}\text{(aq)} + 2\text{H}^+\text{(aq)}$

A

Equilibrium shifts to the right and the $[\text{H}^+]\text{(aq)}$ increases.

B

Equilibrium shifts to the left and the $[\text{CO}_3^{2-}]\text{(aq)}$ increases.

C

Equilibrium shifts to the left and the $[\text{H}_2\text{CO}_3]\text{(aq)}$ increases.

D

Equilibrium shifts to the right and the $[\text{HCO}_3^-]\text{(aq)}$ increases.

Reveal Answer

A

Equilibrium shifts to the right and the $[\text{H}^+]\text{(aq)}$ increases.

According to Le Chatelier's Principle, adding $\text{H}^+$ ions (from the strong acid $\text{HCl}$ ) causes the equilibrium to shift to the left to consume the excess protons, not to the right.

B

Equilibrium shifts to the left and the $[\text{CO}_3^{2-}]\text{(aq)}$ increases.

Adding $\text{H}^+$ shifts the equilibrium to the left toward the protonated forms; this would consume $\text{CO}_3^{2-}$ ions (if present from the second dissociation), causing their concentration to decrease rather than increase.

C

Equilibrium shifts to the left and the $[\text{H}_2\text{CO}_3]\text{(aq)}$ increases.

Correct Answer

The added $\text{H}^+$ ions react with the conjugate base $\text{HCO}_3^-$ , shifting the equilibrium to the left to minimize the disturbance, which results in an increased concentration of carbonic acid ( $\text{H}_2\text{CO}_3$ ).

D

Equilibrium shifts to the right and the $[\text{HCO}_3^-]\text{(aq)}$ increases.

The addition of acid causes the equilibrium to shift to the left to consume the added $\text{H}^+$ , which uses up $\text{HCO}_3^-$ ions, causing their concentration to decrease.

Q33

2025

NESA

7 marks

Q33

7 marks

Chalk is predominantly calcium carbonate. Different brands of chalk vary in their calcium carbonate composition.

The table shows the composition of three different brands of chalk.

	Brand X	Brand Y	Brand Z
$\text{CaCO}_3$ (%)	85.5	83.9	82.4

The following procedure was used to determine the calcium carbonate composition of a chalk sample.

A sample of chalk was crushed in a mortar and pestle.
A 3.00 g sample of the crushed chalk was placed in a conical flask.
100.0 mL of $0.550 \text{ mol L}^{-1} \text{ HCl}(aq)$ was added to the sample and left to react completely, resulting in a clear solution.
Four 20 mL aliquots of this mixture were then titrated with $0.10 \text{ mol L}^{-1} \text{ KOH}$ .

The results of the titrations are recorded.

Burette volume (mL)	Trial 1	Trial 2	Trial 3	Trial 4
Final	7.80	14.90	22.10	29.25
Initial	0.00	7.80	14.90	22.10
Total used	7.80	7.10	7.20	7.15

Determine the brand of the chalk sample. Include a relevant chemical equation in your answer.

Reveal Answer

Exclude the outlier.

Average volume KOH used excluding Trial 1 = $\frac{7.10 + 7.20 + 7.15}{3} = 0.00715$ L

HCl( $aq$ ) + KOH( $aq$ ) $\rightarrow$ KCl( $aq$ ) + H $_2$ O( $l$ )

moles KOH = $0.10 \times 0.00715 = 0.000715$ mol KOH

Ratio HCl : KOH = 1 : 1
$\therefore$ 0.000715 mol HCl for each sample
$\therefore$ $0.000715 \times 5 = 0.003575$ mol total in sampled solution

Initial $n$ HCl = $0.550 \times 0.1000 = 0.0550$ mol HCl

$n$ HCl that reacted with CaCO $_3$ = $0.0550 - 0.003575$ = 0.051425 mol HCl

2HCl( $aq$ ) + CaCO $_3$ ( $s$ ) $\rightarrow$ CaCl $_2$ ( $aq$ ) + H $_2$ O( $l$ ) + CO $_2$ ( $g$ )

Ratio HCl : CaCO $_3$ = 2 : 1
$\therefore$ $\frac{0.051425}{2} = 0.0257125$ mol CaCO $_3$

Mass CaCO $_3$ = $0.0257125 \times MM \text{ CaCO}_3 \, (100.09)$ = 2.5735641 g

% CaCO $_3$ = $\frac{2.5735641}{3.00} \times 100 = 85.7854\% \approx 85.8\%$

Chalk sample has to be Brand X.

