Grade 12 Physical Sciences
Term 2 · Weeks 10–11

Acids and Bases

Paper 2 — Chemistry · Grade 12

Acids and bases are central to chemistry, biology and industry. In this chapter you will quantify acidity using the pH scale, apply the water equilibrium constant Kw to relate [H₃O⁺] and [OH⁻], distinguish strong from weak acids and bases, and perform acid-base titration calculations.

Week 10

8.1 pH, pOH and the Water Equilibrium

Define pH: pH = −log[H₃O⁺]Calculate [H₃O⁺] from pH and vice versaUse Kw = [H₃O⁺][OH⁻] = 1×10⁻¹⁴ at 25°CApply pH + pOH = 14
pH

Definition

pH

pH is a logarithmic measure of the concentration of hydronium ions (H₃O⁺) in a solution: pH = −log[H₃O⁺]. The pH scale typically ranges from 0 to 14 at 25°C: pH < 7 is acidic, pH = 7 is neutral (pure water at 25°C), pH > 7 is basic (alkaline). Each unit change in pH corresponds to a 10-fold change in [H₃O⁺]. A decrease in pH by 1 means [H₃O⁺] has increased tenfold.

Definition

Auto-protolysis of water

Auto-protolysis (self-ionisation) of water is the reaction in which two water molecules react together — one acting as an acid and one as a base: H₂O + H₂O ⇌ H₃O⁺ + OH⁻. The ion product of water is Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C. In any aqueous solution at 25°C, [H₃O⁺] × [OH⁻] = 10⁻¹⁴. In pure water [H₃O⁺] = [OH⁻] = 10⁻⁷ mol·dm⁻³, giving pH = 7.

Formula

pH and pOH

pH=−log⁡[H3O+]pOH=−log⁡[OH−]pH+pOH=14\text{pH} = -\log[\text{H}_3\text{O}^+] \qquad \text{pOH} = -\log[\text{OH}^-] \qquad \text{pH} + \text{pOH} = 14

[H₃O⁺] = concentration of hydronium ions (mol·dm⁻³), [OH⁻] = concentration of hydroxide ions (mol·dm⁻³)

SI unit: dimensionless

Formula

Water equilibrium constant

Kw=[H3O+][OH−]=1,0×10−14K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1{,}0 \times 10^{-14}

Kw = ion product of water (constant at 25°C)

SI unit: mol²·dm⁻⁶

01234567891011121314ACIDNEUTRALBASEHCl(1)Vinegar(3)Water(7)Baking(9)NaOH(14)
Figure 8.1 — The pH scale from 0 (strongly acidic) to 14 (strongly basic). At pH 7 the solution is neutral ([H₃O⁺] = [OH⁻] = 1×10⁻⁷ mol·dm⁻³ at 25 °C).

Water autoionises slightly: 2H₂O ⇌ H₃O⁺ + OH⁻. At 25 °C, Kw = [H₃O⁺][OH⁻] = 1×10⁻¹⁴. In pure water, [H₃O⁺] = [OH⁻] = 1×10⁻⁷ mol·dm⁻³, giving pH = 7 (neutral). Adding an acid increases [H₃O⁺] above 1×10⁻⁷, so pH < 7 (acidic). Adding a base decreases [H₃O⁺] (increases [OH⁻]), so pH > 7 (basic). The relationship pH + pOH = 14 follows directly from taking −log of the Kw expression.

Worked Example

Calculate the pH of a solution where [H₃O⁺] = 5,0×10⁻⁴ mol·dm⁻³.

Given

  • [H₃O⁺] = 5,0×10⁻⁴ mol·dm⁻³

Find

pH

Solution

  1. 1pH = −log[H₃O⁺]
  2. 2pH = −log(5,0×10⁻⁴)
  3. 3pH = −(log 5,0 + log 10⁻⁴)
  4. 4pH = −(0,699 − 4)
  5. 5pH = 3,30
Answer: pH = 3,30 (acidic solution)

Worked Example

A solution has pH = 9,4. Calculate [H₃O⁺] and [OH⁻].

Given

  • pH = 9,4

Find

[H₃O⁺] and [OH⁻]

Solution

  1. 1[H₃O⁺] = 10^(−pH) = 10^(−9,4) = 3,98×10⁻¹⁰ mol·dm⁻³
  2. 2pOH = 14 − 9,4 = 4,6
  3. 3[OH⁻] = 10^(−4,6) = 2,51×10⁻⁵ mol·dm⁻³
  4. 4Check: [H₃O⁺][OH⁻] = 3,98×10⁻¹⁰ × 2,51×10⁻⁵ ≈ 1,0×10⁻¹⁴ ✓
Answer: [H₃O⁺] ≈ 3,98×10⁻¹⁰ mol·dm⁻³; [OH⁻] ≈ 2,51×10⁻⁵ mol·dm⁻³
⚠

Watch Out

Common mistake: forgetting the negative sign. pH = −log[H₃O⁺] (negative log). Since [H₃O⁺] < 1 for typical solutions, log[H₃O⁺] is negative and pH is positive.

