Acids, Bases and Buffers

This lesson covers Brønsted-Lowry acid-base theory, pH calculations for strong acids, strong bases, weak acids, and buffer solutions, the ionic product of water (Kw), polyprotic acids, titration curves for all four acid-base combinations, buffer calculations, and the Henderson-Hasselbalch equation. A thorough understanding of acid-base chemistry is essential for A-Level Chemistry and has applications in biology, medicine, and industry. This material aligns with the AQA and OCR A specifications for A-Level Chemistry.

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Brønsted-Lowry Theory

Key Definition: A Brønsted-Lowry acid is a proton (H⁺) donor. A Brønsted-Lowry base is a proton (H⁺) acceptor.

In every acid-base reaction, there is a transfer of a proton from the acid to the base. The species formed when an acid loses a proton is called its conjugate base. The species formed when a base gains a proton is called its conjugate acid. Every acid-base reaction involves two conjugate pairs.

For example: CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺

Species	Role	Conjugate partner
CH₃COOH	Acid (donates H⁺)	CH₃COO⁻ (conjugate base)
H₂O	Base (accepts H⁺)	H₃O⁺ (conjugate acid)

Water is amphoteric — it can act as both an acid and a base. In the above example it acts as a base; in the reaction of NH₃ with water (NH₃ + H₂O ⇌ NH₄⁺ + OH⁻), water acts as an acid.

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Strong and Weak Acids

Key Definition: A strong acid is one that is completely (fully) dissociated in aqueous solution. A weak acid is one that is partially dissociated in aqueous solution, establishing a dynamic equilibrium between the undissociated acid and its ions.

• Strong acids (e.g. HCl, H₂SO₄, HNO₃) are fully dissociated: HCl → H⁺ + Cl⁻. The concentration of H⁺ ions equals the initial concentration of the acid (for monoprotic strong acids).
• Weak acids (e.g. CH₃COOH, HF, H₂CO₃) are partially dissociated: CH₃COOH ⇌ CH₃COO⁻ + H⁺. Only a small fraction of molecules dissociate; the equilibrium lies far to the left.

The extent of dissociation of a weak acid is quantified by the acid dissociation constant, Ka:

Ka = [H⁺][A⁻] / [HA]

Units of Ka are mol dm⁻³. A larger Ka means a stronger weak acid (greater degree of dissociation). The pKa is defined as:

pKa = −log₁₀(Ka) and Ka = 10⁻ᵖᴷᵃ

A smaller pKa indicates a stronger acid. For example, ethanoic acid has Ka = 1.74 × 10⁻⁵ mol dm⁻³ and pKa = 4.76.

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The Ionic Product of Water, Kw

Water undergoes a small degree of autoionisation (self-ionisation):

H₂O ⇌ H⁺ + OH⁻

Key Definition: The ionic product of water, Kw, is defined as: Kw = [H⁺][OH⁻]. At 298 K (25 °C), Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

In pure water at 298 K: [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³, giving pH = 7.00.

Kw Variation with Temperature

The autoionisation of water is endothermic (ΔH > 0). By Le Chatelier's principle, increasing the temperature shifts the equilibrium to the right, increasing [H⁺] and [OH⁻] and therefore increasing Kw.

Temperature / K	Kw / mol² dm⁻⁶	pH of pure water
273	1.14 × 10⁻¹⁵	7.47
298	1.00 × 10⁻¹⁴	7.00
310 (body temp.)	2.42 × 10⁻¹⁴	6.81
373	5.13 × 10⁻¹³	6.14

Exam Tip: At temperatures above 298 K, the pH of pure water is less than 7, but the water is still neutral because [H⁺] = [OH⁻]. A neutral solution is defined as one where the concentration of H⁺ equals the concentration of OH⁻, not one where pH = 7.

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pH Calculations

pH = −log₁₀[H⁺] and [H⁺] = 10⁻ᵖᴴ

pH of a Strong Acid

For a strong monoprotic acid: [H⁺] = concentration of acid (since it is fully dissociated).

Worked Example 1: Calculate the pH of 0.050 mol dm⁻³ HCl.

[H⁺] = 0.050 mol dm⁻³ pH = −log₁₀(0.050) = 1.30

Worked Example 2: Calculate the pH of 0.0010 mol dm⁻³ HNO₃.

[H⁺] = 0.0010 = 1.0 × 10⁻³ mol dm⁻³ pH = −log₁₀(1.0 × 10⁻³) = 3.00

pH of a Strong Base

For a strong monovalent base (e.g. NaOH, KOH): [OH⁻] = concentration of base.

Use Kw to find [H⁺]: [H⁺] = Kw / [OH⁻], then calculate pH.

Worked Example 3: Calculate the pH of 0.20 mol dm⁻³ NaOH at 298 K.

[OH⁻] = 0.20 mol dm⁻³ [H⁺] = Kw / [OH⁻] = (1.00 × 10⁻¹⁴) / 0.20 = 5.0 × 10⁻¹⁴ mol dm⁻³ pH = −log₁₀(5.0 × 10⁻¹⁴) = 13.30

For a divalent base like Ba(OH)₂: [OH⁻] = 2 × concentration of base (because each formula unit produces 2 OH⁻ ions).

Worked Example 4: Calculate the pH of 0.050 mol dm⁻³ Ba(OH)₂ at 298 K.

