Worked Solutions
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.
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.