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Inquiry Question 3: It is all about hydrogen ions

Analyse titration curves for strong-strong, strong-weak and weak-strong combinations to identify the equivalence point, distinguish it from the end point, and justify indicator selection

A focused answer to the HSC Chemistry Module 6 dot point on titration curves. The four curve shapes, equivalence point vs end point, pH at equivalence for each combination, indicator selection rules with pKa matching, and worked HSC past exam questions.

Generated by Claude Opus 4.811 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
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In plain English

Imagine you are slowly filling a bathtub (the acid) with a hose (the base) and watching a "how sour is it" meter on the wall. If the bath started full of a strong, harsh acid, the meter barely moves for ages and then suddenly leaps from "very sour" to "very not sour" the instant the bath is exactly balanced, because there is nothing left to soak up the change. If instead the bath started with a mild, weak acid, the meter creeps up gently for a while (that gentle creep is a natural buffer, where leftover acid and its partner base share the load) before it finally makes its own jump near the balance point, just not from as sour a starting spot and not necessarily landing on "neutral". The exact instant the bath is chemically balanced is the equivalence point, a fact about the plumbing. The instant your meter or dye actually visibly changes colour is the end point, a fact about the meter you chose. A good study choice is a meter whose "changes here" window sits right on the steep part of the jump, so the two moments happen at essentially the same splash of the hose.

What this dot point is asking

NESA wants you to recognise and sketch the four titration curve shapes (strong-strong, strong-weak, weak-strong, weak-weak), identify the equivalence point on each, calculate or estimate the pH at equivalence, distinguish equivalence point from end point, and choose an indicator whose colour-change range falls on the steep portion of the curve. This consolidates the chemistry from reactions of acids, strong vs weak acids and bases, and conjugate acid-base pairs, and links directly to the Module 5 dot point on titrations and indicators.

The answer

Equivalence point vs end point

The equivalence point is the point at which stoichiometrically equivalent amounts of acid and base have been combined. It is a property of the chemistry.
The end point is the point at which the indicator changes colour. It is a property of the indicator chosen.

A good indicator has an end point that coincides (within experimental tolerance) with the equivalence point. Indicator selection is the art of matching colour-change range to equivalence pH.

Equivalence pH and indicator-matching checkpoints

Strong + strong: equivalence pH $= 7.00$ (spectator ions only); any indicator with a range inside about pH 4 to 10 works.
Weak acid + strong base: equivalence pH is basic (conjugate base $A^-$ hydrolyses water); typically pH 8 to 9, so phenolphthalein (8.3 to 10.0) is standard.
Strong acid + weak base: equivalence pH is acidic (conjugate acid $BH^+$ hydrolyses water); typically pH 5 to 5.5, so methyl orange (3.1 to 4.4) or methyl red (4.4 to 6.2) is standard.
Half-equivalence point (weak titrations only): $pH = pK_a$ of the weak acid, since $[\text{HA}] = [\text{A}^-]$ there - the fastest way to read $K_a$ off a curve.
Rule of thumb: choose an indicator whose stated pH range sits ENTIRELY within the steep, near-vertical section of the curve, not just touching one edge of it.

The four curves at a glance

For a 0.100 mol/L analyte titrated with 0.100 mol/L titrant from a burette, with 25.0 mL of analyte:

Combination	pH at start	pH at equivalence	Shape of curve	Best indicator
Strong acid + strong base	~1	7.00	Very steep jump pH 4 to 10	Any in range 4 to 10 (methyl orange, bromothymol blue, phenolphthalein)
Weak acid + strong base	~3	~8.5 to 9	Gradual buffer rise, then steep jump pH 7 to 11	Phenolphthalein (8.3 to 10.0)
Strong acid + weak base	~1	~5 to 5.5	Steep jump pH 3 to 7, then gradual flattening	Methyl orange (3.1 to 4.4) or methyl red (4.4 to 6.2)
Weak acid + weak base	~3	~7 (depends on relative $K_a$ , $K_b$ )	No sharp jump	No single indicator is reliable; use a pH meter

Strong acid + strong base

Example: 0.100 mol/L $HCl$ titrated with 0.100 mol/L $NaOH$ .

