Cambridge AS & A Level9701

Acids and bases

Chemistry 9701 Chapter Notes

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Acids and basesPartition coefficients
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In this chapter

  1. 1Fundamental Definitions of Acids and Bases
  2. 2The Brønsted-Lowry Theory
  3. 3Strong and Weak Acids and Bases
  4. 4The pH Scale and Strong Acid Calculations
  5. 5pH of Strong Bases and Kw
  6. 6Weak Acids and the Dissociation Constant (Ka)

1. Fundamental Definitions of Acids and Bases

At its simplest, an acid is a substance that neutralises a base, typically forming a salt and water. A more precise chemical definition, known as the Arrhenius definition, is that an acid is a substance that produces hydrogen ions (H+) when dissolved in water. For example, hydrogen chloride gas dissolves in water to form hydrochloric acid: HCl(aq) → H+(aq) + Cl-(aq). Bases are substances that neutralise acids. Many are metal oxides or metal hydroxides. A base that is soluble in water is called an alkali. Alkalis produce hydroxide ions (OH-) in solution. For example, sodium hydroxide: NaOH(aq) → Na+(aq) + OH-(aq).

Acid: HA(aq) → H+(aq) + A-(aq)

Alkali: BOH(aq) → B+(aq) + OH-(aq)

Neutralisation: Acid + Base → Salt + Water

Key term

Alkali: A base that is soluble in water and produces hydroxide ions (OH-) in solution.

Common pitfall

All alkalis are bases, but not all bases are alkalis. Students often use the terms interchangeably, but a base is only an alkali if it dissolves in water.

Fun fact

The sour taste of lemons and limes comes from citric acid, while the sharp taste of vinegar is due to ethanoic acid.

Worked example 12 marks

Write a balanced chemical equation, including state symbols, for the neutralisation reaction between aqueous sulfuric acid (H₂SO₄) and solid copper(II) oxide (CuO).

  1. 1

    Step 1: Identify the reactants and products. The reactants are sulfuric acid (H₂SO₄) and copper(II) oxide (CuO). The products of a neutralisation reaction are a salt and water.

  2. 2

    Step 2: Determine the formula of the salt. The salt is formed from the cation of the base (Cu²⁺) and the anion of the acid (SO₄²⁻). The formula is CuSO₄.

  3. 3

    Step 3: Write the unbalanced equation: H₂SO₄ + CuO → CuSO₄ + H₂O.

  4. 4

    Step 4: Check the balancing. Left side: 2 H, 1 S, 5 O, 1 Cu. Right side: 1 Cu, 1 S, 5 O, 2 H. The equation is already balanced.

  5. 5

    Step 5: Add the state symbols as given in the question: H₂SO₄(aq) + CuO(s) → CuSO₄(aq) + H₂O(l).

Recap

  • An acid is a substance that produces H+ ions in aqueous solution.
  • A base is a substance that neutralises an acid.
  • An alkali is a base that is soluble in water, producing OH- ions.
  • The reaction between an acid and a base is called neutralisation.
  • The products of neutralising a metal oxide or hydroxide with an acid are a salt and water.

Quick check

  1. State the difference between a base and an alkali.1 mark
  2. Name the two products formed when nitric acid (HNO₃) reacts with potassium hydroxide (KOH).1 mark

2. The Brønsted-Lowry Theory

The Brønsted-Lowry theory provides a more general definition of acids and bases that is not limited to aqueous solutions. It defines an acid as a proton (H+) donor and a base as a proton acceptor. In any acid-base reaction, a proton is transferred from the acid to the base. For example, in the reaction HCl + H₂O → H₃O⁺ + Cl⁻, HCl donates a proton and is the acid, while H₂O accepts the proton and is the base. When an acid donates its proton, it becomes a conjugate base. When a base accepts a proton, it becomes a conjugate acid. This creates 'conjugate acid-base pairs'.

Acid (HA) + Base (B) ⇌ Conjugate Base (A⁻) + Conjugate Acid (HB⁺)

Key term

Amphoteric: A substance that can act as either a Brønsted-Lowry acid (proton donor) or a Brønsted-Lowry base (proton acceptor).

Examiner insight

Examiners frequently ask you to identify conjugate acid-base pairs in a given equilibrium. Practice by looking at the change in the number of hydrogen atoms and the charge for each species.

Fun fact

Even in pure water, a tiny fraction of molecules react with each other in a Brønsted-Lowry reaction (H₂O + H₂O ⇌ H₃O⁺ + OH⁻), a process called autoionisation.

Worked example 13 marks

In the following equilibrium, identify the Brønsted-Lowry acid and base on the left-hand side, and their corresponding conjugate species on the right-hand side. NH₃(g) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

  1. 1

    Step 1: Analyse the forward reaction (left to right). The H₂O molecule has donated a proton (H⁺) to become OH⁻. Therefore, H₂O is the Brønsted-Lowry acid.

