TASC TCE 2021-style practice: A statement found online claims the pH scale extends only from 0 to 14. Hydrochloric acid, HCl(aq) (M = 36.46\ g mol^-1), with a concentration of 320\ g L^-1, is sold at a hardware shop. By determining the pH of this solution, show that the statement is incorrect.

Convert to molar concentration: c = 320 / 36.46 = 8.78\ mol L^-1. (1 mark) HCl is a strong monoprotic acid, so it fully ionises: [H^+] = 8.78\ mol L^-1. (1 mark) $pH = -_10[H^+] = -_10(8.78) = -0.94.$ Because the pH is below 0 (about -0.9), the claim that the scale runs only from 0 to 14 is incorrect: concentrated strong acids can have a negative pH. (1 mark)

TASC TCE 2021-style practice: When 1.08 × 10^-2\ mol of sodium is placed in 500\ mL of water, the reaction 2Na(s) + 2H_2O(l) → 2NaOH(aq) + H_2(g) occurs. Calculate the pH of the resulting solution at 25\ ^C, assuming negligible change in water volume. (K_w = 1.0 × 10^-14.)

From the 2:2 ratio, n(NaOH) = n(Na) = 1.08 × 10^-2\ mol. (1 mark) NaOH is a strong base, so [OH^-] = 1.08 × 10^-2 / 0.500 = 2.16 × 10^-2\ mol L^-1. (1 mark) pOH = -_10(2.16 × 10^-2) = 1.67, so at 25\ ^C, pH = 14.00 - 1.67 = 12.33. (1 mark)

TASC TCE 2022-style practice: For the equilibrium NH_3(g) + H_2O(l) NH_4^+(aq) + OH^-(aq), which reactant is the Bronsted-Lowry acid? Justify your answer using conjugate pairs.

A Bronsted-Lowry acid is a proton (H^+) donor. In the forward reaction water gives a proton to ammonia: H_2O loses H^+ to become OH^-, while NH_3 accepts that proton to become NH_4^+. So water is the Bronsted-Lowry acid (proton donor) and ammonia is the base (proton acceptor). The conjugate pairs are H_2O/OH^- and NH_4^+/NH_3. (2 marks: identify water as the acid and justify by proton donation.)

§-Syllabus dot point

TASChemistrySyllabus dot point

How do we describe acid and base strength and calculate the pH of solutions?

Apply the Bronsted-Lowry model, distinguish strong and weak acids, and calculate pH, pOH and Kw relationships.

The Bronsted-Lowry model, conjugate acid-base pairs, strong versus weak acids and bases, the self-ionisation of water, and pH, pOH and Kw calculations with fully worked TASC-style examples.

Generated by Claude Opus 4.89 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

TASC expects you to apply the Bronsted-Lowry model, identify conjugate pairs, separate strength from concentration, and carry out pH, pOH and $K_w$ calculations for strong acids and bases.

The Bronsted-Lowry model

A Bronsted-Lowry acid donates a proton; a base accepts one. When an acid donates a proton it forms its conjugate base, and when a base accepts a proton it forms its conjugate acid. The two members of a conjugate pair differ by exactly one $\text{H}^+$ .

For $\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-$ , $\text{HCl}$ is the acid and $\text{Cl}^-$ its conjugate base; water is the base and $\text{H}_3\text{O}^+$ its conjugate acid. Species like water that can act as either acid or base are amphiprotic.

Strong versus weak (not the same as concentration)

Strength is about the degree of ionisation:

A strong acid ( $\text{HCl}$ , $\text{HNO}_3$ , $\text{H}_2\text{SO}_4$ ) ionises essentially completely, so we write a single arrow.
A weak acid (ethanoic acid $\text{CH}_3\text{COOH}$ ) ionises only partly and sits at equilibrium, so we write $\rightleftharpoons$ .

Concentration is how many moles of acid are dissolved per litre. A solution can be dilute but strong, or concentrated but weak. Keep the two ideas separate; this distinction is a perennial exam discriminator.

Self-ionisation of water and Kw

Water ionises slightly: $2\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$ .

Because self-ionisation is endothermic, raising the temperature increases $K_w$ , so the pH of neutral water falls below $7$ at higher temperatures (it is still neutral because $[\text{H}^+] = [\text{OH}^-]$ ).

pH and pOH

\text{pH} = -\log_{10}[\text{H}^+] \qquad \text{pOH} = -\log_{10}[\text{OH}^-] \qquad \text{pH} + \text{pOH} = 14 \text{ at } 25\ ^\circ\text{C}

A neutral solution at $25\ ^\circ\text{C}$ has $\text{pH} = 7$ ; acidic solutions are below $7$ and basic above $7$ . To reverse a log, $[\text{H}^+] = 10^{-\text{pH}}$ .

pH of a strong base

Find the pH of $0.0050\ \text{mol L}^{-1}$ sodium hydroxide solution at $25\ ^\circ\text{C}$ .

Step 1: Identify the ion concentration from complete ionisation

Sodium hydroxide is a strong base, so it dissociates completely in water. Every formula unit produces one hydroxide ion, meaning the hydroxide concentration equals the stated concentration with no further calculation needed:

[\text{OH}^-] = 0.0050\ \text{mol L}^{-1}.