Acidic, Neutral and Basic Solutions at 25°C

PropertySolution typeConcentrations
Acidic[H₃O⁺] > 1×10⁻⁷ mol·dm⁻³pH < 7
Neutral[H₃O⁺] = [OH⁻] = 1×10⁻⁷ mol·dm⁻³pH = 7
Basic (alkaline)[OH⁻] > 1×10⁻⁷ mol·dm⁻³pH > 7
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Practice Question

Calculate the pH of a solution with [H₃O⁺] = 2,5×10⁻² mol·dm⁻³.

(3 marks)

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Practice Question

A solution has [OH⁻] = 4,0×10⁻³ mol·dm⁻³. Calculate (a) [H₃O⁺] and (b) pH.

(5 marks)

Week 10

8.2 Strong and Weak Acids and Bases

Distinguish strong from weak acids/bases

Definition

Strong acid/base

One that almost completely dissociates to form ions in solution. HCl is a strong acid.

Definition

Weak acid/base

One where only a small percentage of molecules dissociate. HF is a weak acid.

A strong acid such as HCl dissociates completely in water: HCl → H⁺ + Cl⁻ (virtually 100%). This means for 0,1 mol·dm⁻³ HCl, [H₃O⁺] = 0,1 mol·dm⁻³ and pH = 1. A weak acid like acetic acid (CH₃COOH) only partially dissociates (< 5%): CH₃COOH ⇌ CH₃COO⁻ + H⁺. The same 0,1 mol·dm⁻³ acetic acid solution has [H₃O⁺] much less than 0,1 mol·dm⁻³ and a higher pH (about 2,9). Strength refers to degree of ionisation, NOT concentration. A concentrated weak acid is still weak.

Strong vs Weak Acids and Bases

PropertyStrongWeak
Degree of dissociationNearly 100%Small percentage (<< 100%)
Equation typeSingle arrow (→): completeDouble arrow (⇌): partial equilibrium
Examples (acid)HCl, HNO₃, H₂SO₄, HBrCH₃COOH, HF, H₂CO₃, H₂SO₃
Examples (base)NaOH, KOH, Ca(OH)₂NH₃, Ca₃(PO₄)₂
[H₃O⁺] for 0,1 mol·dm⁻³= 0,1 mol·dm⁻³ (equal to initial)< 0,1 mol·dm⁻³ (less than initial)

Worked Example

Calculate the pH of 0,050 mol·dm⁻³ HNO₃ (strong acid) and compare it qualitatively to 0,050 mol·dm⁻³ acetic acid (weak acid).

Given

  • c(HNO₃) = 0,050 mol·dm⁻³
  • c(CH₃COOH) = 0,050 mol·dm⁻³

Find

pH of HNO₃; qualitative comparison

Solution

  1. 1HNO₃ is strong: [H₃O⁺] = 0,050 mol·dm⁻³
  2. 2pH = −log(0,050) = −log(5,0×10⁻²) = 1,30
  3. 3CH₃COOH is weak: only partial ionisation, [H₃O⁺] < 0,050 mol·dm⁻³
  4. 4Therefore pH of acetic acid > 1,30 (less acidic for same concentration)
Answer: HNO₃: pH = 1,30. Acetic acid at same concentration has pH > 1,30 because it is weak and partially dissociates.
⚠

Watch Out

Do NOT confuse acid strength with acid concentration. A 'strong acid' means it fully dissociates — this has nothing to do with how concentrated the solution is.

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Practice Question

Calculate the pH of 0,01 mol·dm⁻³ HCl.

(3 marks)

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Practice Question

Two solutions both have pH = 3. Solution A is HCl and Solution B is acetic acid (CH₃COOH). Which solution has a higher molar concentration? Explain.

(4 marks)

✓

Exam Tip

Exam tip: for strong acid pH calculations at Grade 12, assume complete ionisation. Only treat weak acid equilibria qualitatively unless Ka is provided.