  2. 2

    Step 2: The NH₃ molecule has accepted that proton to become NH₄⁺. Therefore, NH₃ is the Brønsted-Lowry base.

  3. 3

    Step 3: Identify the conjugate pairs. The acid H₂O becomes its conjugate base, OH⁻. The base NH₃ becomes its conjugate acid, NH₄⁺.

  4. 4

    Step 4: State the answer clearly. Acid: H₂O. Base: NH₃. Conjugate acid: NH₄⁺. Conjugate base: OH⁻.

Recap

  • A Brønsted-Lowry acid is a proton (H⁺) donor.
  • A Brønsted-Lowry base is a proton (H⁺) acceptor.
  • A conjugate acid-base pair consists of two species that differ by a single proton.
  • Amphoteric substances, like water, can act as both an acid and a base.

Quick check

  1. What is the conjugate base of the acid H₂SO₄?1 mark
  2. What is the conjugate acid of the base H₂PO₄⁻?1 mark

3. Strong and Weak Acids and Bases

The strength of an acid or base depends on its degree of ionisation (or dissociation) in water. Strong acids, like HCl, completely ionise in solution. This means every molecule of the acid donates its proton. We use a forward arrow (→) to show this goes to completion: HCl(aq) → H⁺(aq) + Cl⁻(aq). Weak acids, like ethanoic acid (CH₃COOH), only partially ionise. Only a small fraction of molecules donate their protons, so the solution contains a mixture of intact acid molecules and ions. We use an equilibrium arrow (⇌) for this: CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq). The same principle applies to bases. Strong bases (e.g., NaOH) fully dissociate, while weak bases (e.g., NH₃) only partially react with water.

Strong Acid: HA → H⁺ + A⁻

Weak Acid: HA ⇌ H⁺ + A⁻

Key term

Degree of dissociation: The fraction of molecules that have dissociated (split into ions) in a solution at equilibrium.

Common pitfall

Confusing the terms 'strong' and 'concentrated'. A strong acid (e.g., HCl) is always strong, even when dilute. A weak acid (e.g., CH₃COOH) is always weak, even when concentrated. Strength refers to dissociation, concentration refers to amount per volume.

Worked example 14 marks

A 0.1 mol dm⁻³ solution of hydrochloric acid has a pH of 1, while a 0.1 mol dm⁻³ solution of ethanoic acid has a pH of 2.88. Explain this difference in terms of acid strength and ionisation.

  1. 1

    Step 1: Define strong and weak acids. Hydrochloric acid is a strong acid, meaning it dissociates completely in water.

  2. 2

    Step 2: Relate concentration to ion concentration for HCl. For 0.1 mol dm⁻³ HCl, the hydrogen ion concentration [H⁺] is also 0.1 mol dm⁻³.

  3. 3

    Step 3: Define ethanoic acid's strength. Ethanoic acid is a weak acid, meaning it only partially dissociates in water in a reversible reaction.

  4. 4

    Step 4: Relate this to its ion concentration. For 0.1 mol dm⁻³ CH₃COOH, the [H⁺] is much lower than 0.1 mol dm⁻³ because most of the acid remains as undissociated molecules.

  5. 5

    Step 5: Link [H⁺] to pH. Since pH = -log[H⁺], a lower [H⁺] (in ethanoic acid) results in a higher pH value compared to hydrochloric acid.

Recap

  • Strong acids and bases completely ionise in water.
  • Weak acids and bases only partially ionise in water.
  • Equations for strong acids use a '→' arrow; weak acids use a '⇌' arrow.
  • For the same concentration, a strong acid will have a much higher [H⁺] and lower pH than a weak acid.

Quick check

  1. Which arrow, → or ⇌, should be used in the equation for the dissociation of the weak base ammonia, NH₃, in water?1 mark

4. The pH Scale and Strong Acid Calculations

The pH scale is a convenient way to express the concentration of hydrogen ions, [H⁺], in a solution. It is a logarithmic scale, meaning a change of 1 pH unit represents a 10-fold change in [H⁺]. The scale typically runs from 0 to 14. A pH less than 7 is acidic, a pH of 7 is neutral, and a pH greater than 7 is alkaline. The pH is calculated using the formula: pH = -log₁₀[H⁺]. For a strong monoprotic acid (an acid that donates one proton), the dissociation is complete. Therefore, the concentration of hydrogen ions is equal to the concentration of the acid itself.

pH = -log₁₀[H⁺]

[H⁺] = 10⁻ᵖᴴ

For a strong monoprotic acid, [H⁺] = [Acid]

Key term

pH: A measure of acidity or alkalinity, defined as the negative logarithm (base 10) of the hydrogen ion concentration in mol dm⁻³.

