SCSA WACE 2021-style practice: Calculate the pH of a solution made by dissolving 0.80\ g of solid NaOH in water and making the volume up to 250\ mL at 25^C. (M(NaOH) = 40.0\ g mol^-1.)

A 5 mark calculation rewards moles, concentration, the K_w step and a correctly rounded pH. **Moles of NaOH.** n = mM = 0.8040.0 = 0.020\ mol. **Concentration.** c = nV = 0.0200.250 = 0.080\ mol L^-1. Sodium hydroxide is a strong base and fully dissociates, so [OH^-] = 0.080\ mol L^-1. **Convert to [H^+] via K_w.** [H^+] = K_w[OH^-] = 1.010^-140.080 = 1.2510^-13\ mol L^-1. **pH.** pH = -_10(1.2510^-13) = 12.90. Markers reward the mole and concentration working, recognising full dissociation, the K_w conversion and a pH above 7 quoted to two decimal places.

§-Syllabus dot point

WAChemistrySyllabus dot point

What makes a substance an acid or a base, and how do we measure and calculate the pH of a solution?

Apply the Bronsted-Lowry theory, identify conjugate acid-base pairs, distinguish strong from weak acids and bases, and calculate pH using Kw

A focused answer to the WACE Year 12 Chemistry dot point on Bronsted-Lowry acids and bases, conjugate pairs, strong versus weak, Kw and pH calculations, with worked examples and common mistakes.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

The Bronsted-Lowry theory defines acid-base behaviour in terms of proton transfer. An acid is a proton ( $\text{H}^+$ ) donor; a base is a proton acceptor. This is broader than the older idea that acids produce $\text{H}^+$ and bases produce $\text{OH}^-$ , because it works for any proton transfer, including reactions in which ammonia acts as a base without any hydroxide being involved.

Conjugate acid-base pairs

When an acid donates a proton it becomes its conjugate base. When a base accepts a proton it becomes its conjugate acid. The two species in a pair differ by exactly one $\text{H}^+$ . Consider:

\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+

Here ethanoic acid donates a proton to water. The conjugate pairs are $\text{CH}_3\text{COOH}/\text{CH}_3\text{COO}^-$ and $\text{H}_2\text{O}/\text{H}_3\text{O}^+$ . Water acts as a base in this reaction. A species like water that can act as either an acid or a base, depending on what it reacts with, is amphiprotic (for example $\text{HCO}_3^-$ and $\text{H}_2\text{O}$ ).

Strong versus weak

The strength of an acid or base describes how completely it ionises in water, not how concentrated it is.

Strong acids (such as $\text{HCl}$ , $\text{HNO}_3$ , $\text{H}_2\text{SO}_4$ ) ionise essentially completely. Their reaction with water is written with a single arrow: $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$ .

Weak acids (such as $\text{CH}_3\text{COOH}$ ) only partially ionise, so they sit at equilibrium with the un-ionised molecule and are written with an equilibrium arrow. At any moment most molecules are not ionised.

The same distinction applies to bases. $\text{NaOH}$ is a strong base (fully dissociated), while aqueous ammonia is a weak base because the equilibrium $\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$ lies well to the left. Note the difference between strength (degree of ionisation) and concentration (mol per litre). A concentrated weak acid can have a higher pH than a dilute strong acid.

Self-ionisation of water and Kw

Water ionises very slightly: $2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-$ . The ionic product is

K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\ \text{mol}^2\,\text{L}^{-2}\ (\text{at}\ 25^{\circ}\text{C})

In pure water $[\text{H}^+] = [\text{OH}^-] = 1.0 \times 10^{-7}\ \text{mol L}^{-1}$ , giving pH 7 (neutral). Because $K_w$ comes from an endothermic ionisation, it increases with temperature, so neutral pH is below 7 at higher temperatures, though the solution is still neutral.

pH calculations

The pH scale measures hydrogen ion concentration:

\text{pH} = -\log_{10}[\text{H}^+] \qquad [\text{H}^+] = 10^{-\text{pH}}

For a basic solution, find $[\text{OH}^-]$ , use $K_w$ to get $[\text{H}^+]$ , then take the pH.

Calculating the pH of a strong base

Question

Find the pH of $0.0500\ \text{mol L}^{-1}$ sodium hydroxide at $25^{\circ}\text{C}$ .

Solution

$\text{NaOH}$ is a strong base, so it is fully dissociated: $[\text{OH}^-] = 0.0500\ \text{mol L}^{-1}$ .

Use $K_w$ to find $[\text{H}^+]$ :

[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.0500} = 2.0 \times 10^{-13}\ \text{mol L}^{-1}

Then $\text{pH} = -\log_{10}(2.0 \times 10^{-13}) = 12.7$ .

The answer is above 7, as expected for a base, to two significant figures (two decimal places in pH reflect the two significant figures in the concentration).

When you answer pH questions in the WACE examination, always check whether the acid or base is strong (full ionisation, so the concentration equals the ion concentration) or weak (an equilibrium that needs $K_a$ , beyond a simple calculation), and quote pH to the correct number of decimal places.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20215 marksCalculate the pH of a solution made by dissolving

0.80\ \text{g}

of solid

\text{NaOH}

in water and making the volume up to

250\ \text{mL}

25^{\circ}\text{C}

. (

M(\text{NaOH}) = 40.0\ \text{g mol}^{-1}

Show worked answer →

A 5 mark calculation rewards moles, concentration, the $K_w$ step and a correctly rounded pH.

Moles of $\text{NaOH}$: $n = \dfrac{m}{M} = \dfrac{0.80}{40.0} = 0.020\ \text{mol}$ .
Concentration: $c = \dfrac{n}{V} = \dfrac{0.020}{0.250} = 0.080\ \text{mol L}^{-1}$ . Sodium hydroxide is a strong base and fully dissociates, so $[\text{OH}^-] = 0.080\ \text{mol L}^{-1}$ .
Convert to $[\text{H}^+]$ via $K_w$: $[\text{H}^+] = \dfrac{K_w}{[\text{OH}^-]} = \dfrac{1.0\times10^{-14}}{0.080} = 1.25\times10^{-13}\ \text{mol L}^{-1}$ .
pH: $\text{pH} = -\log_{10}(1.25\times10^{-13}) = 12.90$ .

Markers reward the mole and concentration working, recognising full dissociation, the $K_w$ conversion and a pH above 7 quoted to two decimal places.

WACE 20224 marksA student claims that because ethanoic acid and hydrochloric acid are both

0.10\ \text{mol L}^{-1}

, they must have the same pH. Explain why this claim is incorrect, referring to the Bronsted-Lowry theory and the difference between acid strength and concentration.

Show worked answer →

A 4 mark explain answer needs the strong-versus-weak ionisation distinction tied to proton donation.

Both are Bronsted-Lowry acids (proton donors) but they ionise to different extents. $\text{HCl}$ is a strong acid and ionises completely: $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$ , so $[\text{H}^+] = 0.10\ \text{mol L}^{-1}$ and $\text{pH} = 1.0$ .

Ethanoic acid is a weak acid and only partially ionises: $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$ , so $[\text{H}^+] \ll 0.10\ \text{mol L}^{-1}$ and the pH is around $2.9$ , much higher than for $\text{HCl}$ .

Conclusion. Strength (extent of ionisation) and concentration (mol per litre) are different properties, so equal concentrations of a strong and a weak acid give different $[\text{H}^+]$ and therefore different pH.

Markers reward the single-arrow versus equilibrium-arrow distinction, the resulting difference in $[\text{H}^+]$ , and a clear statement that strength is not concentration.