Marking Criteria

Descriptor	Marks
Correctly identifies the brand AND Provides all correct calculations AND Provides a balanced chemical equation including states	7
Provides substantially correct calculations AND Provides a balanced chemical equation	6
Provides the main steps in the calculation AND Provides a balanced chemical equation	5
Provides the main steps in the calculation AND Provides a balanced chemical equation	4
Provides some steps in the calculation	3
Provides some steps in the calculation	2
Provides some relevant information	1
None of the above	0

Q38

2022

SCSA

15 marks

Q38

A tablet used to reduce the effects of indigestion contained a mixture of sodium hydrogencarbonate and sodium carbonate.

Five tablets were crushed and dissolved in distilled water, which was added to a volumetric flask and the volume made up to 250.0 mL.

Aliquots (25.00 mL) of the solution were transferred to conical flasks and titrated against a 0.0955 mol L $^{-1}$ solution of hydrochloric acid.

The masses of sodium hydrogencarbonate and sodium carbonate in each tablet were found to be:

sodium hydrogencarbonate – 106.5 mg
sodium carbonate – 187.5 mg.

Q38a

8 marks

Calculate the average titre that would have been obtained to produce these results. Use the following molar masses in your calculation:

$M(NaHCO_3) = 84.008 \text{ g mol}^{-1}$
$M(Na_2CO_3) = 105.99 \text{ g mol}^{-1}$ .

Reveal Answer

$n(\text{NaHCO}_3) = \frac{0.1065}{84.008} \times 5 = 6.339 \times 10^{-3} \text{ mol in 250 mL}$

$n(\text{Na}_2\text{CO}_3) = \frac{0.1875}{105.99} \times 5 = 8.845 \times 10^{-3} \text{ mol in 250 mL}$

$n(\text{NaHCO}_3) = \frac{25}{250} \times 6.339 \times 10^{-3} = 6.339 \times 10^{-4} \text{ mol in one aliquot}$

$n(\text{Na}_2\text{CO}_3) = \frac{25}{250} \times 8.845 \times 10^{-3} = 8.845 \times 10^{-4} \text{ mol in one aliquot}$

$n(\text{H}^+) = n(\text{NaHCO}_3)$ in titration $= 6.339 \times 10^{-4} \text{ mol}$

$n(\text{H}^+) = 2 n(\text{Na}_2\text{CO}_3)$ in titration $= 2(8.845 \times 10^{-4}) = 1.769 \times 10^{-3} \text{ mol}$

$n(\text{H}^+ \text{ total}) = 6.339 \times 10^{-4} + 1.769 \times 10^{-3} = 2.403 \times 10^{-3} \text{ mol}$

$V(\text{HCl required}) = \frac{2.403 \times 10^{-3}}{0.0955} = 0.02516 \text{ L (25.16 mL)}$

Marking Criteria

Descriptor	Marks
Calculates $n(\text{NaHCO}_3) = \frac{0.1065}{84.008} \times 5 = 6.339 \times 10^{-3} \text{ mol in 250 mL}$	1
Calculates $n(\text{Na}_2\text{CO}_3) = \frac{0.1875}{105.99} \times 5 = 8.845 \times 10^{-3} \text{ mol in 250 mL}$	1
Calculates $n(\text{NaHCO}_3) = \frac{25}{250} \times 6.339 \times 10^{-3} = 6.339 \times 10^{-4} \text{ mol in one aliquot}$	1
Calculates $n(\text{Na}_2\text{CO}_3) = \frac{25}{250} \times 8.845 \times 10^{-3} = 8.845 \times 10^{-4} \text{ mol in one aliquot}$	1
Recognises that $n(\text{H}^+) = n(\text{NaHCO}_3)$ in titration $= 6.339 \times 10^{-4} \text{ mol}$	1
Recognises that $n(\text{H}^+) = 2 n(\text{Na}_2\text{CO}_3)$ in titration $= 2(8.845 \times 10^{-4}) = 1.769 \times 10^{-3} \text{ mol}$	1
Calculates $n(\text{H}^+ \text{ total}) = 6.339 \times 10^{-4} + 1.769 \times 10^{-3} = 2.403 \times 10^{-3} \text{ mol}$	1
Calculates $V(\text{HCl required}) = \frac{2.403 \times 10^{-3}}{0.0955} = 0.02516 \text{ L (25.16 mL)}$	1

Q38b

3 marks

Hydrochloric acid must be standardised against a primary standard before it can be used in titrations such as the one described in part (a). List three properties of substances suitable for use as primary standards.