Examiner insight

Always show your working clearly, stating the assumption that for a strong acid, [H⁺] = [Acid]. Answers for pH should be given to two decimal places.

Worked example 12 marks

Calculate the pH of a 0.050 mol dm⁻³ solution of nitric acid (HNO₃).

  1. 1

    Step 1: Identify the acid type. Nitric acid (HNO₃) is a strong monoprotic acid.

  2. 2

    Step 2: Determine the hydrogen ion concentration. Because it is a strong acid, it dissociates completely. Therefore, [H⁺] = [HNO₃] = 0.050 mol dm⁻³.

  3. 3

    Step 3: Apply the pH formula. pH = -log₁₀[H⁺].

  4. 4

    Step 4: Substitute the value and calculate. pH = -log₁₀(0.050) = 1.30. (pH is usually given to 2 decimal places).

Worked example 22 marks

Calculate the concentration of a hydrochloric acid solution which has a pH of 1.50.

  1. 1

    Step 1: State the formula to find [H⁺] from pH. [H⁺] = 10⁻ᵖᴴ.

  2. 2

    Step 2: Substitute the pH value. [H⁺] = 10⁻¹·⁵⁰ = 0.0316 mol dm⁻³.

  3. 3

    Step 3: Relate [H⁺] to acid concentration. Since HCl is a strong acid, [HCl] = [H⁺].

  4. 4

    Step 4: State the final answer with units. [HCl] = 0.032 mol dm⁻³ (to 2 significant figures).

Recap

  • pH is the negative base-10 logarithm of the hydrogen ion concentration.
  • Low pH means high [H⁺] and high acidity.
  • A 1-unit change in pH corresponds to a 10-fold change in [H⁺].
  • For a strong monoprotic acid, [H⁺] is equal to the initial concentration of the acid.

Quick check

  1. What is the pH of a 0.010 mol dm⁻³ solution of HCl?1 mark

5. pH of Strong Bases and Kw

To find the pH of a basic solution, we first need to consider the autoionisation of water. Water itself slightly dissociates: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). The equilibrium constant for this is the ionic product of water, Kw. At 298 K (25 °C), Kw = [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ mol² dm⁻⁶. This relationship holds true for any aqueous solution. For a strong base like NaOH, it dissociates completely to produce OH⁻ ions. We can calculate [OH⁻] from the base's concentration. Then, we use the Kw expression to find the corresponding [H⁺], and finally, use the pH formula to calculate the pH.

Kw = [H⁺][OH⁻]

Kw = 1.0 x 10⁻¹⁴ mol² dm⁻⁶ (at 298 K)

Key term

Ionic product of water (Kw): The equilibrium constant for the autoionisation of water, given by the product of the hydrogen and hydroxide ion concentrations.

Common pitfall

A very common mistake is to calculate pOH (-log[OH⁻]) and report this as the pH. You must convert [OH⁻] to [H⁺] using Kw before calculating pH.

Worked example 13 marks

Calculate the pH of a 0.0200 mol dm⁻³ solution of sodium hydroxide (NaOH) at 298 K. (Kw = 1.0 x 10⁻¹⁴ mol² dm⁻⁶)

  1. 1

    Step 1: Identify the base type. Sodium hydroxide (NaOH) is a strong monoprotic base.

  2. 2

    Step 2: Determine the hydroxide ion concentration. As it's a strong base, it dissociates completely. [OH⁻] = [NaOH] = 0.0200 mol dm⁻³.

  3. 3

    Step 3: Use Kw to find [H⁺]. Kw = [H⁺][OH⁻], so [H⁺] = Kw / [OH⁻].

  4. 4

    Step 4: Substitute values. [H⁺] = (1.0 x 10⁻¹⁴) / 0.0200 = 5.0 x 10⁻¹³ mol dm⁻³.

  5. 5

    Step 5: Calculate pH. pH = -log₁₀[H⁺] = -log₁₀(5.0 x 10⁻¹³) = 12.30.

Recap

  • Kw is the ionic product of water, linking [H⁺] and [OH⁻] in any aqueous solution.
  • At 298 K, Kw has a constant value of 1.0 x 10⁻¹⁴ mol² dm⁻⁶.
  • To find the pH of a strong base, first find [OH⁻], then use Kw to find [H⁺], then calculate pH.
  • Solutions with high [OH⁻] have low [H⁺] and therefore a high pH.

