← Module 6: Acid/Base Reactions

NSWChemistrySyllabus dot point

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 OpusReviewed by Better Tuition Academy9 min answer

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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.

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 KaK_a, KbK_b) No sharp jump No single indicator is reliable; use a pH meter

Strong acid + strong base

Example: 0.100 mol/L HClHCl titrated with 0.100 mol/L NaOHNaOH.

  • Start: [H+]=0.100[H^+] = 0.100, pH=1.00pH = 1.00.
  • Pre-equivalence: pH rises slowly as excess H+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+Na^+ and Clβˆ’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 CH3COOHCH_3COOH titrated with 0.100 mol/L NaOHNaOH.

  • Start: weak acid alone, pHβ‰ˆ2.87pH \approx 2.87 (from Kac\sqrt{K_a c}).
  • After a small addition of NaOHNaOH, a buffer forms (CH3COOHCH_3COOH + CH3COOβˆ’CH_3COO^-). pH rises slowly through the buffer plateau.
  • Half-equivalence point (half the titrant volume to equivalence): [CH3COOH]=[CH3COOβˆ’][CH_3COOH] = [CH_3COO^-], so pH=pKa=4.74pH = pK_a = 4.74. This is the most reliable way to read pKapK_a off a curve.
  • Equivalence: all weak acid has been converted to its conjugate base. The conjugate base hydrolyses water, giving pHβ‰ˆ8.5pH \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 HClHCl titrated with 0.100 mol/L NH3NH_3.

  • Start: pH=1.00pH = 1.00.
  • Pre-equivalence: H+H^+ is consumed; pH rises gradually.
  • Equivalence: all H+H^+ has reacted to form NH4+NH_4^+ at about 0.050 mol/L. The conjugate acid hydrolyses, giving pHβ‰ˆ5.1pH \approx 5.1 (acidic). Calculation: Ka(NH4+)=Kw/Kb=5.56Γ—10βˆ’10K_a(NH_4^+) = K_w / K_b = 5.56 \times 10^{-10}, [H+]=Kac=5.3Γ—10βˆ’6[H^+] = \sqrt{K_a c} = 5.3 \times 10^{-6}, pH=5.28pH = 5.28.
  • Post-equivalence: buffer of NH4+NH_4^+/NH3NH_3 forms; pH rises to a plateau near pKa(NH4+)=9.26pK_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 KaK_a of the cation and KbK_b of the anion: if Ka>KbK_a > K_b, slightly acidic; if Kb>KaK_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: HInβ‡ŒH++Inβˆ’HIn \rightleftharpoons H^+ + In^-, with the two forms different colours. Its colour change (HInHIn to Inβˆ’In^-) is half-complete at pH=pKa,HInpH = pK_{a,HIn} and effectively complete over Β±1\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

Worked example

A 25.0 mL portion of 0.100 mol/L formic acid (HCOOHHCOOH, Ka=1.8Γ—10βˆ’4K_a = 1.8 \times 10^{-4}) is titrated with 0.100 mol/L NaOHNaOH. 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+]β‰ˆKac=1.8Γ—10βˆ’5=4.24Γ—10βˆ’3[H^+] \approx \sqrt{K_a c} = \sqrt{1.8 \times 10^{-5}} = 4.24 \times 10^{-3} mol/L, pH=2.37pH = 2.37.

(b) Half-equivalence. [HCOOH]=[HCOOβˆ’][HCOOH] = [HCOO^-], so pH=pKa=βˆ’log⁑10(1.8Γ—10βˆ’4)=3.74pH = pK_a = -\log_{10}(1.8 \times 10^{-4}) = 3.74.

(c) Equivalence. All acid converted to HCOOβˆ’HCOO^-, total volume 50.0 mL, so c(HCOOβˆ’)=0.0500c(HCOO^-) = 0.0500 mol/L. Kb=Kw/Ka=5.56Γ—10βˆ’11K_b = K_w/K_a = 5.56 \times 10^{-11}. [OHβˆ’]β‰ˆKbc=1.67Γ—10βˆ’6[OH^-] \approx \sqrt{K_b c} = 1.67 \times 10^{-6}, pOH=5.78pOH = 5.78, pH=8.22pH = 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.

In one sentence

A titration curve plots pH against titrant volume; the steep vertical region brackets the equivalence point (pH 7 for strong-strong, basic for weak acid plus strong base, acidic for strong acid plus weak base, no clear jump for weak-weak), and a usable indicator is one whose colour-change range falls inside that steep region, so that the end point (where the indicator changes colour) coincides with the stoichiometric equivalence point.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

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 β†’

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=pKa+log⁑10([Aβˆ’]/[HA])pH = pK_a + \log_{10}([A^-]/[HA]), passing through the half-equivalence point at pH=pKa=4.74pH = pK_a = 4.74 (where [CH3COOH]=[CH3COOβˆ’][CH_3COOH] = [CH_3COO^-]). Steepens sharply near the equivalence point at 25.0 mL added, where pHβ‰ˆ8.7pH \approx 8.7 (basic, because the conjugate base CH3COOβˆ’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 CH3COOβˆ’CH_3COO^- at concentration 0.05000.0500 mol/L (total volume now 50.0 mL). Kb=Kw/Ka=5.56Γ—10βˆ’10K_b = K_w / K_a = 5.56 \times 10^{-10}, so [OHβˆ’]β‰ˆKbc=5.27Γ—10βˆ’6[OH^-] \approx \sqrt{K_b c} = 5.27 \times 10^{-6}, pOH=5.28pOH = 5.28, pH=8.72pH = 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 β†’

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 Aβˆ’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.

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