Module 6: Acid/Base Reactions — Worked Solutions (HSC Chemistry)
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Worked examples for HSC Chemistry Module 6: Acid/Base Reactions. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Chemistry — Module 6: Acid/Base Reactions. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — pH of a strong acid
Question
A $25.0\ \text{mL}$ sample of $0.0500\ \text{mol L}^{-1}$ hydrochloric acid, $\text{HCl}_{(aq)}$, is diluted with water to a final volume of $250.0\ \text{mL}$. Calculate the pH of the diluted solution. Give your answer to two decimal places.
Solution
Dilution first: moles of $\text{HCl}$ don't change, so use $c_1 V_1 = c_2 V_2$.
$$c_2 = \frac{(0.0500)(25.0)}{250.0} = 5.00 \times 10^{-3}\ \text{mol L}^{-1}$$
$\text{HCl}$ is a strong, monoprotic acid — it fully ionises, so $[\text{H}^+] = 5.00 \times 10^{-3}\ \text{mol L}^{-1}$.
$$\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(5.00 \times 10^{-3}) = 2.30$$
Keep two decimal places — only the digits after the point count as significant figures in a pH.
There are two ideas stitched together here: a dilution, then a pH. Let's take them in order.
Diluting doesn't create or destroy any $\text{HCl}$ — it just spreads the same moles through more water — so we use $c_1 V_1 = c_2 V_2$:
$$c_2 = \frac{(0.0500)(25.0)}{250.0} = 5.00 \times 10^{-3}\ \text{mol L}^{-1}$$
Now, why can we jump straight to $[\text{H}^+]$? Because $\text{HCl}_{(aq)}$ is a strong acid, it ionises completely ($\text{HCl}_{(aq)} \rightarrow \text{H}^+{}_{(aq)} + \text{Cl}^-{}_{(aq)}$), and each $\text{HCl}$ gives one $\text{H}^+$. So $[\text{H}^+] = 5.00 \times 10^{-3}\ \text{mol L}^{-1}$.
Finally, pH is just a log scale of that concentration:
$$\text{pH} = -\log_{10}(5.00 \times 10^{-3}) = 2.30$$
We quote two decimal places because, in a logarithm, the decimals carry the significant figures of the original concentration.
Step 1 — dilute ($c_1 V_1 = c_2 V_2$):
$$c_2 = \frac{(0.0500)(25.0)}{250.0} = 5.00 \times 10^{-3}\ \text{mol L}^{-1}$$
Step 2 — strong monoprotic acid, full ionisation:
- $[\text{H}^+] = 5.00 \times 10^{-3}\ \text{mol L}^{-1}$
Step 3 — pH:
$$\text{pH} = -\log_{10}(5.00 \times 10^{-3}) = 2.30$$
Where the marks go
- 1 mark: Correct diluted concentration $5.00 \times 10^{-3}\ \text{mol L}^{-1}$ via $c_1 V_1 = c_2 V_2$
- 1 mark: Recognises HCl as a strong acid so $[\text{H}^+]$ equals the acid concentration
- 1 mark: Correct $\text{pH} = 2.30$ to two decimal places
Key idea
Dilute with $c_1 V_1 = c_2 V_2$; a strong monoprotic acid fully ionises so $[\text{H}^+]$ equals its concentration, then $\text{pH} = -\log_{10}[\text{H}^+]$.
Example 2 — Titration / neutralisation
Question
In a titration, $20.00\ \text{mL}$ of a sodium hydroxide solution, $\text{NaOH}_{(aq)}$, of unknown concentration is exactly neutralised by $18.50\ \text{mL}$ of $0.100\ \text{mol L}^{-1}$ sulfuric acid, $\text{H}_2\text{SO}_4{}_{(aq)}$. Write the balanced neutralisation equation and calculate the concentration of the $\text{NaOH}_{(aq)}$ solution.
Solution
Balanced equation — sulfuric acid is diprotic, so two $\text{NaOH}$ per acid:
$$2\text{NaOH}_{(aq)} + \text{H}_2\text{SO}_4{}_{(aq)} \rightarrow \text{Na}_2\text{SO}_4{}_{(aq)} + 2\text{H}_2\text{O}_{(l)}$$
Moles of acid: $n(\text{H}_2\text{SO}_4) = 0.100 \times 0.01850 = 1.85 \times 10^{-3}\ \text{mol}$.
Mole ratio is $2:1$, so $n(\text{NaOH}) = 2 \times 1.85 \times 10^{-3} = 3.70 \times 10^{-3}\ \text{mol}$.
$$c(\text{NaOH}) = \frac{3.70 \times 10^{-3}}{0.02000} = 0.185\ \text{mol L}^{-1}$$
The $2:1$ ratio is the whole question — miss it and you halve your answer.
Let's build the balanced equation first, because the mole ratio inside it drives everything. Sulfuric acid is diprotic — it can donate two protons — so it needs two hydroxides to neutralise it:
$$2\text{NaOH}_{(aq)} + \text{H}_2\text{SO}_4{}_{(aq)} \rightarrow \text{Na}_2\text{SO}_4{}_{(aq)} + 2\text{H}_2\text{O}_{(l)}$$
Now we work with the side we fully know, the acid. Remembering to convert millilitres to litres:
$$n(\text{H}_2\text{SO}_4) = c \times V = 0.100 \times 0.01850 = 1.85 \times 10^{-3}\ \text{mol}$$
The equation tells us 2 moles of $\text{NaOH}$ react per 1 mole of acid, so we double the acid moles to find the base:
$$n(\text{NaOH}) = 2 \times 1.85 \times 10^{-3} = 3.70 \times 10^{-3}\ \text{mol}$$
Finally, concentration is moles over volume (in litres):
$$c(\text{NaOH}) = \frac{3.70 \times 10^{-3}}{0.02000} = 0.185\ \text{mol L}^{-1}$$
Equation (diprotic acid → $2:1$):
$$2\text{NaOH}_{(aq)} + \text{H}_2\text{SO}_4{}_{(aq)} \rightarrow \text{Na}_2\text{SO}_4{}_{(aq)} + 2\text{H}_2\text{O}_{(l)}$$
- $n(\text{H}_2\text{SO}_4) = 0.100 \times 0.01850 = 1.85 \times 10^{-3}\ \text{mol}$
- ratio $2:1$ → $n(\text{NaOH}) = 3.70 \times 10^{-3}\ \text{mol}$
- $c(\text{NaOH}) = \dfrac{3.70 \times 10^{-3}}{0.02000} = 0.185\ \text{mol L}^{-1}$
Where the marks go
- 1 mark: Correct balanced equation with $2:1$ NaOH to $\text{H}_2\text{SO}_4$ ratio
- 1 mark: Correct moles of $\text{H}_2\text{SO}_4 = 1.85 \times 10^{-3}\ \text{mol}$
- 1 mark: Applies the $2:1$ ratio to get $n(\text{NaOH}) = 3.70 \times 10^{-3}\ \text{mol}$
- 1 mark: Correct concentration $c(\text{NaOH}) = 0.185\ \text{mol L}^{-1}$
Key idea
In a titration, find moles from the known solution, apply the balanced mole ratio (sulfuric acid is diprotic → $2:1$), then divide by volume in litres.
Frequently asked questions
Step-by-step solutions to exam-style questions on Acid/Base Reactions in HSC Chemistry, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Chemistry syllabus for Acid/Base Reactions, using the methods and notation expected in the exam.