How many SCSA Chemistry questions cover Acids and bases?

AusGrader has 164 SCSA Chemistry questions on Acids and bases, all with instant AI grading and detailed marking feedback.

SCSA (WACE) Chemistry — Acids and bases Questions & Answers | AusGrader

Q12

2024

SCSA

1 mark

Q12

1 mark

One limitation of the Brønsted-Lowry Theory of acids and bases is that it

A

does not explain reactions between acidic and basic oxides, such as $SO_3(g) + MgO(s) \rightarrow MgSO_4(s)$ , as they do not involve the transfer of protons.

B

does not explain the production of a neutral salt solution resulting from the reaction between a strong acid and strong base.

C

links acids and bases into conjugate acid-base pairs rather than accounting for the transfer of protons.

D

cannot explain the acidity and basicity of acidic and basic salts.

Q15

2020

SCSA

1 mark

Q15

1 mark

The reaction of aniline ( $\text{C}_6\text{H}_5\text{NH}_2$ ) with water is an equilibrium process:

$\text{C}_6\text{H}_5\text{NH}_2(\ell) + \text{H}_2\text{O}(\ell) \rightleftharpoons \text{C}_6\text{H}_5\text{NH}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$

A conjugate acid-base pair in this process is

A

$\text{C}_6\text{H}_5\text{NH}^-$ and $\text{H}_2\text{O}$

B

$\text{C}_6\text{H}_5\text{NH}_2$ and $\text{C}_6\text{H}_5\text{NH}^-$

C

$\text{C}_6\text{H}_5\text{NH}^-$ and $\text{H}_3\text{O}^+$

D

$\text{H}_3\text{O}^+$ and $\text{C}_6\text{H}_5\text{NH}_2$

Q34

2023

SCSA

9 marks

Q34

Sorbic acid is a monoprotic weak acid that occurs widely in nature and is used as a food preservative due to its antimicrobial properties. The ionisation of sorbic acid in water to the sorbate ion and hydronium ion is shown in the equation below:

\text{CH}_3\text{(CH)}_4\text{COOH(aq)} + \text{H}_2\text{O}(\ell) \rightleftharpoons \text{CH}_3\text{(CH)}_4\text{COO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}

Q34a

2 marks

Write the equilibrium constant K expression for the ionisation of sorbic acid in water.

\text{K} = \frac{[\text{CH}_3(\text{CH})_4\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3(\text{CH})_4\text{COOH}]}

Marking Criteria

Descriptor	Marks
K = [CH3(CH)4COO-][H3O+] / [CH3(CH)4COOH]	2
Partially correct equilibrium constant expression	1
None of the above	0

Q34b

4 marks

Under certain conditions, a $0.250\text{ mol L}^{-1}$ aqueous solution of sorbic acid has a pH of 2.23. Calculate the concentration of $\text{H}_3\text{O}^+$ to determine the percentage yield of the sorbate ion at equilibrium in $1.00\text{ L}$ of the solution.

$[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-2.23} = 5.89 \times 10^{-3} \text{ mol L}^{-1}$

For 1.00 L solution, $n(\text{H}_3\text{O}^+) = n(\text{CH}_3(\text{CH})_4\text{COO}^-) = 5.89 \times 10^{-3} \text{ mol}$
For 1.00 L solution, $n(\text{CH}_3(\text{CH})_4\text{COOH}) = 0.250 \text{ mol}$

\begin{align*} \text{\% yield sorbate ion} &= \frac{n(\mathrm{CH_3CH(OH)COO^-})}{n(\mathrm{CH_3CH(OH)COOH})}\times 100\\ &= \frac{5.89\times 10^{-3}}{0.250}\times 100\\ &= 2.36\% \end{align*}

Marking Criteria

Descriptor	Marks
$[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-2.23} = 5.89 \times 10^{-3} \text{ mol L}^{-1}$	1
For 1.00 L solution, $n(\text{H}_3\text{O}^+) = n(\text{CH}_3(\text{CH})_4\text{COO}^-) = 5.89 \times 10^{-3} \text{ mol}$	1
For 1.00 L solution, $n(\text{CH}_3(\text{CH})_4\text{COOH}) = 0.250 \text{ mol}$	1
Percentage yield is 2.36%	1

Q34c

3 marks

Explain the classification of sorbic acid as a weak acid with reference to both your answer to part (b) above and its acidity constant value $\text{K}_\text{a} = 1.73 \times 10^{-5}$ (20 °C).

Weak acids undergo partial or incomplete ionisation in water. The answer to part (b) is numerically small, indicating that only a small percentage of sorbic acid in solution is ionised. The value of $K_a$ is less than one, which indicates a greater proportion of reactants compared to products.

Marking Criteria

Descriptor	Marks
Recognises that weak acids undergo partial/incomplete ionisation in water	1
Explains that the answer to part (b) is numerically small, indicating that only a small percentage of sorbic acid in solution is ionised	1
Recognises that the value of $K_a$ is less than one, which indicates a greater proportion of reactants compared to products	1

Q28

2022

SCSA

7 marks

Q28

7 marks

Explain why potassium hydrogensulfite, $\mathrm{KHSO_3}$ , produces an acidic solution when dissolved in water, while potassium hydrogencarbonate, $\mathrm{KHCO_3}$ , produces a basic solution when dissolved in water. Use equations to illustrate your explanation.

