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pH and pOH calculations explained: HSC Chemistry Module 5

Syllabus dot point

“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”

9 min read

Quick answer

To find pH, use pH=−log⁡10[H+]pH = -\log_{10}[H^+] directly for strong acids (and pH=14−pOHpH = 14 - pOH for strong bases), and for weak acids and bases use Ka or Kb with an ICE table to find [H+][H^+] or [OH−][OH^-], applying the 5% approximation only when justified.

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Worked example 2: when the 5% rule fails
  4. Examples in context
  5. Try this
  6. Exam-style questions
  7. Practice questions
In plain English

Think of pH as a way of counting how "crowded" a solution is with hydrogen ions, but on a scale that shrinks huge numbers down to something manageable. Because the actual concentration of H+H^+ can range from about 1 mol/L down to a tiny 0.00000000000001 mol/L, chemists take the negative log so the scale runs neatly from 0 to 14 instead of stretching across fourteen powers of ten. A strong acid like HCl is a "generous donor": every single molecule hands over its H+H^+, so what you weighed out is exactly what you get in solution. A weak acid like ethanoic acid (the acid in vinegar) is a "reluctant donor": most molecules keep their H+H^+ to themselves, and only a small fraction let go at any moment, which is why you need an ICE table and a KaK_a value to work out how many actually did. Water itself is constantly handing H+H^+ back and forth between its own molecules, which is why even pure water has a tiny but non-zero [H+][H^+], and why pH and pOH are always locked together through pH+pOH=14pH + pOH = 14 at 25 degrees C, like two ends of a see-saw.

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 (KwK_w), and apply the relationships pH=−log⁡10[H+]pH = -\log_{10}[H^+] and pH+pOH=14pH + 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.

pH scale with common substances A horizontal scale from pH 0 on the left to pH 14 on the right. Below pH 7 is acidic, at pH 7 is neutral, above pH 7 is basic. Substances are pinned to the scale at their typical pH: battery acid at 0, lemon at 2, vinegar at 3, coffee at 5, milk at 6, pure water at 7, baking soda at 9, ammonia at 11, bleach at 13. The pH scale 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Acidic Neutral Basic battery acid lemon vinegar coffee milk pure water baking soda ammonia bleach pH = −log₁₀[H⁺] and pOH = −log₁₀[OH⁻] In water at 25 °C, pH + pOH = 14. Each pH unit is a tenfold change in [H⁺].

The pH scale

pH=−log⁡10[H+]pH = -\log_{10}[H^+]

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

pH [H+][H^+] Description
1 10−110^{-1} strongly acidic
4 10−410^{-4} weakly acidic
7 10−710^{-7} neutral at 25°C
10 10−1010^{-10} weakly basic
13 10−1310^{-13} strongly basic

The auto-ionisation of water

Water self-ionises:

2H2O(l)⇌H3O(aq)++OH(aq)−2H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)}

At 25°C:

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

Taking −log⁡10-\log_{10} of both sides gives:

pH+pOH=14(at 25°C)pH + pOH = 14 \quad \text{(at 25°C)}

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

The four relationships you must know cold
  • pH=−log⁡10[H+]pH = -\log_{10}[H^+] and pOH=−log⁡10[OH−]pOH = -\log_{10}[OH^-].
  • pH+pOH=14pH + pOH = 14 at 25 degrees C.
  • Kw=[H+][OH−]=1.0×10−14K_w = [H^+][OH^-] = 1.0 \times 10^{-14} at 25 degrees C (increases at higher temperature, since auto-ionisation is endothermic).
  • For a weak acid, Ka=[H+][A−][HA]K_a = \dfrac{[H^+][A^-]}{[HA]}, and [H+]≈Ka⋅C0[H^+] \approx \sqrt{K_a \cdot C_0} only when the 5 percent rule holds.

