What are significant figures for logs?

Only the digits after the decimal point in a log are significant. A [H^+] of 1.64 × 10^-3 (3 sig fig) gives pH = 2.79 (2 decimal places).

What are wrong sig figs?

For pH, the number of decimal places equals the sig figs of the concentration. pH = 2.79 (3 sig fig), not 2.794 or 2.8.

What are dilution errors?

When diluting a strong acid 10-fold, pH increases by 1 (more dilute, less acidic). When diluting a weak acid 10-fold, pH increases by about 0.5 because dissociation increases as concentration falls.

NESA 2022 HSC-style practice: Calculate the pH of a 0.150 mol/L solution of ethanoic acid (Ka = 1.8 × 10⁻⁵).

A 4 mark answer requires the dissociation equation, an ICE table, the approximation step, and the final pH. **Step 1: Dissociation.** $CH_3COOH_(aq) CH_3COO^-_(aq) + H^+_(aq)$ **Step 2: ICE table.** | | CH_3COOH | CH_3COO^- | H^+ | |---|---|---|---| | Initial | 0.150 | 0 | 0 | | Change | -x | +x | +x | | Equilibrium | 0.150 - x | x | x | **Step 3: Ka expression and approximation.** $K_a = (x^2)/(0.150 - x) ≈ (x^2)/(0.150) = 1.8 × 10^-5$ (Approximation valid because Ka is small.) $x^2 = 2.7 × 10^-6$ $x = [H^+] = 1.64 × 10^-3 mol/L$ **Step 4: pH.** $pH = -_10(1.64 × 10^-3) = 2.79$ **Check** the 5% rule: x / 0.150 = 1.1\%, so the approximation is valid. Markers reward (1) the dissociation equation, (2) the ICE setup, (3) the Ka substitution with approximation justified, (4) pH to two decimal places (correct sig figs for log).

NSWChemistrySyllabus dot point

Inquiry Question 5: How are acids and bases defined and how do they behave in aqueous solution?

Conduct investigations and perform calculations to determine the pH and pOH of strong and weak acids and bases, applying the formulae pH equals negative log of hydrogen ion concentration, and pH plus pOH equals 14

A focused answer to the HSC Chemistry Module 5 dot point on pH and pOH. The pH and pOH formulae, the auto-ionisation of water, strong vs weak acid/base calculations using ICE tables, dilution effects, and worked HSC past exam questions.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-22

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

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What this dot point is asking
The answer
Worked example 2: when the 5% rule fails
Examples in context
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What this dot point is asking

NESA wants you to calculate the pH and pOH of strong and weak acid and base solutions, use the auto-ionisation constant of water ( $K_w$ ), and apply the relationships $pH = -\log_{10}[H^+]$ and $pH + pOH = 14$ . The chemistry of proton donation and acceptance is set up in the Brønsted-Lowry dot point. Expect a calculation question every year, with weak-acid problems carrying the highest marks.

The answer

The diagram below places common substances on the 0-14 pH scale.

The pH scale

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

A lower pH means a higher $[H^+]$ and a more acidic solution. Each unit of pH corresponds to a tenfold change in $[H^+]$ .

pH	$[H^+]$	Description
1	$10^{-1}$	strongly acidic
4	$10^{-4}$	weakly acidic
7	$10^{-7}$	neutral at 25°C
10	$10^{-10}$	weakly basic
13	$10^{-13}$	strongly basic

The auto-ionisation of water

Water self-ionises:

2H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)}

At 25°C:

K_w = [H^+][OH^-] = 1.0 \times 10^{-14}

Taking $-\log_{10}$ of both sides gives:

pH + pOH = 14 \quad \text{(at 25°C)}

For pure water, $[H^+] = [OH^-] = 10^{-7}$ , so pH = pOH = 7. Water is neutral.

Strong acids and bases

Strong acids (HCl, $HNO_3$ , $H_2SO_4$ , $HClO_4$ ) dissociate completely. $[H^+]$ equals the acid concentration (for monoprotic acids).

Strong bases (NaOH, KOH, $Ca(OH)_2$ , $Ba(OH)_2$ ) dissociate completely. Be careful with diprotic bases: $[OH^-] = 2 \times$ concentration of $Ca(OH)_2$ .

Calculation is one step:

pH = -\log_{10}[H^+] \quad \text{or} \quad pOH = -\log_{10}[OH^-]

Weak acids and bases

Weak acids dissociate only partially. Use the dissociation constant Ka and an ICE table.