Step 2: Calculate pOH using the definition

The pOH is the negative base-ten logarithm of the hydroxide concentration, just as pH is the negative log of the hydrogen ion concentration. Applying that definition:

\text{pOH} = -\log_{10}(0.0050) = -\log_{10}(5.0 \times 10^{-3}) = 2.30.

Step 3: Convert pOH to pH using the water relationship

At $25\ ^\circ\text{C}$ the ionic product of water gives $\text{pH} + \text{pOH} = 14.00$ . Rearranging to find pH:

\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.30 = 11.70.

Step 4: Verify using $K_w$ directly

As a check, calculate $[\text{H}^+]$ from $K_w = [\text{H}^+][\text{OH}^-]$ and take its negative log:

[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.0050} = 2.0 \times 10^{-12}\ \text{mol L}^{-1},

\text{pH} = -\log_{10}(2.0 \times 10^{-12}) = 11.70.

Final answer: pH = 11.70

Polyprotic acids and diprotic bases

Some acids donate more than one proton. Sulfuric acid, $\text{H}_2\text{SO}_4$ , is diprotic: the first ionisation is essentially complete (strong), while the second, $\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-}$ , is only moderately strong. For a dilute strong diprotic acid you can often treat both protons as fully released, so $[\text{H}^+] \approx 2c$ . Phosphoric acid, $\text{H}_3\text{PO}_4$ , releases three protons in successive, progressively weaker steps. When a question gives a diprotic species, check whether each ionisation is strong or weak before assuming complete release.

The logarithmic nature of pH

Because pH is a base-ten logarithm, a change of one pH unit corresponds to a tenfold change in $[\text{H}^+]$ . A solution of pH $3$ is ten times more acidic than pH $4$ and a hundred times more acidic than pH $5$ . This is why diluting a strong acid tenfold raises its pH by exactly one unit (until the concentration approaches that of water itself, where the self-ionisation of water must be included). Keep significant figures in mind: only the digits after the decimal point in a pH value are significant, because the integer part fixes only the power of ten.

In the exam, always decide first whether the acid or base is strong or weak, then choose the method. State the temperature assumption when you use $K_w = 1.0 \times 10^{-14}$ , and show the link between $[\text{H}^+]$ , $[\text{OH}^-]$ and $K_w$ when a question gives you one and asks for the other.

Exam-style practice questions

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

TCE 20213 marksA statement found online claims the pH scale extends only from

0

14

. Hydrochloric acid,

\text{HCl}(aq)

(

M = 36.46\ \text{g mol}^{-1}

), with a concentration of

320\ \text{g L}^{-1}

, is sold at a hardware shop. By determining the pH of this solution, show that the statement is incorrect.

Show worked answer →

Convert to molar concentration: $c = 320 / 36.46 = 8.78\ \text{mol L}^{-1}$ . (1 mark)

$\text{HCl}$ is a strong monoprotic acid, so it fully ionises: $[\text{H}^+] = 8.78\ \text{mol L}^{-1}$ . (1 mark)

\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(8.78) = -0.94.

Because the pH is below

0

(about

-0.9

), the claim that the scale runs only from

0

14

is incorrect: concentrated strong acids can have a negative pH. (1 mark)

TCE 20213 marksWhen

1.08 \times 10^{-2}\ \text{mol}

of sodium is placed in

500\ \text{mL}

of water, the reaction

2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)

occurs. Calculate the pH of the resulting solution at

25\ ^\circ\text{C}

, assuming negligible change in water volume. (

K_w = 1.0 \times 10^{-14}

Show worked answer →

From the $2:2$ ratio, $n(\text{NaOH}) = n(\text{Na}) = 1.08 \times 10^{-2}\ \text{mol}$ . (1 mark)

$\text{NaOH}$ is a strong base, so $[\text{OH}^-] = 1.08 \times 10^{-2} / 0.500 = 2.16 \times 10^{-2}\ \text{mol L}^{-1}$ . (1 mark)

$\text{pOH} = -\log_{10}(2.16 \times 10^{-2}) = 1.67$ , so at $25\ ^\circ\text{C}$ , $\text{pH} = 14.00 - 1.67 = 12.33$ . (1 mark)

TCE 20222 marksFor the equilibrium

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

, which reactant is the Bronsted-Lowry acid? Justify your answer using conjugate pairs.

Show worked answer →

A Bronsted-Lowry acid is a proton ( $\text{H}^+$ ) donor. In the forward reaction water gives a proton to ammonia: $\text{H}_2\text{O}$ loses $\text{H}^+$ to become $\text{OH}^-$ , while $\text{NH}_3$ accepts that proton to become $\text{NH}_4^+$ .

So water is the Bronsted-Lowry acid (proton donor) and ammonia is the base (proton acceptor). The conjugate pairs are $\text{H}_2\text{O}/\text{OH}^-$ and $\text{NH}_4^+/\text{NH}_3$ . (2 marks: identify water as the acid and justify by proton donation.)