An illustrative titration curve shows how the shape of the pH-versus-volume graph tells you whether the original acid was strong or weak, and where the equivalence point sits:

Illustrative titration curve: weak monoprotic acid titrated with 0.100 mol/L NaOH An illustrative graph of pH against volume of 0.100 mol per litre sodium hydroxide added to 25.0 millilitres of a weak monoprotic acid. The curve starts near pH 3, rises gently through a buffering plateau, then rises steeply through an equivalence point at 20.0 millilitres and pH 8.5, above pH 7 because the conjugate base hydrolyses water, before levelling off at high pH with excess base. 14 10.5 7 3.5 0 equivalence point 20.0 mL, pH 8.5 (above 7) buffering region gradual rise, weak acid 0 10 20 30 40 Volume of 0.100 mol/L NaOH added / mL (illustrative ExamExplained curve, not lab data) pH vs volume of NaOH added

The initial pH of 3.0 (rather than close to pH 1, as a strong acid of similar concentration would show) and the equivalence-point pH of 8.5 (above 7, from hydrolysis of the conjugate base) are the two signatures that identify a weak acid from a titration curve alone, without knowing KaK_a in advance.

Strong acids and bases

Strong acids (HCl, HNO3HNO_3, H2SO4H_2SO_4, HClO4HClO_4) dissociate completely. [H+][H^+] equals the acid concentration (for monoprotic acids).

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

Calculation is one step:

pH=−log⁡10[H+]orpOH=−log⁡10[OH−]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⇌H++A−HA \rightleftharpoons H^+ + A^-:

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

If the initial concentration is C0C_0 and the extent of dissociation is xx:

Ka=x2C0−xK_a = \frac{x^2}{C_0 - x}

When KaK_a is small and C0C_0 is reasonable (the 5% rule), approximate C0−x≈C0C_0 - x \approx C_0:

x=[H+]≈Ka⋅C0x = [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+][H^+] of 1.64×10−31.64 \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+H^+ and the hydrogen ion concentration equals the acid concentration, [H+]=0.025[H^+] = 0.025 mol/L. The pH then comes straight from the definition:

pH=−log⁡10(0.025)=1.60pH = -\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[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)=1.60pOH = -\log_{10}(0.025) = 1.60

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

pH=14−1.60=12.40pH = 14 - 1.60 = 12.40

Step 3: (c) 0.025 mol/L CH3COOHCH_3COOH (weak acid, Ka=1.8×10−5K_a = 1.8 \times 10^{-5}).

A weak acid only partially dissociates, so [H+][H^+] is no longer the acid concentration. With a small KaK_a we use the approximation [H+]≈Ka⋅C0[H^+] \approx \sqrt{K_a \cdot C_0}, which assumes the amount that dissociates is tiny compared with C0C_0:

x=[H+]≈Ka⋅C0=(1.8×10−5)(0.025)x = [H^+] \approx \sqrt{K_a \cdot C_0} = \sqrt{(1.8 \times 10^{-5})(0.025)}

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

pH=−log⁡10(6.71×10−4)=3.17pH = -\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%: 6.71×10−4/0.025=2.7%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 (HCOOHHCOOH, Ka=1.8×10−4K_a = 1.8 \times 10^{-4}).

Step 1: Try the approximation.

x≈Ka⋅C0=(1.8×10−4)(0.0010)=4.24×10−4 mol/Lx \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. 4.24×10−4/0.0010=42%4.24 \times 10^{-4} / 0.0010 = 42\%, well above 5%. The approximation fails. Solve the quadratic.

Step 3: Set up and solve.

x20.0010−x=1.8×10−4\frac{x^2}{0.0010 - x} = 1.8 \times 10^{-4}

x2+(1.8×10−4)x−(1.8×10−7)=0x^2 + (1.8 \times 10^{-4})x - (1.8 \times 10^{-7}) = 0

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

x=−1.8×10−4+7.524×10−72=−1.8×10−4+8.674×10−42=3.44×10−4x = \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+]=3.44×10−4[H^+] = 3.44 \times 10^{-4} mol/L.

Step 4: pH and percent dissociation.

pH=−log⁡10(3.44×10−4)=3.46pH = -\log_{10}(3.44 \times 10^{-4}) = 3.46

Percent dissociation=[H+]eqC0×100%=3.44×10−40.0010×100%=34.4%\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+][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.