Start: $[H^+] = 0.100$ , $pH = 1.00$ .
Pre-equivalence: pH rises slowly as excess $H^+$ is consumed.
Near equivalence: pH jumps almost vertically from about 4 to about 10 within one drop of titrant.
Equivalence: pH = 7.00. Only $Na^+$ and $Cl^-$ remain (both spectators).
Post-equivalence: pH rises gradually, approaching the pH of the excess strong base.

The very wide vertical section (about 6 pH units) tolerates almost any common indicator with a range between 4 and 10.

Weak acid + strong base

Example: 0.100 mol/L $CH_3COOH$ titrated with 0.100 mol/L $NaOH$ .

Start: weak acid alone, $pH \approx 2.87$ (from $\sqrt{K_a c}$ ).
After a small addition of $NaOH$ , a buffer forms ( $CH_3COOH$ + $CH_3COO^-$ ). pH rises slowly through the buffer plateau.
Half-equivalence point (half the titrant volume to equivalence): $[CH_3COOH] = [CH_3COO^-]$ , so $pH = pK_a = 4.74$ . This is the most reliable way to read $pK_a$ off a curve.
Equivalence: all weak acid has been converted to its conjugate base. The conjugate base hydrolyses water, giving $pH \approx 8.5$ to 9 (basic).
Post-equivalence: pH approaches that of the excess strong base.

The steep section here spans roughly pH 7 to pH 11, so phenolphthalein (8.3 to 10.0) is the standard choice. Methyl orange would change too early.

Strong acid + weak base

Example: 0.100 mol/L $HCl$ titrated with 0.100 mol/L $NH_3$ .

Start: $pH = 1.00$ .
Pre-equivalence: $H^+$ is consumed; pH rises gradually.
Equivalence: all $H^+$ has reacted to form $NH_4^+$ at about 0.050 mol/L. The conjugate acid hydrolyses, giving $pH \approx 5.1$ (acidic). Calculation: $K_a(NH_4^+) = K_w / K_b = 5.56 \times 10^{-10}$ , $[H^+] = \sqrt{K_a c} = 5.3 \times 10^{-6}$ , $pH = 5.28$ .
Post-equivalence: buffer of $NH_4^+$ / $NH_3$ forms; pH rises to a plateau near $pK_a(NH_4^+) = 9.26$ at the half-past-equivalence point if you continue adding base. (Most HSC questions stop at equivalence.)

The steep section spans roughly pH 3 to pH 7. Methyl orange (3.1 to 4.4) or methyl red (4.4 to 6.2) work. Phenolphthalein would change too late.

Weak acid + weak base

The steep section is short or absent because both sides resist pH change. The equivalence pH depends on the relative $K_a$ of the cation and $K_b$ of the anion: if $K_a > K_b$ , slightly acidic; if $K_b > K_a$ , slightly basic; if equal, exactly 7.

For accuracy, do not use indicators here. Use a pH probe and read the inflection point off the curve.

Why indicator choice matters

An indicator is itself a weak acid: $HIn \rightleftharpoons H^+ + In^-$ , with the two forms different colours. Its colour change ( $HIn$ to $In^-$ ) is half-complete at $pH = pK_{a,HIn}$ and effectively complete over $\pm 1$ unit. For a sharp end point, this range must lie within the steep vertical portion of the titration curve. Outside that portion, the colour change is gradual and the end point is imprecise.

Indicator	pH range	Acid colour	Base colour
Methyl orange	3.1 to 4.4	red	yellow
Methyl red	4.4 to 6.2	red	yellow
Bromothymol blue	6.0 to 7.6	yellow	blue
Phenolphthalein	8.3 to 10.0	colourless	pink

An owned illustrative curve for the weak acid + strong base case shows exactly where these landmarks sit:

Worked example

A 25.0 mL portion of 0.100 mol/L formic acid ( $HCOOH$ , $K_a = 1.8 \times 10^{-4}$ ) is titrated with 0.100 mol/L $NaOH$ . Calculate the pH (a) at the start, (b) at the half-equivalence point, (c) at equivalence, and (d) recommend an indicator.

