Grade 12 · Chemistry · Lesson 4

Acids & Bases — Quantitative

Calculate pH, pOH and Ka of weak acids, perform acid-base titration calculations, and interpret titration curves.

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Brønsted-Lowry Theory (Recap)

An acid is a proton (H+) donor; a base is a proton acceptor. When an acid donates a proton, it becomes its conjugate base. When a base accepts a proton, it becomes its conjugate acid.

HA + H2O ⇌ H3O+ + A−
(acid + base ⇌ conjugate acid + conjugate base)

Conjugate pairs differ by exactly one H+: HA / A− and H3O+ / H2O are the two conjugate pairs in the above equilibrium.

Self-Ionisation of Water & Kw

Water undergoes self-ionisation (autoprotolysis):

H2O + H2O ⇌ H3O+ + OH−
Kw = [H3O+][OH−] = 1.0 × 10−14 at 25°C

In pure water at 25°C: [H3O+] = [OH−] = 1.0 × 10−7 mol·dm−3. Kw increases with temperature (water self-ionisation is endothermic).

pH and pOH

pH = −log[H3O+]      [H3O+] = 10−pH
pOH = −log[OH−]      [OH−] = 10−pOH
pH + pOH = 14   (at 25°C)
Solution type[H3O+]pH
Acidic> 1.0 × 10−7< 7
Neutral= 1.0 × 10−7= 7
Basic (alkaline)< 1.0 × 10−7> 7

Strong vs Weak Acids

Strong acids (HCl, H2SO4, HNO3) dissociate completely in water. [H3O+] = concentration of acid.

0.10 mol·dm−3 HCl: [H3O+] = 0.10 → pH = −log(0.10) = 1.00

Weak acids (CH3COOH, HF, H2CO3) dissociate partially. The acid dissociation constant Ka describes the equilibrium:

HA ⇌ H+ + A−
Ka = [H+][A−] / [HA]

Worked example — weak acid pH: Calculate the pH of 0.10 mol·dm−3 ethanoic acid. Ka(CH3COOH) = 1.8 × 10−5.

Let x = [H+] = [CH3COO−]
Ka = x2 / (0.10 − x) ≈ x2 / 0.10   (x << 0.10)
x2 = 1.8 × 10−5 × 0.10 = 1.8 × 10−6
x = [H+] = 1.34 × 10−3 mol·dm−3
pH = −log(1.34 × 10−3) = 2.87

Acid-Base Indicators

An indicator is a weak acid (HIn) whose conjugate base (In−) has a different colour:

HIn ⇌ H+ + In−    (colour 1 ⇌ colour 2)
IndicatorAcid colourBase colourRange (pH)
Methyl orangeRedYellow3.1 – 4.4
PhenolphthaleinColourlessPink8.2 – 10.0
Universal indicatorRed → orangeGreen → blue/violet1 – 14

Titrations & Calculations

In a titration, a solution of known concentration (titrant in the burette) is added to a measured volume of analyte in the conical flask until the equivalence point is reached (stoichiometric amounts have reacted).

At equivalence point: n(acid) = n(base)    (for 1:1 reactions)
caVa = cbVb      (for monoprotic acid and monobasic base)

Worked example: 25.0 cm3 of NaOH solution is neutralised by 20.0 cm3 of 0.100 mol·dm−3 HCl. Find c(NaOH).

n(HCl) = c × V = 0.100 × 0.0200 = 2.00 × 10−3 mol
n(NaOH) = n(HCl) = 2.00 × 10−3 mol    (1:1 ratio)
c(NaOH) = n / V = 2.00 × 10−3 / 0.0250 = 0.0800 mol·dm−3

Titration Curves

A titration curve is a graph of pH (y-axis) against volume of titrant added (x-axis). The shape depends on the strength of the acid and base:

Titration typeEquivalence point pHBest indicatorCurve shape
Strong acid + Strong base~7.0Phenolphthalein or methyl orangeSharp vertical jump at EP
Weak acid + Strong base> 7 (basic)Phenolphthalein (8.2–10)Buffer region before EP; more gradual jump
Strong acid + Weak base< 7 (acidic)Methyl orange (3.1–4.4)Sharp jump lower on curve
Half-equivalence point: When exactly half the weak acid has been neutralised, [HA] = [A−]. Substituting into Ka = [H+][A−]/[HA] gives [H+] = Ka, so pH = pKa at the half-equivalence point. This is the centre of the buffer region.
⭐ IEB Extension — Buffer Solutions & Henderson-Hasselbalch

A buffer solution resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable concentrations.