Exam technique for pH calculations

Always show your working in this order: (1) write the dissociation/auto-ionisation equation, (2) state whether the species is strong (complete) or weak (needs Ka/Kb and an ICE table), (3) substitute into the correct formula, (4) check the 5 percent rule if you used the square-root approximation for a weak acid, (5) state the final pH to the correct number of decimal places (matching the significant figures of the given concentration). For a diprotic strong base like Ca(OH)2Ca(OH)_2 or Ba(OH)2Ba(OH)_2, write the doubling step explicitly, "[OH−]=2×c[OH^-] = 2 \times c", rather than doing it in your head, since markers cannot award a mark for an unstated step.

Common traps
Forgetting to double for Ca(OH)2Ca(OH)_2 or Ba(OH)2Ba(OH)_2
Each formula unit produces two OH−OH^-.
Mixing H2SO4H_2SO_4 assumptions
The first dissociation is complete; the second is actually a weak-acid equilibrium (Ka2 ≈ 1.2×10−21.2 \times 10^{-2}), not a further complete dissociation. HSC problems commonly approximate the second proton as fully dissociating too, but only because dilute solutions push the equilibrium far enough to the right to make that a reasonable simplification, so a careful answer states the assumption explicitly.
Skipping the 5% check for weak acids
If xx exceeds 5% of C0C_0, you must solve the quadratic. The approximation fails for concentrations below ~10−310^{-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≈1.3×10−8[H^+] = 10^{-7.9} \approx 1.3 \times 10^{-8} mol L−1^{-1}. After a heavy storm in 2024 the river ran with [H+]=5.0×10−7[H^+] = 5.0 \times 10^{-7} mol L−1^{-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=14pH + pOH = 14, confirming the calibrated electrode reading. The same pH=−log⁡[H+]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^{-1}, with Ka=1.8×10−5K_a = 1.8 \times 10^{-5}. Because the acid is weak, an ICE table yields [H+]=Ka×c=1.8×10−5×0.83=3.9×10−3[H^+] = \sqrt{K_a \times c} = \sqrt{1.8 \times 10^{-5} \times 0.83} = 3.9 \times 10^{-3} mol L−1^{-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+][H^+] and [OH−][OH^-] at 25 degrees C. [2 marks]

  • Cue. pH=−log⁡10[H+]pH = -\log_{10}[H^+], pOH=−log⁡10[OH−]pOH = -\log_{10}[OH^-], pH+pOH=14pH + pOH = 14, [H+][OH−]=1.0×10−14[H^+][OH^-] = 1.0 \times 10^{-14}.

Q2. Calculate the pH of a 0.020 mol L−1^{-1} solution of Ba(OH)2Ba(OH)_2, a strong diprotic base. [3 marks]

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

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

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

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2022 HSC-style4 marks
Calculate 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.

CH3COOH(aq)⇌CH3COO(aq)−+H(aq)+CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}

Step 2: ICE table.

CH3COOHCH_3COOH CH3COO−CH_3COO^- H+H^+
Initial 0.150 0 0
Change −x-x +x+x +x+x
Equilibrium 0.150−x0.150 - x xx xx

Step 3: Ka expression and approximation.

Ka=x20.150−x≈x20.150=1.8×10−5K_a = \frac{x^2}{0.150 - x} \approx \frac{x^2}{0.150} = 1.8 \times 10^{-5}

(Approximation valid because Ka is small.)

x2=2.7×10−6x^2 = 2.7 \times 10^{-6}

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

Step 4: pH.

pH=−log⁡10(1.64×10−3)=2.79pH = -\log_{10}(1.64 \times 10^{-3}) = 2.79

Check the 5% rule: x/0.150=1.1%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 HSC-style3 marks
A 0.020 mol/L solution of Ca(OH)₂ is prepared. Calculate the pH of this solution at 25°C, assuming complete dissociation.
Show worked answer →

Ca(OH)2Ca(OH)_2 is a strong base. Each formula unit produces 2 hydroxide ions.