(a) Start: Weak acid alone. $[H^+] \approx \sqrt{K_a c} = \sqrt{(1.8 \times 10^{-4})(0.100)} = 4.24 \times 10^{-3}$ mol/L, $pH = 2.37$ .
(b) Half-equivalence: $[HCOOH] = [HCOO^-]$ , so $pH = pK_a = -\log_{10}(1.8 \times 10^{-4}) = 3.74$ .
(c) Equivalence: All acid converted to $HCOO^-$ , total volume 50.0 mL, so $c(HCOO^-) = 0.0500$ mol/L. $K_b = K_w/K_a = 5.56 \times 10^{-11}$ . $[OH^-] \approx \sqrt{K_b c} = 1.67 \times 10^{-6}$ , $pOH = 5.78$ , $pH = 8.22$ .
(d) Indicator: Equivalence pH 8.22 lies just inside the phenolphthalein range (8.3 to 10.0). Phenolphthalein is acceptable; thymol blue (8.0 to 9.6) would also work and might give a slightly sharper transition.

Common traps

Equating equivalence point with pH 7: True only for strong-strong. Weak acid plus strong base equivalence is basic; weak base plus strong acid equivalence is acidic.
Mixing up equivalence point and end point: They should coincide, but they are conceptually distinct. Equivalence is set by stoichiometry; end point is set by the indicator.
Using phenolphthalein on a strong acid plus weak base titration: The equivalence pH is around 5; phenolphthalein would change at pH 8 to 10, far past equivalence. Use methyl orange or methyl red.
Forgetting the dilution at equivalence: When 25.0 mL of acid is neutralised by 25.0 mL of base, the total volume is 50.0 mL. The conjugate ion concentration is half the original acid concentration.
Calling the buffer plateau "flat": It is gradually rising, not flat. The buffer resists pH change but does not freeze it.
Trying to titrate a weak acid against a weak base: No sharp jump means no reliable indicator end point. Use a pH probe.

Exam technique for titration-curve questions

When asked to SKETCH a curve, always mark and label FOUR features even if the question only asks for one: the starting pH, the buffer region (if a weak species is present), the equivalence point (with its numerical pH if you can calculate it), and the shape after equivalence. When asked to JUSTIFY an indicator choice, never just name the indicator - state the indicator's range, state the equivalence pH (calculated or estimated), and explicitly say the range lies inside the steep section while a rejected indicator's range does not. For a calculation embedded in a titration question, always show the relevant $K_a$ or $K_b$ expression, substitute values with units, and round the final pH to 2 decimal places (matching the precision of $-\log_{10}$ of a value with 2 to 3 significant figures).

Examples in context

Example 1. Standardising oxalic acid for HSC Chemistry exams. Most NSW schools use oxalic acid dihydrate as the primary standard for standardising NaOH stock. Students titrate 0.0500 mol L $^{-1}$ oxalic acid against the unknown NaOH using phenolphthalein. Oxalic acid is diprotic ( $K_{a1} = 5.6 \times 10^{-2}$ , $K_{a2} = 1.5 \times 10^{-4}$ ), producing a curve with only one usable jump because the two $pK_a$ values are within four units. The equivalence pH is 8.4 (conjugate-base hydrolysis), so phenolphthalein straddles it cleanly. The exercise illustrates why diprotic weak acid curves often show one apparent equivalence in HSC practical work.

Example 2. Sydney Water dosing of caustic into the trunk main from Prospect. Caustic dosing of finished drinking water raises pH from 7.4 to 7.8 to inhibit copper-pipe corrosion. The dosing pump is calibrated against an inline titration that emulates a strong base / strong acid curve. Operators record pH every two minutes against $0.0500 \text{ mol L}^{-1}$ HCl titre and look for the steep transition through pH 7 to flag end of equivalence. Any drift in the curve shape, particularly if the steep region migrates above pH 8, indicates the alkalinity of the source water has changed and the dose must be recalculated to prevent under-treatment.

Try this

Q1. Sketch and label a titration curve for a strong acid added to a strong base, noting the equivalence pH and one suitable indicator. [3 marks]

Cue. Curve starts high pH, drops sharply through pH 7 at equivalence, levels at low pH; bromothymol blue (6.0 to 7.6) is suitable.