How it works:

  • Add strong acid (H+): reacts with A− → HA. pH barely changes.
  • Add strong base (OH−): reacts with HA → A− + H2O. pH barely changes.
  • Buffer capacity is greatest when [HA] = [A−] (i.e. at the half-equivalence point, pH = pKa).

Henderson-Hasselbalch equation:

pH = pKa + log([A−] / [HA])

Example: A buffer contains 0.10 mol CH3COOH and 0.15 mol CH3COO− in 1 L. Ka = 1.8×10−5, pKa = 4.74.

pH = 4.74 + log(0.15/0.10) = 4.74 + 0.176 = 4.92

Biological buffers: blood pH is maintained at 7.35–7.45 by the CO2/HCO3− buffer system (pKa = 6.1) together with the haemoglobin buffer and kidney regulation.

Interactive Titration Curve

Titration Setup
0 mL
Current pH Readout
Volume added
0.0 mL
pH
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Stage
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Equivalence
25.0 mL
0/6
Quiz complete! Review the explanations to strengthen your understanding.
IEB Extension Questions
Show all calculations with units. For pH problems, always state your assumption (strong = complete dissociation; weak = use Ka). For titration questions, state what is in the flask and what is in the burette.
Question 1 — pH Calculations
Calculate the pH of the following solutions at 25°C:

(a) 0.050 mol·dm−3 HCl (strong acid)
(b) 0.20 mol·dm−3 NaOH (strong base) — first find pOH, then pH
(c) 0.050 mol·dm−3 methanoic acid (HCOOH); Ka = 1.8 × 10−4
(d) A solution where [H3O+] = 3.5 × 10−9 mol·dm−3. Is it acidic, neutral, or basic?

(8 marks)
Question 2 — Titration Calculation
In a titration, 25.0 cm3 of H2SO4 solution is placed in a conical flask. It is titrated against 0.200 mol·dm−3 NaOH solution. The average volume of NaOH used is 30.0 cm3.

The reaction is: H2SO4 + 2NaOH → Na2SO4 + 2H2O

(a) Calculate the moles of NaOH used.
(b) Determine the moles of H2SO4 in the flask. (Note the 1:2 ratio.)
(c) Calculate the concentration of the H2SO4 solution.
(d) Calculate the pH of the H2SO4 solution, assuming complete dissociation of both H+ ions.

(8 marks)
Question 3 — Titration Curves
(a) Sketch (or describe) the titration curve for the titration of 25 cm3 of 0.10 M CH3COOH with 0.10 M NaOH. Label: initial pH, buffer region, half-equivalence point, equivalence point pH, and the region after the equivalence point.

(b) Explain why the equivalence point pH is greater than 7 for this titration.
(c) Which indicator — methyl orange or phenolphthalein — would be more suitable for this titration? Justify your answer.
(d) If pKa(CH3COOH) = 4.74, what is the pH at the half-equivalence point?

(8 marks)
Question 4 — Indicators & Conjugate Pairs
(a) Write the conjugate base of each of the following acids: HNO3; H2PO4−; NH4+; H2O.

(b) Explain why water is amphoteric (can act as both acid and base) using Brønsted-Lowry theory.

(c) Phenolphthalein (pKa = 9.1) is added to a solution with pH = 7. What colour will it appear? Explain using the HIn equilibrium.

(6 marks)
Question 5 — Reading a Titration Data Table
A 25.0 cm3 sample of a weak monoprotic acid HA of unknown concentration is titrated with 0.100 mol·dm−3 NaOH. The pH is recorded after each addition:
V(NaOH) / cm³0.05.010.015.020.025.030.035.040.0
pH2.834.054.444.745.055.448.7411.9212.19
(a) Using the table, identify the equivalence point volume. Justify your choice by referring to how the pH changes between consecutive readings around that volume.
(b) At V = 15.0 cm3, exactly half the acid has been neutralised, so [HA] = [A−] at this point. Using the pH reading at V = 15.0 cm3 and Ka = [H+][A−]/[HA], calculate Ka for this acid.
(c) Using your equivalence point volume from (a) and c(NaOH) = 0.100 mol·dm−3, calculate n(NaOH) used, then n(HA) originally present, then c(HA) in the original 25.0 cm3 sample.
(d) Explain, using the size of the pH jump between V = 25.0 cm3 and V = 35.0 cm3 in the table, why phenolphthalein (range 8.2–10.0) is a suitable indicator for this titration, while methyl orange (range 3.1–4.4) is not.

(9 marks)