Ca(OH)2(aq)→Ca(aq)2++2OH(aq)−Ca(OH)_{2(aq)} \rightarrow Ca^{2+}_{(aq)} + 2OH^-_{(aq)}

[OH−]=2×0.020=0.040[OH^-] = 2 \times 0.020 = 0.040 mol/L.

pOH=−log⁡10(0.040)=1.40pOH = -\log_{10}(0.040) = 1.40

pH=14−pOH=14−1.40=12.60pH = 14 - pOH = 14 - 1.40 = 12.60

Markers reward (1) doubling for the two hydroxides per formula unit, (2) the pOH calculation, (3) using pH+pOH=14pH + pOH = 14 correctly. Forgetting to double is the most common error.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks
State the value of [OH−][OH^-] in a solution with [H+]=1.0×10−9[H^+] = 1.0 \times 10^{-9} mol/L at 25 degrees C, and classify the solution as acidic, neutral or basic.
Show worked solution →

Step 1: apply KwK_w.

[OH−]=Kw[H+]=1.0×10−141.0×10−9=1.0×10−5 mol/L[OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-9}} = 1.0 \times 10^{-5} \text{ mol/L}

Step 2: classify. Since [H+]=1.0×10−9<1.0×10−7[H^+] = 1.0 \times 10^{-9} < 1.0 \times 10^{-7}, the solution has fewer hydrogen ions than pure water, so it is basic (pH 9).

Marking criteria: 1 mark for the correct [OH−][OH^-] using KwK_w, 1 mark for correctly classifying the solution as basic with a reason.

foundation3 marks
Calculate the pH of a 0.050 mol/L solution of hydrochloric acid, HCl, a strong monoprotic acid.
Show worked solution →

Step 1: complete dissociation.

HCl(aq)→H(aq)++Cl(aq)−HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)}

HCl is a strong acid, so [H+]=0.050[H^+] = 0.050 mol/L.

Step 2: apply the pH formula.

pH=−log⁡10(0.050)=1.30pH = -\log_{10}(0.050) = 1.30

Step 3: sig figs. The concentration has 2 significant figures, so pH is reported to 2 decimal places: pH = 1.30.

Marking criteria: 1 mark for recognising complete dissociation and [H+]=0.050[H^+] = 0.050, 1 mark for the correct log calculation, 1 mark for the correct sig figs.

core5 marks
A buffer-development lab prepares 250.0 mL of a solution by dissolving 0.912 g of benzoic acid (C6H5COOHC_6H_5COOH, M=122.1M = 122.1 g mol−1^{-1}, Ka=6.3×10−5K_a = 6.3 \times 10^{-5}) in water. Calculate the pH of the resulting solution to 2 decimal places, checking whether the 5 percent approximation is valid.
Show worked solution →

Step 1: moles of benzoic acid.

n=mM=0.912 g122.1 g mol−1=7.470×10−3 moln = \frac{m}{M} = \frac{0.912\ \text{g}}{122.1\ \text{g mol}^{-1}} = 7.470 \times 10^{-3}\ \text{mol}

Step 2: initial concentration.

C0=nV=7.470×10−3 mol0.2500 L=2.988×10−2 mol L−1C_0 = \frac{n}{V} = \frac{7.470 \times 10^{-3}\ \text{mol}}{0.2500\ \text{L}} = 2.988 \times 10^{-2}\ \text{mol L}^{-1}

Step 3: ICE table and approximation.

C6H5COOH(aq)⇌C6H5COO(aq)−+H(aq)+C_6H_5COOH_{(aq)} \rightleftharpoons C_6H_5COO^-_{(aq)} + H^+_{(aq)}

Ka=x2C0−x≈x2C0K_a = \frac{x^2}{C_0 - x} \approx \frac{x^2}{C_0}

x=[H+]≈Ka⋅C0=(6.3×10−5)(2.988×10−2)=1.882×10−6=1.372×10−3 mol L−1x = [H^+] \approx \sqrt{K_a \cdot C_0} = \sqrt{(6.3 \times 10^{-5})(2.988 \times 10^{-2})} = \sqrt{1.882 \times 10^{-6}} = 1.372 \times 10^{-3}\ \text{mol L}^{-1}

Step 4: check the 5 percent rule.

