VKC ScienceChemistry for written exams Telegram
VKC Notes · Analytical chemistry · Geo-Scientist P-I · Geo-Scientist P-II

17.1 Acid–Base Titrations: Curves, Indicators, Buffers

★★★ 2021–2025: 20 parts, 195 marks (Geo-Scientist Paper I 18 parts / 175 m · Paper II 2 / 20 m). A full 15-mark titration-curve numerical has come every year since 2022.

Syllabus line. Geo-Scientist Paper I (acid–base titrations): “Titration curves for strong acid-strong base, weak acid-strong base and weak base-strong acid titrations, polyprotic acids, poly-equivalent bases, determining the equivalence point: theory of acid-base indicators, pH change range of indicator, selection of proper indicator. Principles used in estimation of mixtures of NaHCO₃ and Na₂CO₃ (by acidimetry).” Buffer numericals also appear in Paper II under chemical equilibrium. Redox titrations are in 17.2.

Two-page short notes for revisionAll VKC notes
Download PDF PLAN
PDF edition, print-ready. It opens with the VKC plan, together with every locked answer, note and test on the site.

CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

17.1.1 Salt hydrolysis and buffers ★★★

  • Salt hydrolysis: the ion of a weak acid or weak base reacts with water and gives back the weak acid or base, so the solution is not neutral. CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ (basic); NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ (acidic).SN
(17.1.1)
(17.1.2)
  • Weak base salt (BH⁺Cl⁻): pH = ½(pKw − pKb − log C). Use the Kw the question gives: with Kw = 1.1 × 10⁻¹⁴, pKw = 13.96, not 14.SN
  • Buffer: a weak acid with its salt (or weak base with its salt). Added H⁺ is taken up by A⁻, added OH⁻ by HA, so only the ratio [A⁻]/[HA] changes. Best near pH = pKa ± 1; buffer capacity is greatest at [A⁻] = [HA].SN

Solved PYQs Geo 2025 Q3(b) 15m Geo 2022 Q12(a) 10m Geo 2024 Q6 10m

  1. 0.01 M CH₃COONa, Ka = 1.8 × 10⁻⁵, Kw = 1.1 × 10⁻¹⁴: Kh = 1.1 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 6.11 × 10⁻¹⁰; h = √(6.11 × 10⁻¹⁰ / 0.01) = 2.47 × 10⁻⁴ = 0.0247 %; [OH⁻] = Ch = 2.47 × 10⁻⁶ M, pOH = 5.61, pH = 13.96 − 5.61 = 8.35.
  2. Buffer 0.20 M CH₃COOH + 0.25 M CH₃COOK (1 L), Ka = 1.75 × 10⁻⁵; add 0.5 mL of 1 M HCl = 5.0 × 10⁻⁴ mol H⁺. pH₁ = 4.757 + log(0.25/0.20) = 4.854; acetate → 0.2495, acid → 0.2005; pH₂ = 4.757 + log(0.2495/0.2005) = 4.852. ΔpH = −0.002 (the same acid in 1 L of water: ΔpH = −3.70).SN
  3. 0.5 mol m⁻³ each of CH₃COOH and CH₃COONa + 0.02 mol m⁻³ NaOH: acid → 0.48, salt → 0.52; pH = 4.745 + log(0.52/0.48) = 4.78, a rise of only 0.035.

17.1.2 Strong acid with strong base ★★

methyl orange 3.1–4.4phenolphthalein 8.3–10.0025507510002468101214volume of 0.100 M NaOH added / mLpHpKₐ = 4.74 (½-eq.)equivalence pH 8.72pH 7.00buffer regionHClCH3COOH
Fig. 17.1.1 Computed curves: 50 mL of 0.100 M HCl (black) and of 0.100 M CH₃COOH (red) with 0.100 M NaOH.
Draw it in 60 s: pH 0–14 vs mL; black S jumping at 50 mL through pH 7; red curve from 2.9, flat near 4.7, jump through 8.7; shade 3.1–4.4, 8.3–10.
  1. Before equivalence: [H⁺] = (moles HCl left)/(total volume). After: [OH⁻] = (excess NaOH)/(total volume). At equivalence: pH 7 (NaCl does not hydrolyse).SN
  2. The jump at 0.1 M runs from about pH 4 to 10 within ±0.1 mL, so methyl orange, methyl red, bromothymol blue and phenolphthalein all work; at 0.001 M the jump shrinks to pH 6–8 and only bromothymol blue fits.SN

Solved PYQ Geo 2022 Q5(a) 10m

50 mL of 0.1 M HCl titrated with 0.1 M NaOH: pH at 0, 10, 50 and 60 mL.