Quick check

  1. A solution has a hydroxide ion concentration of 1.0 x 10⁻⁴ mol dm⁻³. What is its hydrogen ion concentration at 298 K?1 mark

6. Weak Acids and the Dissociation Constant (Ka)

Since weak acids only partially dissociate, we cannot assume [H⁺] is equal to the acid concentration. Instead, we use an equilibrium constant called the acid dissociation constant, Ka. For a generic weak acid HA dissociating as HA ⇌ H⁺ + A⁻, the Ka expression is Ka = ([H⁺][A⁻])/[HA]. The larger the Ka value, the more the acid dissociates and the stronger the acid. For pH calculations, we make two key assumptions: 1) The dissociation of the acid is so small that the equilibrium concentration of HA is the same as its initial concentration. 2) The concentration of H⁺ from the autoionisation of water is negligible. These assumptions simplify the expression to Ka ≈ [H⁺]² / [HA], which can be rearranged to find [H⁺] and then the pH.

Ka = ([H⁺][A⁻]) / [HA]

For weak acid calculations: Ka ≈ [H⁺]² / [HA]initial

[H⁺] ≈ √(Ka × [HA])

Key term

Acid dissociation constant (Ka): An equilibrium constant that measures the extent to which a weak acid dissociates in solution; a larger Ka indicates a stronger acid.

Examiner insight

Examiners expect you to state the two assumptions made in a weak acid pH calculation if asked to describe the method. Marks can be lost for omitting them.

Worked example 13 marks

Calculate the pH of a 0.100 mol dm⁻³ solution of ethanoic acid at 298 K. (Ka for ethanoic acid = 1.74 x 10⁻⁵ mol dm⁻³)

  1. 1

    Step 1: Write the expression for Ka. For CH₃COOH ⇌ H⁺ + CH₃COO⁻, Ka = ([H⁺][CH₃COO⁻])/[CH₃COOH].

  2. 2

    Step 2: Apply simplifying assumptions. Assume [H⁺] = [CH₃COO⁻] and [CH₃COOH] remains 0.100 mol dm⁻³. The expression becomes Ka ≈ [H⁺]² / [CH₃COOH].

  3. 3

    Step 3: Rearrange the formula to solve for [H⁺]. [H⁺]² ≈ Ka × [CH₃COOH], so [H⁺] ≈ √(Ka × [CH₃COOH]).

  4. 4

    Step 4: Substitute the values. [H⁺] ≈ √(1.74 x 10⁻⁵ × 0.100) = √(1.74 x 10⁻⁶) = 1.319 x 10⁻³ mol dm⁻³.

  5. 5

    Step 5: Calculate the pH. pH = -log₁₀(1.319 x 10⁻³) = 2.88.

Recap

  • Ka is the equilibrium constant for the dissociation of a weak acid.
  • The larger the Ka value, the stronger the weak acid.
  • Two key assumptions are made when calculating the pH of a weak acid.
  • The simplified formula [H⁺] ≈ √(Ka × [HA]) is used to find the hydrogen ion concentration.

Quick check

  1. Acid X has a Ka of 1x10⁻⁴ and Acid Y has a Ka of 1x10⁻⁶. Which is the stronger acid?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define an acid and a base according to the Brønsted-Lowry theory.2 marks
  2. A solution of hydrochloric acid has a pH of 0.70. Calculate the concentration of the acid in mol dm⁻³.2 marks
  3. Calculate the pH of a 0.150 mol dm⁻³ solution of barium hydroxide, Ba(OH)₂, at 298 K. Assume it fully dissociates. (Kw = 1.0 x 10⁻¹⁴ mol² dm⁻⁶)3 marks
  4. Explain, with the aid of an equation, why a 0.5 mol dm⁻³ solution of ethanoic acid is a poorer electrical conductor than a 0.5 mol dm⁻³ solution of sulfuric acid.4 marks
  5. For the reaction: H₂SO₄ + HNO₃ ⇌ HSO₄⁻ + H₂NO₃⁺. Identify the two conjugate acid-base pairs.2 marks
  6. A 0.080 mol dm⁻³ solution of a weak monoprotic acid, HA, has a pH of 3.10. Calculate the value of Ka for this acid.4 marks
  7. Explain why water is described as being amphoteric. Use two different chemical equations to support your answer.3 marks
  8. A student dissolves 2.50 g of sodium hydroxide in water and makes the solution up to 250 cm³. Calculate the pH of this solution at 298 K. (Mr of NaOH = 40.0; Kw = 1.0 x 10⁻¹⁴ mol² dm⁻⁶)4 marks
  9. pKa is defined as -log₁₀(Ka). Calculate the pH of a 0.200 mol dm⁻³ solution of propanoic acid, which has a pKa of 4.87.4 marks
  10. Distinguish clearly between a 'strong acid' and a 'concentrated acid', giving an example of a solution for each of the four possible combinations (strong/dilute, strong/concentrated, weak/dilute, weak/concentrated).5 marks

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