1.372×10−32.988×10−2×100%=4.59%\frac{1.372 \times 10^{-3}}{2.988 \times 10^{-2}} \times 100\% = 4.59\%

This is just under 5 percent, so the approximation is valid (no quadratic needed).

Step 5: pH.

pH=−log⁡10(1.372×10−3)=2.86pH = -\log_{10}(1.372 \times 10^{-3}) = 2.86

Marking criteria: 1 mark for correct moles, 1 mark for correct C0C_0, 1 mark for the correct approximation substitution, 1 mark for checking and confirming the 5 percent rule, 1 mark for pH to the correct 2 decimal places.

core5 marks
The graph below shows an illustrative titration curve for 25.0 mL of a monoprotic acid titrated with 0.100 mol/L NaOH, reaching the equivalence point at 20.0 mL of titrant with pH 8.5 at equivalence. (a) State whether the acid is strong or weak, justifying your answer using the initial pH and the equivalence point pH shown. (b) Identify a suitable indicator for this titration from the curve.
Show worked solution →
Reading the curve
The curve starts at pH 3.0 (not pH close to 1, which a strong acid of similar concentration would show), rises gradually through a buffering region, then shows a steep vertical jump around 20.0 mL, and the equivalence point sits at pH 8.5 (above 7), not pH 7.
(a) Strong or weak
The acid is WEAK. Two pieces of evidence from the curve: (1) the initial pH (3.0) is higher than expected for a strong acid of comparable concentration, indicating only partial dissociation; (2) the equivalence point pH is above 7 (8.5), because the conjugate base of a weak acid hydrolyses water to produce a basic salt solution, which only happens when the original acid was weak.
(b) Indicator choice
Phenolphthalein (colour change range approximately pH 8.2 to 10.0) is suitable, because its colour-change range brackets the equivalence point pH of 8.5, sitting within the steep vertical region of the curve. Methyl orange (range about pH 3.1 to 4.4) would change colour far too early, well before equivalence, and would not be suitable.

Marking criteria: (a) 1 mark for identifying "weak", 1 mark for citing the elevated initial pH as evidence, 1 mark for citing the equivalence point pH above 7 (with the hydrolysis reasoning) as evidence. (b) 1 mark for naming phenolphthalein, 1 mark for justifying it against the equivalence point pH shown on the curve.

core4 marks
A 0.0100 mol/L solution of hydrochloric acid is diluted 100-fold. (a) Calculate the pH before and after dilution. (b) A student diluted a 0.0100 mol/L solution of ethanoic acid (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}) by the same factor and found the pH changed by only about 0.9 of a unit rather than the full 2 units seen for HCl. Explain this observation.
Show worked solution →

(a) HCl before and after dilution.

Before: HCl is a strong acid, so [H+]=0.0100[H^+] = 0.0100 mol/L.

pHbefore=−log⁡10(0.0100)=2.00pH_{\text{before}} = -\log_{10}(0.0100) = 2.00

After 100-fold dilution, [H+]=0.0100/100=1.00×10−4[H^+] = 0.0100 / 100 = 1.00 \times 10^{-4} mol/L.

pHafter=−log⁡10(1.00×10−4)=4.00pH_{\text{after}} = -\log_{10}(1.00 \times 10^{-4}) = 4.00

The pH increased by exactly 2 units, matching the 100-fold (two powers of ten) dilution, because a strong acid's [H+][H^+] scales directly with concentration.

(b) Explanation for the weak acid. For a weak acid, [H+]≈Ka⋅C0[H^+] \approx \sqrt{K_a \cdot C_0}, so [H+][H^+] scales with the SQUARE ROOT of concentration, not concentration directly. Diluting C0C_0 by a factor of 100 only reduces [H+][H^+] by a factor of 100=10\sqrt{100} = 10, which corresponds to a pH change of log⁡10(10)=1\log_{10}(10) = 1 unit from the square-root relationship alone; the observed change is a little less than 1 (about 0.9) because percent dissociation also rises slightly as the solution becomes more dilute, partially buffering the pH shift. This is why weak acids resist pH change on dilution far more than strong acids do.

