How many NESA Chemistry questions cover Using Brønsted–Lowry Theory?

AusGrader has 79 NESA Chemistry questions on Using Brønsted–Lowry Theory, all with instant AI grading and detailed marking feedback.

NESA (HSC) Chemistry — Using Brønsted–Lowry Theory Questions & Answers | AusGrader

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

Q6

2024

NESA

1 mark

Q6

1 mark

What is the hydroxide ion concentration of a solution of potassium hydroxide with a pH of 11?

A

$10^{-11} \text{ mol L}^{-1}$

B

$10^{-3} \text{ mol L}^{-1}$

C

$10^{3} \text{ mol L}^{-1}$

D

$10^{11} \text{ mol L}^{-1}$

Q28

2024

NESA

3 marks

Q28

3 marks

Iodic acid and sulfamic acid are monoprotic acids. A $0.100\text{ mol L}^{-1}$ solution of iodic acid has a pH of 1.151, as does a $0.120\text{ mol L}^{-1}$ solution of sulfamic acid.

Show that neither iodic acid nor sulfamic acid dissociates completely in water, and determine which is the stronger acid.

For both acids, $[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-1.151} = 0.0706 \text{ mol L}^{-1}$ , which is less than the acid concentration. Accordingly, both acids do not dissociate completely in water and are weak.

Since a smaller concentration of iodic acid is required to give a pH of 1.151 compared to sulfamic acid, iodic acid must ionise to a greater extent than sulfamic acid, ie iodic acid must be a stronger acid than sulfamic acid.

Marking Criteria

Descriptor	Marks
Shows that both acids do not dissociate completely in water and that iodic acid is stronger	3
Provides some information relating to of the dissociation of one or both acids	2
Provides some relevant information	1
None of the above	0

Q22

2023

NESA

4 marks

Q22

4 marks

Explain how the following substances would be classified under the Arrhenius and Brønsted–Lowry definitions of acids. Support your answer with relevant equations.

$\text{HCl}(aq)$
$\text{NH}_4\text{Cl}(aq)$

According to Arrhenius, acids are hydrogen-containing compounds that dissociate in water to give $\text{H}^+$ ions. $\text{HCl}(aq)$ would be considered an acid by Arrhenius as it produces $\text{H}^+$ ions in water.

\text{HCl}(aq) \rightarrow \text{H}^+(aq) + \text{Cl}^-(aq)

Arrhenius would not recognise the salt $\text{NH}_4\text{Cl}$ as an acid, as the predominant ions present in aqueous solution are ammonium and chloride.

In Brønsted–Lowry theory, acids are defined as proton donors. $\text{HCl}(aq)$ is a proton donor and therefore a Brnsted–Lowry acid.

Ammonium chloride ( $\text{NH}_4\text{Cl}$ ) is classified as a Brnsted–Lowry acid as the ammonium ion donates a proton to water and forms a hydronium ion.

\text{NH}_4^+(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_3(aq) + \text{H}_3\text{O}^+(aq)

Marking Criteria

Descriptor	Marks
Classifies both substances with respect to Arrhenius and Brønsted–Lowry theories Provides two relevant chemical equations	4
Classifies one substance with respect to Arrhenius and Brønsted–Lowry theories Provides a relevant chemical equation OR Classifies both substances with respect to Arrhenius OR Brønsted–Lowry theory Provides a relevant chemical equation OR Classifies both substances with respect to Arrhenius and Brønsted–Lowry theories	3
Classifies one substance with respect to Arrhenius and Brønsted–Lowry theories OR Classifies both substances with respect to Arrhenius OR Brønsted–Lowry theory	2
• Provides some relevant information	1
None of the above	0

Q29

2024

NESA

4 marks

Q29

4 marks

$150\text{ mL}$ of a $0.20\text{ mol L}^{-1}$ sodium hydroxide solution is added to $100\text{ mL}$ of a $0.10\text{ mol L}^{-1}$ sulfuric acid solution.

Calculate the pH of the resulting solution, assuming that the volume of the resulting solution is $250\text{ mL}$ and that its temperature is $25^\circ\text{C}$ .

$n(\text{NaOH}) = 0.150 \text{ L} \times 0.20 \text{ mol L}^{-1} = 0.030 \text{ mol}$

$n(\text{H}_2\text{SO}_4) = 0.100 \text{ L} \times 0.10 \text{ mol L}^{-1} = 0.010 \text{ mol}$

$\text{H}_2\text{SO}_4(aq) + 2\text{NaOH}(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l)$

$n(\text{NaOH} \text{ required for reaction with } \text{H}_2\text{SO}_4) = 2 \times 0.010 \text{ mol} = 0.020 \text{ mol}$

$n(\text{NaOH} \text{ in excess}) = 0.030 \text{ mol} - 0.020 \text{ mol} = 0.010 \text{ mol}$

$[\text{NaOH}] = \frac{0.010 \text{ mol}}{0.250 \text{ L}} = 0.040 \text{ mol L}^{-1}$

$\text{p(OH)} = -\log_{10}(0.040) = 1.40$

$\text{pH} = 14.00 - 1.40 = 12.60$

Marking Criteria

Descriptor	Marks
Calculates the pH of the resulting solution	4
Provides most relevant steps of the calculation	3
Provides some relevant steps of the calculation	2
Provides some relevant information	1
None of the above	0

NESA Chemistry Using Brønsted–Lowry Theory

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