[OH⁻] = 2 × 0.050 = 0.10 mol dm⁻³ [H⁺] = (1.00 × 10⁻¹⁴) / 0.10 = 1.0 × 10⁻¹³ mol dm⁻³ pH = −log₁₀(1.0 × 10⁻¹³) = 13.00

pH of a Weak Acid

For a weak acid HA, the equilibrium is: HA ⇌ H⁺ + A⁻

We make two assumptions:

1. [H⁺] ≈ [A⁻] (because almost all H⁺ comes from the weak acid, not from water autoionisation)
2. [HA]_equilibrium ≈ [HA]_initial (because so little dissociates)

Therefore: Ka = [H⁺]² / [HA]

Rearranging: [H⁺] = √(Ka × [HA])

Then: pH = −log₁₀[H⁺]

Worked Example 5: Calculate the pH of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³).

[H⁺] = √(Ka × [HA]) = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³ pH = −log₁₀(1.32 × 10⁻³) = 2.88

Check: the assumption is valid because [H⁺] (1.32 × 10⁻³) is much smaller than [HA] (0.100), so only ~1.3% has dissociated.

Exam Tip: Always state the two assumptions when calculating the pH of a weak acid: (1) [H⁺] = [A⁻] and (2) [HA]_eqm ≈ [HA]_initial. If the degree of dissociation exceeds ~5%, the assumptions become less valid and you may need to solve the full quadratic equation.

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Polyprotic Acids

A polyprotic acid can donate more than one proton per molecule.

Key Definition: A diprotic acid donates two protons per molecule (e.g. H₂SO₄, H₂CO₃). A triprotic acid donates three protons (e.g. H₃PO₄).

Sulfuric acid (H₂SO₄) is diprotic:

• First dissociation: H₂SO₄ → H⁺ + HSO₄⁻ (strong — fully dissociates)
• Second dissociation: HSO₄⁻ ⇌ H⁺ + SO₄²⁻ (weak — partially dissociates, Ka₂ = 0.012 mol dm⁻³)

For dilute H₂SO₄ solutions (≤ 0.5 mol dm⁻³), both dissociations are essentially complete, so [H⁺] ≈ 2 × [H₂SO₄].

For example: 0.050 mol dm⁻³ H₂SO₄ gives [H⁺] ≈ 0.10 mol dm⁻³, so pH ≈ 1.00.

For concentrated H₂SO₄ solutions, the second dissociation is incomplete and [H⁺] < 2 × [H₂SO₄].

Carbonic acid (H₂CO₃) is a weak diprotic acid where both dissociations are weak: Ka₁ = 4.3 × 10⁻⁷, Ka₂ = 4.7 × 10⁻¹¹ mol dm⁻³. In pH calculations, only the first dissociation is normally significant.

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Dilution and Mixing of Strong Acids and Bases

When a strong acid and a strong base are mixed, calculate the moles of H⁺ and OH⁻, determine which is in excess, and calculate the concentration of the excess ion in the total volume of the mixed solution.

Worked Example 6: Multi-Step pH Calculation

50.0 cm³ of 0.100 mol dm⁻³ HCl is mixed with 30.0 cm³ of 0.100 mol dm⁻³ NaOH. Calculate the pH of the resulting solution at 298 K.

Step 1: Calculate moles. n(H⁺) = 0.100 × (50.0/1000) = 5.00 × 10⁻³ mol n(OH⁻) = 0.100 × (30.0/1000) = 3.00 × 10⁻³ mol

Step 2: Determine excess. HCl + NaOH → NaCl + H₂O (1:1 ratio) Excess H⁺ = 5.00 × 10⁻³ − 3.00 × 10⁻³ = 2.00 × 10⁻³ mol

Step 3: Calculate concentration in total volume. Total volume = 50.0 + 30.0 = 80.0 cm³ = 0.0800 dm³ [H⁺] = 2.00 × 10⁻³ / 0.0800 = 0.0250 mol dm⁻³

Step 4: Calculate pH. pH = −log₁₀(0.0250) = 1.60

Worked Example 7: Excess Base After Mixing

25.0 cm³ of 0.200 mol dm⁻³ NaOH is mixed with 25.0 cm³ of 0.100 mol dm⁻³ HCl. Calculate the pH at 298 K.

n(OH⁻) = 0.200 × 0.0250 = 5.00 × 10⁻³ mol n(H⁺) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol Excess OH⁻ = 5.00 × 10⁻³ − 2.50 × 10⁻³ = 2.50 × 10⁻³ mol Total volume = 50.0 cm³ = 0.0500 dm³ [OH⁻] = 2.50 × 10⁻³ / 0.0500 = 0.0500 mol dm⁻³ [H⁺] = Kw / [OH⁻] = (1.00 × 10⁻¹⁴) / 0.0500 = 2.00 × 10⁻¹³ mol dm⁻³ pH = −log₁₀(2.00 × 10⁻¹³) = 12.70

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Buffer Solutions

Key Definition: A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it.

There are two types:

Acidic Buffer

Made from a weak acid and its conjugate base (usually added as a sodium or potassium salt). For example, ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COONa).

The solution contains large reserves of both the weak acid (HA) and its conjugate base (A⁻).

How it works:

• When H⁺ is added: the conjugate base reacts with the H⁺ to form the undissociated weak acid. CH₃COO⁻ + H⁺ → CH₃COOH This removes most of the added H⁺, so pH barely decreases.