For a weak acid $HA \rightleftharpoons H^+ + A^-$ :

K_a = \frac{[H^+][A^-]}{[HA]}

If the initial concentration is $C_0$ and the extent of dissociation is $x$ :

K_a = \frac{x^2}{C_0 - x}

When $K_a$ is small and $C_0$ is reasonable (the 5% rule), approximate $C_0 - x \approx C_0$ :

x = [H^+] \approx \sqrt{K_a \cdot C_0}

Significant figures for logs

Only the digits after the decimal point in a log are significant. A $[H^+]$ of $1.64 \times 10^{-3}$ (3 sig fig) gives pH = 2.79 (2 decimal places).

Worked example

Calculate the pH of three solutions at 25°C. The three parts deliberately compare a strong acid, a strong base and a weak acid at the same concentration, so the method changes with the type of species rather than the number.

Step 1: (a) 0.025 mol/L HCl (strong acid).

HCl dissociates completely, so every mole of acid releases a mole of $H^+$ and the hydrogen ion concentration equals the acid concentration, $[H^+] = 0.025$ mol/L. The pH then comes straight from the definition:

pH = -\log_{10}(0.025) = 1.60

Step 2: (b) 0.025 mol/L NaOH (strong base).

NaOH is a strong base, so it fully releases hydroxide and $[OH^-] = 0.025$ mol/L. We work out pOH first because the concentration we know is of hydroxide, not hydrogen ions:

pOH = -\log_{10}(0.025) = 1.60

Because water links the two scales through $pH + pOH = 14$ at 25°C, we convert pOH to pH:

pH = 14 - 1.60 = 12.40

Step 3: (c) 0.025 mol/L $CH_3COOH$ (weak acid, $K_a = 1.8 \times 10^{-5}$ ).

A weak acid only partially dissociates, so $[H^+]$ is no longer the acid concentration. With a small $K_a$ we use the approximation $[H^+] \approx \sqrt{K_a \cdot C_0}$ , which assumes the amount that dissociates is tiny compared with $C_0$ :

x = [H^+] \approx \sqrt{K_a \cdot C_0} = \sqrt{(1.8 \times 10^{-5})(0.025)}

= \sqrt{4.5 \times 10^{-7}} = 6.71 \times 10^{-4} \text{ mol/L}

pH = -\log_{10}(6.71 \times 10^{-4}) = 3.17

Step 4: Check the approximation. The 5% rule confirms the shortcut was valid, since the dissociated fraction is well under 5%: $6.71 \times 10^{-4} / 0.025 = 2.7\%$ .

Notice: the strong and weak acid had the same concentration but very different pH values (1.60 vs 3.17), because the weak acid is only partially dissociated.

Worked example 2: when the 5% rule fails

Calculate the pH and percent dissociation of 0.0010 mol/L formic acid ( $HCOOH$ , $K_a = 1.8 \times 10^{-4}$ ).

Step 1: Try the approximation.

x \approx \sqrt{K_a \cdot C_0} = \sqrt{(1.8 \times 10^{-4})(0.0010)} = 4.24 \times 10^{-4} \text{ mol/L}

Step 2: Check the 5% rule. $4.24 \times 10^{-4} / 0.0010 = 42\%$ , well above 5%. The approximation fails. Solve the quadratic.

Step 3: Set up and solve.

\frac{x^2}{0.0010 - x} = 1.8 \times 10^{-4}

x^2 + (1.8 \times 10^{-4})x - (1.8 \times 10^{-7}) = 0

x = \frac{-1.8 \times 10^{-4} + \sqrt{(1.8 \times 10^{-4})^2 + 4(1.8 \times 10^{-7})}}{2}

x = \frac{-1.8 \times 10^{-4} + \sqrt{7.524 \times 10^{-7}}}{2} = \frac{-1.8 \times 10^{-4} + 8.674 \times 10^{-4}}{2} = 3.44 \times 10^{-4}

So $[H^+] = 3.44 \times 10^{-4}$ mol/L.

Step 4: pH and percent dissociation.

pH = -\log_{10}(3.44 \times 10^{-4}) = 3.46

\text{Percent dissociation} = \frac{[H^+]_{\text{eq}}}{C_0} \times 100\% = \frac{3.44 \times 10^{-4}}{0.0010} \times 100\% = 34.4\%

The approximate $[H^+]$ (4.24 × 10⁻⁴) overestimated the true value (3.44 × 10⁻⁴) by ~23%. As a weak acid becomes more dilute, percent dissociation rises and the 5% rule begins to fail. Always check.

Common traps

Forgetting to double for $Ca(OH)_2$ or $Ba(OH)_2$: Each formula unit produces two $OH^-$ .
Mixing $H_2SO_4$ assumptions: The first dissociation is complete; the second has Ka ≈ $1.2 \times 10^{-2}$ . HSC usually treats $H_2SO_4$ as fully dissociating both protons for dilute solutions, but a careful answer states the assumption.
Skipping the 5% check for weak acids: If $x$ exceeds 5% of $C_0$ , you must solve the quadratic. The approximation fails for concentrations below ~ $10^{-3}$ mol/L of typical weak acids.
Wrong sig figs: For pH, the number of decimal places equals the sig figs of the concentration. pH = 2.79 (3 sig fig), not 2.794 or 2.8.
Dilution errors: When diluting a strong acid 10-fold, pH increases by 1 (more dilute, less acidic). When diluting a weak acid 10-fold, pH increases by about 0.5 because dissociation increases as concentration falls.