Q2. A 25.0 mL sample of 0.0500 mol L $^{-1}$ acetic acid is titrated with 0.0500 mol L $^{-1}$ NaOH. Calculate the pH at (a) the start, (b) the half-equivalence point, (c) the equivalence point. [3 marks]

Cue. (a) Weak acid pH = $\frac{1}{2}(pK_a - \log c)$ = 3.04. (b) pH = $pK_a$ = 4.74. (c) Salt $CH_3COO^-$ hydrolyses, pH approximately 8.7.

Q3. A weak base is titrated with HCl. (a) State whether the equivalence pH is above, equal to or below 7 and justify. (b) Recommend a suitable indicator from methyl orange (3.1 to 4.4), methyl red (4.2 to 6.3) or phenolphthalein (8.3 to 10.0). (c) Sketch the curve, marking the buffer region. [2+1+2 marks]

Cue. (a) Below 7: conjugate acid of weak base hydrolyses. (b) Methyl orange or methyl red. (c) High pH start, gentle drop through buffer plateau at $pH = 14 - pK_b$ , sharp drop through equivalence in acidic region.

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.

2023 HSC5 marksSketch the titration curve obtained when 0.100 mol/L NaOH is added from a burette to 25.0 mL of 0.100 mol/L CH3COOH in a conical flask. On the curve, label the equivalence point, the buffer region, the half-equivalence point, and the pH at equivalence. Justify the choice of phenolphthalein (range 8.3 to 10.0) rather than methyl orange (range 3.1 to 4.4) as the indicator.

Show worked answer →

A 5 mark answer needs the shape, four labels, the pH at equivalence (around 8.7), and explicit indicator reasoning.

Shape: Starts at pH around 2.87 (weak acid alone). Rises gradually through a buffer region where $pH = pK_a + \log_{10}([A^-]/[HA])$ , passing through the half-equivalence point at $pH = pK_a = 4.74$ (where $[CH_3COOH] = [CH_3COO^-]$ ). Steepens sharply near the equivalence point at 25.0 mL added, where $pH \approx 8.7$ (basic, because the conjugate base $CH_3COO^-$ hydrolyses). Past equivalence the curve approaches the pH of the excess strong base.
Equivalence pH calculation: At equivalence, all acid has been converted to $CH_3COO^-$ at concentration $0.0500$ mol/L (total volume now 50.0 mL). $K_b = K_w / K_a = 5.56 \times 10^{-10}$ , so $[OH^-] \approx \sqrt{K_b c} = 5.27 \times 10^{-6}$ , $pOH = 5.28$ , $pH = 8.72$ .
Indicator choice: Phenolphthalein changes colour over 8.3 to 10.0, which lies within the steep vertical portion of the curve and brackets the equivalence pH (8.72). Methyl orange changes over 3.1 to 4.4, well below the equivalence pH, so it would change colour long before equivalence and give a large titration error.

Markers reward (1) correct curve shape with all four labels, (2) the buffer/half-equivalence insight, (3) numerical pH at equivalence, (4) explicit comparison of indicator range to equivalence pH.

2020 HSC3 marksExplain why bromothymol blue (range 6.0 to 7.6) is a suitable indicator for a strong acid - strong base titration but not for a weak acid - strong base titration.

Show worked answer →

In a strong acid - strong base titration, the equivalence point is at pH 7. The pH curve has a very steep vertical section spanning roughly pH 4 to pH 10, comfortably containing the bromothymol blue range (6.0 to 7.6). Any indicator changing colour in that range gives a sharp, accurate end point.

In a weak acid - strong base titration, the equivalence point is basic (around pH 8.5 to 9 for typical weak acids like ethanoic acid, because the conjugate base $A^-$ hydrolyses). The steep portion of the curve is shifted upward and spans roughly pH 7 to pH 11. Bromothymol blue would change colour at pH 6 to 7.6, which is before the equivalence point, on the flat buffer plateau. The end point would arrive too early, giving an underestimate of the acid concentration.

Markers reward (1) noting the equivalence pH for each combination, (2) describing where the steep section sits, (3) explicitly linking the indicator range to the equivalence pH.

Practice questions

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

foundation2 marksState whether the equivalence point of each titration is acidic, neutral or basic, and give one reason: (a)

\text{HCl}

titrated with

\text{NaOH}

; (b)

\text{CH}_3\text{COOH}

titrated with

\text{NaOH}

Show worked solution →

(a) HCl + NaOH. Neutral, pH 7.00. Both product ions ( $\text{Na}^+$ and $\text{Cl}^-$ ) are spectators that do not hydrolyse water.