  1. 0 mL: [H⁺] = 0.100 M, pH 1.00.
  2. 10 mL: HCl left = 5.0 − 1.0 = 4.0 mmol in 60 mL; [H⁺] = 0.0667 M, pH 1.18.
  3. 50 mL: equivalence, NaCl only, pH 7.00.
  4. 60 mL: excess NaOH 1.0 mmol in 110 mL; [OH⁻] = 9.09 × 10⁻³ M, pOH 2.04, pH 11.96.

17.1.3 Weak acid with strong base ★★★

RegionSpecies presentpH from
(1) startHA only[H⁺] = √(KaC)
(2) before equivalenceHA + A⁻: a bufferHenderson–Hasselbalch; at ½-equivalence pH = pKa
(3) equivalenceA⁻ onlyhydrolysis, eq. (17.1.1): pH > 7
(4) afterA⁻ + excess OH⁻excess OH⁻ decides; A⁻ adds almost nothing
  • Reading a curve: the pH at half the equivalence volume gives pKa, which identifies the acid. A curve with ½-equivalence pH near 4.7–4.8 is acetic acid (Ka 1.69 × 10⁻⁵, pKa 4.77); benzoic acid would sit at 4.20, the pyridinium ion at 5.23, HCN at 9.30 Geo 2024 Q3(a) 15m.SN
  • The weaker the acid, the higher the start, the shorter the jump; below Ka ≈ 10⁻⁸ (HCN, pKa 9.3) there is no usable jump Geo 2023 Q5(b).

Solved PYQ Geo 2023 Q5(b) 15m

100 mL of 0.1 M CH₃COOH (Ka = 1.82 × 10⁻⁵) with 0.1 M NaOH, Kw = 1.1 × 10⁻¹⁴: pH at 0, 50, 100 and 100.1 mL; indicator.

  1. 0 mL: [H⁺] = √(1.82 × 10⁻⁶) = 1.35 × 10⁻³ M, pH 2.87.
  2. 50 mL (half-equivalence): [A⁻] = [HA], pH = pKa = 4.74.
  3. 100 mL: 0.050 M CH₃COONa; [OH⁻] = √(KwC/Ka) = 5.50 × 10⁻⁶ M; pH = 13.96 − 5.26 = 8.70.
  4. 100.1 mL: excess OH⁻ = 0.01 mmol in 200.1 mL = 5.0 × 10⁻⁵ M; pOH 4.30, pH 9.66. Indicator: phenolphthalein (8.3–10.0).SN

17.1.4 Weak base with strong acid; a diprotic base ★★

  • NH₃ + HCl: NH₃ + H⁺ → NH₄⁺. The curve falls; start [OH⁻] = √(KbC); ½-equivalence pH = pKa(NH₄⁺) = 9.25; equivalence acidic (NH₄⁺ hydrolyses, pH ≈ 5.3 at 0.05 M); use methyl orange or methyl red Geo 2022 Q5(c) 10m.SN
  • Weak diprotic base B with strong acid: two buffer regions (B/BH⁺, then BH⁺/BH₂²⁺) and two equivalence points. At 1 equivalent, BH⁺ is amphiprotic: pH ≈ ½(pKa1 + pKa2); at 0.5 and 1.5 equivalents pH ≈ pKa2 and pKa1; at 2 equivalents [H⁺] = √(Ka1C). Na₂CO₃ (17.1.6) is the classic case Geo 2021 Q6(c) 10m.SN

Part of VKC Basic

Read all 11 sections of this topic, its short notes and the PDF with VKC Basic

VKC Basic (₹999 for 6 months) opens every solved answer, the notes, the handbook and the daily answer archive. Every Mains test series includes it.

See what is included Already a member? Sign in

Solved PYQs, short notes and books: vkcscience.comTelegram: t.me/upsc_chemistry_vkc