Reveal Answer

Answers could include any 3 of:

high molar mass
not hygroscopic
not deliquescent
available with a known purity
does not react with substances in the atmosphere (e.g. CO₂)
(highly) soluble
predictive reactivity.

Marking Criteria

Descriptor	Marks
Lists three properties of substances used as primary standards	3
Lists two properties of substances used as primary standards	2
States one property of substances used as primary standards	1
None of the above	0

Q38c

4 marks

Methyl orange, which changes colour between a pH of 3.1 and a pH of 4.4, was chosen as the indicator for this reaction. Justify, with the aid of an equation, the selection of this indicator for the titration.

Reveal Answer

$\text{CO}_2$ is produced in the reaction with acid, and $\text{CO}_2$ reacts with water to produce $\text{H}_3\text{O}^+$ . Therefore, at the equivalence point there will be a greater concentration of $\text{H}_3\text{O}^+$ than $\text{OH}^-$ , so the solution will have a pH of less than 7.

A relevant equation is (any reasonable answer accepted, such as):

$\text{CO}_2\text{(aq)} + 2 \text{H}_2\text{O(l)} \rightleftharpoons \text{HCO}_3^- + \text{H}_3\text{O}^+\text{(aq)}$
$\text{H}_2\text{CO}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HCO}_3^- + \text{H}_3\text{O}^+\text{(aq)}$ .

Marking Criteria

Descriptor	Marks
Recognises that $\text{CO}_2$ is produced in the reaction with acid	1
Recognises that $\text{CO}_2$ reacts with water to produce $\text{H}_3\text{O}^+$	1
States that at the equivalence point there will be a greater concentration of $\text{H}_3\text{O}^+$ than $\text{OH}^-$ , so the solution will have a pH of less than 7	1
Provides an appropriate equation (accept any reasonable, correct equation) e.g. $\text{CO}_2\text{(aq)} + 2 \text{H}_2\text{O(l)} \rightleftharpoons \text{HCO}_3^- + \text{H}_3\text{O}^+\text{(aq)}$ or $\text{H}_2\text{CO}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HCO}_3^- + \text{H}_3\text{O}^+\text{(aq)}$	1

Q39

2024

NESA

4 marks

Q39

4 marks

Water and octan-1-ol do not mix. When an aqueous solution of bromoacetic acid ( $\text{BrCH}_2\text{COOH}$ ) is shaken with octan-1-ol, an equilibrium system is established between bromoacetic acid dissolved in the octan-1-ol and in the water.

\text{BrCH}_2\text{COOH}(aq) \rightleftharpoons \text{BrCH}_2\text{COOH}(octan\text{-}1\text{-}ol)

The equilibrium constant expression for this system is

K_{eq} = \frac{[\text{BrCH}_2\text{COOH}(octan\text{-}1\text{-}ol)]}{[\text{BrCH}_2\text{COOH}(aq)]}.

An aqueous solution of bromoacetic acid with an initial concentration of $0.1000 \text{ mol L}^{-1}$ is shaken with an equal volume of octan-1-ol. Bromoacetic acid does not dissociate in octan-1-ol but does dissociate in water, with $K_a = 1.29 \times 10^{-3}$ . When the system has reached equilibrium, the $[\text{H}^+]$ is $9.18 \times 10^{-3} \text{ mol L}^{-1}$ .

Calculate the equilibrium concentration of aqueous bromoacetic acid and hence, or otherwise, calculate the $K_{eq}$ for the octan-1-ol and water system.