Week 11

8.3 Acid-Base Titrations

Perform titration calculations: caVa/cbVb = na/nbExplain the choice of indicator for a titration

Definition

Equivalence point

The equivalence point is the point in a titration at which the number of moles of the added substance (titrant) is stoichiometrically equivalent to the number of moles of the substance being titrated. At the equivalence point, the mole ratio of acid to base matches the ratio in the balanced equation. For a strong acid–strong base titration, the equivalence point is at pH = 7.

V NaOH (cm³)pHEquivalencepointpH 702468101214Strong acid + strong base — equivalence at pH 7
Figure 8.1 — Titration curve: pH vs volume of NaOH added to HCl. The steep rise shows the equivalence point (pH = 7 for strong–strong). The choice of indicator depends on the pH range of the equivalence point.

Formula

Titration calculation

caVacbVb=nanb\frac{c_a V_a}{c_b V_b} = \frac{n_a}{n_b}

ca = concentration of acid (mol·dm⁻³), Va = volume of acid (dm³), cb = concentration of base (mol·dm⁻³), Vb = volume of base (dm³), na = moles of acid in balanced equation, nb = moles of base in balanced equation

SI unit: mol·dm⁻³

A titration determines the unknown concentration of an acid or base by reacting it with a standard solution of known concentration. The equivalence point is when exactly the right amount of each has been added to react completely. An indicator is chosen so that its colour change (end point) occurs as close as possible to the equivalence point. The choice depends on the nature of the acid-base pair: strong acid-strong base → pH 7 at equivalence (use bromothymol blue or phenolphthalein); strong acid-weak base → pH < 7 (use methyl orange); weak acid-strong base → pH > 7 (use phenolphthalein).

Choosing the Right Indicator

PropertyTitration typeEquivalence point pH and indicator
Strong acid + Strong basepH = 7 at equivalencePhenolphthalein or bromothymol blue
Strong acid + Weak basepH < 7 at equivalenceMethyl orange (changes colour in acid range, pH 3,1–4,4)
Weak acid + Strong basepH > 7 at equivalencePhenolphthalein (changes colour in basic range, pH 8,3–10,0)

Worked Example

25,0 cm³ of H₂SO₄ of unknown concentration requires 30,0 cm³ of 0,200 mol·dm⁻³ NaOH to reach the equivalence point. Calculate the concentration of the H₂SO₄.

Given

  • Va = 25,0 cm³ = 0,0250 dm³
  • Vb = 30,0 cm³ = 0,0300 dm³
  • cb = 0,200 mol·dm⁻³
  • H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O ∴ na : nb = 1 : 2

Find

ca (concentration of H₂SO₄)

Solution

  1. 1caVa / cbVb = na / nb
  2. 2ca × 0,0250 / (0,200 × 0,0300) = 1/2
  3. 3ca × 0,0250 = 0,200 × 0,0300 × (1/2)
  4. 4ca × 0,0250 = 0,003 000
  5. 5ca = 0,003 000 / 0,0250
  6. 6ca = 0,120 mol·dm⁻³
Answer: ca(H₂SO₄) = 0,120 mol·dm⁻³

Worked Example

What volume of 0,150 mol·dm⁻³ HCl is needed to neutralise 20,0 cm³ of 0,100 mol·dm⁻³ NaOH?

Given

  • cb = 0,100 mol·dm⁻³ NaOH
  • Vb = 20,0 cm³ = 0,0200 dm³
  • ca = 0,150 mol·dm⁻³ HCl
  • HCl + NaOH → NaCl + H₂O ∴ na : nb = 1 : 1

Find

Va

Solution

  1. 1caVa / cbVb = 1/1
  2. 2caVa = cbVb
  3. 30,150 × Va = 0,100 × 0,0200
  4. 4Va = 0,002 000 / 0,150
  5. 5Va = 0,01333 dm³ = 13,3 cm³
Answer: Va = 13,3 cm³
✓

Exam Tip

Exam tip: always convert volumes to dm³ before substituting (divide cm³ by 1 000). State the mole ratio from the balanced equation clearly — this is where most marks are lost.

⚠

Watch Out

Common mistake: using na : nb = 1 : 1 for H₂SO₄ and NaOH. The balanced equation H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O gives a 1 : 2 ratio. Always write the balanced equation first.

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Practice Question

20,0 cm³ of NaOH solution is titrated against 0,100 mol·dm⁻³ HCl. The average titre of HCl is 16,0 cm³. Calculate the concentration of the NaOH.

(5 marks)

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Practice Question

Explain why methyl orange is a suitable indicator for a titration of HCl with NH₃ (a weak base), but phenolphthalein is not.

(5 marks)

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Chemical Equilibrium

Acids and Bases Grade 12 Physical Sciences CAPS Notes | MathSciBuddy