(b) CH3COOH + NaOH. Basic, pH above 7. The conjugate base $\text{CH}_3\text{COO}^-$ hydrolyses water ( $\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$ ), producing excess $\text{OH}^-$ .

Marking criteria: 1 mark per part for the correct acidic/neutral/basic classification with a reason that names the relevant species or hydrolysis behaviour (a bare "neutral"/"basic" with no reason earns no mark for that part).

foundation4 marksIn a titration, 20.0 mL of 0.0500 mol/L HCl in a conical flask requires 18.40 mL of NaOH solution from a burette to reach the equivalence point (bromothymol blue end point). Calculate the concentration of the NaOH solution, to 3 significant figures.

Show worked solution →

Step 1: moles of HCl (the known reagent).

n(\text{HCl}) = c \times V = 0.0500\ \text{mol L}^{-1} \times 0.0200\ \text{L} = 1.00 \times 10^{-3}\ \text{mol}

Step 2: mole ratio from the equation.

\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}

1:1 ratio, so $n(\text{NaOH}) = n(\text{HCl}) = 1.00 \times 10^{-3}\ \text{mol}$ at equivalence.

Step 3: concentration of NaOH.

c(\text{NaOH}) = \frac{n}{V} = \frac{1.00 \times 10^{-3}\ \text{mol}}{18.40 \times 10^{-3}\ \text{L}} = 0.054347...\ \text{mol L}^{-1}

Step 4: round to 3 significant figures (matching the 3 s.f. in 0.0500 and 18.40).

c(\text{NaOH}) = 0.0543\ \text{mol L}^{-1}

Marking criteria: 1 mark for correct moles of HCl, 1 mark for the correct 1:1 mole ratio, 1 mark for the concentration calculation, 1 mark for the correct answer to 3 significant figures with units.

core5 marksA 25.0 mL sample of 0.150 mol/L formic acid (

\text{HCOOH}

K_a = 1.8 \times 10^{-4}

) is titrated with 0.150 mol/L NaOH. Calculate the pH (a) at the start, before any NaOH is added, and (b) at the equivalence point, to 2 decimal places.

Show worked solution →

(a) pH at the start (weak acid alone).

[\text{H}^+] \approx \sqrt{K_a c} = \sqrt{(1.8 \times 10^{-4})(0.150)} = \sqrt{2.7 \times 10^{-5}} = 5.196 \times 10^{-3}\ \text{mol L}^{-1}

pH = -\log_{10}(5.196 \times 10^{-3}) = 2.28

(b) pH at equivalence.

Equal concentrations and a 1:1 ratio mean equivalence occurs at $V(\text{NaOH}) = 25.0$ mL, so total volume is 50.0 mL and the salt concentration is halved.

c(\text{HCOO}^-) = \frac{0.150 \times 25.0}{50.0} = 0.0750\ \text{mol L}^{-1}

K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} = 5.56 \times 10^{-11}

[\text{OH}^-] \approx \sqrt{K_b c} = \sqrt{(5.56 \times 10^{-11})(0.0750)} = 2.04 \times 10^{-6}\ \text{mol L}^{-1}

pOH = -\log_{10}(2.04 \times 10^{-6}) = 5.69, \qquad pH = 14.00 - 5.69 = 8.31

Marking criteria: (a) 1 mark for the correct $\sqrt{K_ac}$ setup, 1 mark for the correct pH. (b) 1 mark for correctly halving the concentration at the dilution point, 1 mark for the correct $K_b$ and $[\text{OH}^-]$ , 1 mark for the correct final pH to 2 decimal places.

core5 marksThe owned illustrative curve below shows 25.0 mL of 0.100 mol/L

\text{CH}_3\text{COOH}

titrated with 0.100 mol/L NaOH, with four labelled points: Start (pH 2.87, V = 0), Half-equivalence (pH = pKa = 4.74, V = 12.5 mL), Equivalence (pH 8.72, V = 25.0 mL), and the flattening tail at high volume. (a) Explain why the curve rises only gradually between the Start and Half-equivalence points. (b) State the concentration of

\text{CH}_3\text{COOH}

remaining at the Half-equivalence point relative to

\text{CH}_3\text{COO}^-

, and justify using the label given. (c) Explain why the curve becomes almost flat well past the Equivalence point.