Reveal Answer

$K_a = \frac{[\text{H}^+][\text{BrCH}_2\text{COO}^-]}{[\text{BrCH}_2\text{COOH}]}$

\begin{align*} [\text{BrCH}_2\text{COOH}](aq) &= \frac{[\text{H}^+][\text{BrCH}_2\text{COO}^-]}{K_a}\\ &= \frac{[9.18 \times 10^{-3}][9.18 \times 10^{-3}]}{1.29 \times 10^{-3}}\\ &= 0.0654 \text{ mol L}^{-1} \end{align*}

$[\text{BrCH}_2\text{COOH}](\text{octan-1-ol}) = 0.1000 - 0.0654 - 9.18 \times 10^{-3} = 0.0254 \text{ mol L}^{-1}$

K_{eq} = \frac{[\text{BrCH}_2\text{COOH}(\text{octan-1-ol})]}{[\text{BrCH}_2\text{COOH}(aq)]} = \frac{0.0254 \text{ mol L}^{-1}}{0.0654 \text{ mol L}^{-1}} = 0.390

Marking Criteria

Descriptor	Marks
Calculates the $K_{eq}$	4
Provides most steps of the calculation	3
Provides some steps of the calculation	2
Provides some relevant information	1
None of the above	0

Q5

2022

SCSA

1 mark

Q5

1 mark

Which of the following pairs, in equimolar amounts, would result in an acidic buffer solution?
(i) $\mathrm{CH_3COOH/CH_3COO^-}$
(ii) $\mathrm{H_2CO_3/HCO_3^-}$
(iii) $\mathrm{NH_3/NH_4^+}$
(iv) $\mathrm{H_2SO_4/HSO_4^-}$
(v) $\mathrm{CH_3NH_2/CH_3NH_3^+}$

A

i, ii and iv

B

iii and v

C

ii, iii and v

D

i and ii

Reveal Answer

A

i, ii and iv

While pairs (i) and (ii) form acidic buffers, pair (iv) contains a strong acid ( $\mathrm{H_2SO_4}$ ) which completely dissociates and therefore cannot form a buffer solution.

B

iii and v

Pairs (iii) and (v) consist of weak bases and their conjugate acids. In equimolar amounts, they would form basic buffer solutions with a pH greater than 7.

C

ii, iii and v

Although pair (ii) forms an acidic buffer, pairs (iii) and (v) are weak bases with their conjugate acids, which would result in basic buffer solutions.

D

i and ii

Correct Answer

Both (i) and (ii) consist of a weak acid and its conjugate base with a $pK_a < 7$ . According to the Henderson-Hasselbalch equation, equimolar amounts will result in $pH = pK_a$ , creating an acidic buffer solution.

Q15

2021

SCSA

1 mark

Q15

1 mark

Which of the following processes does not contribute to the building of weaker seashells through ocean acidification?

A

$HCO_3^-(aq) + H_2O(\ell) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)$

B

$2 H^+(aq) + CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_2(aq) + H_2O(\ell)$

C

$CO_2(g) + H_2O(\ell) \rightleftharpoons H_2CO_3(aq)$

D

$H_2CO_3(aq) + H_2O(\ell) \rightleftharpoons HCO_3^-(aq) + H_3O^+(aq)$

Reveal Answer

A

$HCO_3^-(aq) + H_2O(\ell) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)$

This equilibrium determines the concentration of carbonate ions ( $CO_3^{2-}$ ). Increased ocean acidity shifts this reaction to the left, reducing the carbonate available for marine organisms to build strong shells.

B

$2 H^+(aq) + CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_2(aq) + H_2O(\ell)$

Correct Answer

This reaction represents the dissolution of already-formed calcium carbonate ( $CaCO_3$ ) shells by increased acidity. While it destroys existing shells, it is not a process that inhibits the initial building of new shells.

C

$CO_2(g) + H_2O(\ell) \rightleftharpoons H_2CO_3(aq)$

This reaction shows carbon dioxide dissolving in water to form carbonic acid, which is the primary driver of ocean acidification that ultimately impairs shell building.

D

$H_2CO_3(aq) + H_2O(\ell) \rightleftharpoons HCO_3^-(aq) + H_3O^+(aq)$

This reaction shows the dissociation of carbonic acid into bicarbonate and hydronium ions, which increases ocean acidity and leads to a reduction in the carbonate ions needed for shell building.

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