The $K^+$ ions in solution are neutral and do not react with water, whereas the $HSO_3^-$ and $HCO_3^-$ ions undergo hydrolysis reactions. For the hydrolysis reactions for $HSO_3^-$ , the reaction that produces $H_3O^+$ occurs to a greater extent than the reaction that produces $OH^-$ , therefore the solution will be acidic. For the hydrolysis reactions for $HCO_3^-$ , the reaction that produces $OH^-$ occurs to a greater extent than the reaction that produces $H_3O^+$ , therefore the solution is basic. A basic solution has a greater concentration of $OH^-$ ions than $H_3O^+$ ions, and an acidic solution has a greater concentration of $H_3O^+$ ions than $OH^-$ ions.

The relevant equations for $HSO_3^-$ are (any 1 of the following):

$HSO_3^-(aq) + H_2O(l) \rightleftharpoons SO_3^{2-}(aq) + H_3O^+(aq)$
$HSO_3^-(aq) + H_2O(l) \rightleftharpoons H_2SO_3(aq) + OH^-(aq)$ .

The relevant equations for $HCO_3^-$ are (any 1 of the following):

$HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_2CO_3(aq) + OH^-(aq)$
$HCO_3^-(aq) + H_2O(l) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)$ .

Marking Criteria

Descriptor	Marks
Recognises that the $K^+$ ions in solution are neutral/do not react with water.	1
Recognises that the $HSO_3^-$ and $HCO_3^-$ ions undergo hydrolysis reactions.	1
Recognises that for the hydrolysis reactions for $HSO_3^-$ , the reaction that produces $H_3O^+$ occurs to a greater extent than the reaction that produces $OH^-$ , therefore the solution will be acidic.	1
Recognises that for the hydrolysis reactions for $HCO_3^-$ , the reaction that produces $OH^-$ occurs to a greater extent than the reaction that produces $H_3O^+$ , therefore the solution is basic.	1
Recognises that a basic solution has a greater concentration of $OH^-$ ions than $H_3O^+$ ions, or an acidic solution has a greater concentration of $H_3O^+$ ions than $OH^-$ ions.	1
Provides at least one appropriate equation for $HSO_3^-$ , such as $HSO_3^-(aq) + H_2O(l) \rightleftharpoons SO_3^{2-}(aq) + H_3O^+(aq)$ or $HSO_3^-(aq) + H_2O(l) \rightleftharpoons H_2SO_3(aq) + OH^-(aq)$ .	1
Provides at least one appropriate equation for $HCO_3^-$ , such as $HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_2CO_3(aq) + OH^-(aq)$ or $HCO_3^-(aq) + H_2O(l) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)$ .	1

Q39

2025

SCSA

14 marks

Q39

14 marks

Freon-11 is a colourless chlorofluorocarbon that boils at 23.77 °C. Prior to the knowledge of the ozone-depleting potential of chlorofluorocarbons (CFCs) and other possible harmful effects on the environment, it was used as a refrigerant.

The following data was used to determine that Freon-11 is trichlorofluoromethane, with a molecular formula of $CCl_3F$ .

A Freon-11 sample of 4.121 g was combusted in excess oxygen. All the carbon in the compound was converted to carbon dioxide and in a separate process, all its chlorine was converted into hydrochloric acid. The carbon dioxide produced had a mass of 1.320 g and the hydrochloric acid formed, required 85.70 mL of 1.050 mol L $^{-1}$ of ammonia solution for complete neutralisation.
Another sample of the Freon-11 with a mass of 3.721 g occupied a volume of 0.6068 L at a pressure of 120.00 kPa and temperature of 50.6 °C.

Using the same data, use calculations and reasoning to demonstrate that this is the correct molecular formula.

Carbon
$n(C) = n(CO_2) = 1.320/44.01 = 0.02999$ mol
$m(C) = 0.02299 \times 12.01 = 0.3602$ g

Chlorine
$n(Cl) = n(HCl) = n(NH_3) = 1.050 \times 0.08570 = 0.08999$ mol
$m(Cl) = 0.08999 \times 35.45 = 3.1900$ g

Fluorine
$m(F) = 4.121 - (0.3602 + 3.1900) = 0.5708$ g
$n(F) = 0.5708 / 19.00 = 0.03004$ mol

Mole Ratio
C: 0.02999
Cl: 0.08999
F: 0.03004

Simplify
Divide by 0.02999
C: 1
Cl: 3.00
F: 1.002

Empirical Formula
$CCl_3F$

Molecular Formula
Empirical formula mass (EFM) = 137.36 amu/g mol $^{-1}$
$n = PV/RT = (120.0 \times 0.6068)/(8.314 \times 323.75) = 0.027052476$ mol
Molecular formula mass (MFM) = $m/n = 3.721/0.0276503204 = 137.54$ g mol $^{-1}$
Molecular formula = MFM/EFM $\times$ Empirical formula = $137.54/137.36 \times CCl_3F = 1.0013 \times CCl_3F$

Molecular formula = Empirical formula = $CCl_3F$

Marking Criteria

Descriptor	Marks
n(C) calculation	1
m(C) calculation	1
n(Cl) calculation	1
m(Cl) calculation	1
m(F) calculation	1
n(F) calculation	1
Mole Ratio setup	1
Simplify ratio	1
Empirical Formula	1
Empirical formula mass (EFM)	1
n = PV/RT calculation	1
Molecular formula mass (MFM) calculation	1
Molecular formula calculation	1
Molecular formula = Empirical formula statement	1

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