Marking criteria: (a) 1 mark for the correct pH before dilution, 1 mark for the correct pH after dilution (2 marks total for both values with working). (b) 1 mark for identifying the square-root relationship between [H+][H^+] and C0C_0 for a weak acid, 1 mark for correctly linking this to a smaller pH shift than the strong acid case.

exam6 marks
A first-year chemistry student argues that 'because Ca(OH)2Ca(OH)_2 and NaOH are both strong bases, a 0.10 mol/L solution of each must have the same pH.' Evaluate this claim quantitatively, then explain, using Le Chatelier's principle, what would happen to the pH of the Ca(OH)2Ca(OH)_2 solution if solid CaCO3CaCO_3 were added and allowed to reach a new dissolution equilibrium with excess undissolved solid present.
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This is a 6-mark EVALUATE: markers reward a numeric comparison plus reasoned equilibrium argument, not just an assertion.

Band 6 PLAN.

  • Calculate pH of 0.10 mol/L NaOH (monoprotic) and 0.10 mol/L Ca(OH)2Ca(OH)_2 (diprotic) separately.
  • State the claim is FALSE, with the numeric gap as evidence.
  • Address the CaCO3CaCO_3 addition using Le Chatelier on the (separate) CaCO3CaCO_3 dissolution equilibrium, connecting back to the effect on the existing OH−OH^- concentration only if relevant, and reasoning about the new system's own equilibrium.

Model answer.

For 0.10 mol/L NaOH: complete dissociation gives [OH−]=0.10[OH^-] = 0.10 mol/L.

pOH=−log⁡10(0.10)=1.00,pH=14−1.00=13.00pOH = -\log_{10}(0.10) = 1.00, \quad pH = 14 - 1.00 = 13.00

For 0.10 mol/L Ca(OH)2Ca(OH)_2: each formula unit releases two hydroxide ions, so [OH−]=2×0.10=0.20[OH^-] = 2 \times 0.10 = 0.20 mol/L.

pOH=−log⁡10(0.20)=0.70,pH=14−0.70=13.30pOH = -\log_{10}(0.20) = 0.70, \quad pH = 14 - 0.70 = 13.30

The claim is FALSE: although both are strong bases (complete dissociation), the Ca(OH)2Ca(OH)_2 solution has a higher pH (13.30 vs 13.00) because it is diprotic and releases twice the hydroxide ions per mole dissolved. "Strong" describes the extent of dissociation, not the number of hydroxide ions released per formula unit.

For the CaCO3CaCO_3 addition: CaCO3CaCO_3 is only sparingly soluble, and adding solid establishes its own dissolution equilibrium, CaCO3(s)⇌Ca(aq)2++CO3(aq)2−CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO_{3(aq)}^{2-}. This equilibrium is independent of the Ca(OH)2Ca(OH)_2 already dissolved, though the shared Ca2+Ca^{2+} ion (a common-ion effect) would slightly suppress further CaCO3CaCO_3 dissolution compared with dissolving CaCO3CaCO_3 in pure water, by Le Chatelier's principle: the existing Ca2+Ca^{2+} from Ca(OH)2Ca(OH)_2 shifts the CaCO3CaCO_3 dissolution equilibrium to the left, reducing the amount of CaCO3CaCO_3 that dissolves. Since CaCO3CaCO_3 dissolution does not directly add or remove OH−OH^-, the pH of the solution remains governed by the original Ca(OH)2Ca(OH)_2 concentration and is not significantly changed by this common-ion suppression at equilibrium.

Marker's note: top-band answers (1) correctly calculate BOTH pH values with correct doubling for the diprotic base, (2) explicitly reject the claim with the numeric gap as evidence, (3) correctly distinguish "strong" (extent of dissociation) from "concentration of ions released", and (4) apply Le Chatelier to the correct equilibrium (the CaCO3CaCO_3 dissolution, via the common-ion effect) rather than vaguely restating the acid/base equilibrium.

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Based on the official NESA syllabus and published NESA past papers and marking guidelines. Not yet individually reviewed by a person; always check NESA and your school's materials for current rules.

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