Examples in context

Example 1. Macquarie River pH monitoring near Dubbo. WaterNSW field officers sample the Macquarie River monthly above and below the Dubbo wastewater outfall. A typical upstream sample registers pH 7.9 with $[H^+] = 10^{-7.9} \approx 1.3 \times 10^{-8}$ mol L $^{-1}$ . After a heavy storm in 2024 the river ran with $[H^+] = 5.0 \times 10^{-7}$ mol L $^{-1}$ , dropping the pH to 6.3. The shift correlates with first-flush stormwater carrying organic acids from upstream irrigation. Officers cross-check pOH using $pH + pOH = 14$ , confirming the calibrated electrode reading. The same $pH = -\log[H^+]$ relation that HSC candidates use is the equation embedded in every commercial water-quality probe.

Example 2. Calculating the pH of acetic acid in vinegar. A typical commercial vinegar contains acetic acid at 0.83 mol L $^{-1}$ , with $K_a = 1.8 \times 10^{-5}$ . Because the acid is weak, an ICE table yields $[H^+] = \sqrt{K_a \times c} = \sqrt{1.8 \times 10^{-5} \times 0.83} = 3.9 \times 10^{-3}$ mol L $^{-1}$ , giving pH 2.4. NSW Health uses this pH range when issuing food storage advice. The 5 percent approximation is valid here because the dissociated fraction is only 0.47 percent. If a student forgot the ICE table and assumed full dissociation, they would predict pH 0.08, off by more than two units, an error that would lose 3 of 4 marks in Section II.

Try this

Q1. State the relationship between pH, pOH, $[H^+]$ and $[OH^-]$ at 25 degrees C. [2 marks]

Cue. $pH = -\log_{10}[H^+]$ , $pOH = -\log_{10}[OH^-]$ , $pH + pOH = 14$ , $[H^+][OH^-] = 1.0 \times 10^{-14}$ .

Q2. Calculate the pH of a 0.020 mol L $^{-1}$ solution of $Ba(OH)_2$ , a strong diprotic base. [3 marks]

Cue. $[OH^-] = 2 \times 0.020 = 0.040$ mol L $^{-1}$ , $pOH = -\log(0.040) = 1.40$ , $pH = 14 - 1.40 = 12.60$ .

Q3. A 0.10 mol L $^{-1}$ solution of formic acid (HCOOH, $K_a = 1.8 \times 10^{-4}$ ) is prepared. (a) Write the equilibrium expression. (b) Calculate $[H^+]$ using the 5 percent approximation. (c) State the resulting pH and the percent ionisation. [1+2+2 marks]

Cue. (a) $K_a = [H^+][HCOO^-] / [HCOOH]$ . (b) $[H^+] \approx \sqrt{1.8 \times 10^{-4} \times 0.10} = 4.2 \times 10^{-3}$ mol L $^{-1}$ . (c) $pH = 2.37$ , ionisation $= 4.2$ percent (just within the 5 percent rule).

Exam-style practice questions

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

2022 HSC4 marksCalculate the pH of a 0.150 mol/L solution of ethanoic acid (Ka = 1.8 × 10⁻⁵).

Show worked answer →

A 4 mark answer requires the dissociation equation, an ICE table, the approximation step, and the final pH.

Step 1: Dissociation.

CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}

Step 2: ICE table.

	$CH_3COOH$	$CH_3COO^-$	$H^+$
Initial	0.150	0	0
Change	$-x$	$+x$	$+x$
Equilibrium	$0.150 - x$	$x$	$x$

Step 3: Ka expression and approximation.

K_a = \frac{x^2}{0.150 - x} \approx \frac{x^2}{0.150} = 1.8 \times 10^{-5}

(Approximation valid because Ka is small.)

x^2 = 2.7 \times 10^{-6}

x = [H^+] = 1.64 \times 10^{-3} \text{ mol/L}

Step 4: pH.

pH = -\log_{10}(1.64 \times 10^{-3}) = 2.79

Check the 5% rule: $x / 0.150 = 1.1\%$ , so the approximation is valid.

Markers reward (1) the dissociation equation, (2) the ICE setup, (3) the Ka substitution with approximation justified, (4) pH to two decimal places (correct sig figs for log).

2018 HSC3 marksA 0.020 mol/L solution of Ca(OH)₂ is prepared. Calculate the pH of this solution at 25°C, assuming complete dissociation.