Show worked solution →

(a) Gradual rise (buffer action): Between Start and Half-equivalence, the flask contains a mixture of unreacted $\text{CH}_3\text{COOH}$ and its conjugate base $\text{CH}_3\text{COO}^-$ formed by the NaOH added so far. Added $\text{OH}^-$ is consumed by the weak acid ( $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$ ), and the buffer pair resists large pH swings, so the pH creeps up slowly rather than jumping.
(b) Concentrations at Half-equivalence: The label states $pH = pK_a = 4.74$ at this point. Since the Henderson-Hasselbalch relationship gives $pH = pK_a + \log_{10}([\text{CH}_3\text{COO}^-]/[\text{CH}_3\text{COOH}])$ , a pH exactly equal to $pK_a$ means the logarithmic term is zero, i.e. $[\text{CH}_3\text{COO}^-] = [\text{CH}_3\text{COOH}]$ - the acid is exactly half-neutralised, which also matches the volume label (12.5 mL is half of the 25.0 mL equivalence volume).
(c) Flat tail after equivalence: Once all the $\text{CH}_3\text{COOH}$ has reacted, further NaOH simply accumulates as excess strong base in a large, near-constant total volume. Because $\text{OH}^-$ from a strong base is not buffered by any remaining weak acid, its concentration (and hence pH) changes only slowly per additional mL once it is already large, so the curve flattens toward the pH set by the excess strong base.

Marking criteria: (a) 1 mark for identifying the buffer/conjugate pair, 1 mark for linking it to resisting pH change. (b) 1 mark for stating the concentrations are equal, 1 mark for justifying via the pH = pKa label. (c) 1 mark for correctly explaining the flattening as excess strong base with no more buffering weak acid present.

exam7 marksA school laboratory has three indicators available: methyl orange (3.1 to 4.4), bromothymol blue (6.0 to 7.6) and phenolphthalein (8.3 to 10.0). Evaluate which indicator should be selected for EACH of (i) a strong acid - strong base titration, (ii) a weak acid (

K_a = 1.8 \times 10^{-5}

) - strong base titration, and (iii) a strong acid - weak base (

K_b = 1.8 \times 10^{-5}

) titration, justifying each choice with a calculated or estimated equivalence pH, and explain what would go wrong if phenolphthalein were used for titration (iii).

Show worked solution →

This is a 7-mark EVALUATE: markers reward a justified choice with numbers for all three cases plus a specific failure analysis, not just three indicator names.

Band 6 PLAN.

(i) Strong-strong: equivalence pH = 7.00 exactly (spectator ions). All three indicators changing near 6 to 8 would work reasonably, but bromothymol blue (6.0 to 7.6) is the tightest match, centred almost exactly on pH 7.
(ii) Weak acid + strong base: estimate equivalence pH. Using $K_b = K_w/K_a = 5.56 \times 10^{-10}$ and a typical post-dilution salt concentration around 0.05 mol/L, $[\text{OH}^-] \approx \sqrt{(5.56\times10^{-10})(0.05)} = 5.27\times10^{-6}$ , $pOH = 5.28$ , $pH = 8.72$ . Phenolphthalein (8.3 to 10.0) brackets this.
(iii) Strong acid + weak base: using $K_a(\text{conjugate acid}) = K_w/K_b = 5.56\times10^{-10}$ and the same 0.05 mol/L estimate, $[\text{H}^+] \approx \sqrt{(5.56\times10^{-10})(0.05)} = 5.27\times10^{-6}$ , $pH = 5.28$ . Methyl orange (3.1 to 4.4) is closest, though its upper edge (4.4) sits a little below the equivalence pH; bromothymol blue is worse (starts at 6.0, well past equivalence) and phenolphthalein is unusable.
Failure analysis for phenolphthalein in (iii): its range (8.3 to 10.0) lies far ABOVE the estimated equivalence pH of 5.28, on the flat, heavily-buffered tail of the curve where large volumes of titrant change the pH only slightly. The indicator would not change colour until well past equivalence, giving a large positive titration error (an overestimate of titrant volume, and hence an incorrect calculated concentration).
Judgement: no single indicator serves all three titrations; the correct indicator must be selected case by case by matching its stated range to the calculated equivalence pH, with bromothymol blue for neutral equivalence, phenolphthalein for basic equivalence and methyl orange for acidic equivalence.

Model paragraph (excerpt). Indicator selection cannot be generalised across titration types because the equivalence pH itself moves depending on which conjugate species, if any, hydrolyses water. For the strong acid - strong base case the equivalence point sits at exactly pH 7.00, so bromothymol blue (6.0 to 7.6) gives the tightest bracket of the three available indicators. Both remaining titrations involve a weak species: the weak acid case pushes the equivalence pH up to approximately 8.7 because the conjugate base $\text{CH}_3\text{COO}^-$ -type ion hydrolyses to release $\text{OH}^-$ , which is why phenolphthalein, not bromothymol blue, is required there. The weak base case does the opposite, pulling the equivalence pH down to approximately 5.3 through hydrolysis of the conjugate acid, which is why methyl orange is the best available match even though its range does not perfectly centre on the equivalence pH. Using phenolphthalein for this last titration would fail completely: its colour change would not begin until several millilitres of titrant past the true equivalence point, on the flat excess-acid tail of the curve, systematically overstating the titre and any concentration calculated from it.

Marker's note: top-band answers (1) calculate or clearly estimate a numerical equivalence pH for all three cases using the correct hydrolysis species, (2) select a specific indicator for each case with the range compared explicitly to that pH, (3) give a concrete, mechanistic failure description for the phenolphthalein misuse (not just "it would be wrong"), and (4) close with an explicit judgement that indicator choice is case-by-case rather than universal.

exam6 marksA council water lab titrates 25.0 mL of 0.100 mol/L HCl with 0.100 mol/L ammonia solution (

\text{NH}_3

K_b = 1.8 \times 10^{-5}

) to model how residual chlorine dosing interacts with ammonia in drinking water. (a) Calculate the pH at the equivalence point, to 2 decimal places. (b) Assess whether methyl orange (3.1 to 4.4) or phenolphthalein (8.3 to 10.0) is the more appropriate indicator for this titration, and justify your assessment using the value found in (a).

Show worked solution →

(a) pH at equivalence.

At equivalence, all $\text{HCl}$ has reacted with $\text{NH}_3$ to form $\text{NH}_4^+$ . Equal concentrations and a 1:1 ratio mean equivalence occurs at 25.0 mL of titrant, so total volume is 50.0 mL.

c(\text{NH}_4^+) = \frac{0.100 \times 25.0}{50.0} = 0.0500\ \text{mol L}^{-1}

K_a(\text{NH}_4^+) = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}

[\text{H}^+] \approx \sqrt{K_a c} = \sqrt{(5.56 \times 10^{-10})(0.0500)} = 5.27 \times 10^{-6}\ \text{mol L}^{-1}

pH = -\log_{10}(5.27 \times 10^{-6}) = 5.28

(b) Assessment of indicator. The calculated equivalence pH (5.28) lies just above the methyl orange range (3.1 to 4.4) but far below the phenolphthalein range (8.3 to 10.0). Neither indicator perfectly centres on pH 5.28, but methyl orange is the far more defensible choice: its upper limit (4.4) is only about 0.9 pH units from the true equivalence point and lies on the steep part of the jump for this strong acid - weak base combination, whereas phenolphthalein would not change colour until roughly 3 to 5 pH units past equivalence, on the flat buffered tail, causing a severe overestimate of the titre. A more precise alternative not offered here would be methyl red (4.4 to 6.2), which straddles 5.28 almost exactly.

Marking criteria: (a) 1 mark for correctly halving the concentration at the dilution point, 1 mark for the correct $K_a(\text{NH}_4^+)$ , 1 mark for the correct $[\text{H}^+]$ , 1 mark for the correct final pH to 2 decimal places. (b) 1 mark for correctly assessing methyl orange as preferable with a reasoned distance-from-equivalence argument, 1 mark for explicitly explaining why phenolphthalein would fail (colour change far past equivalence on the buffered tail).