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Acids, Bases & Electrochemistry

Acids & Bases

5th Year · 6th Year (Leaving Cert)

  • ✓By the end of this lesson students will be able to define acids and bases according to the Arrhenius and Brønsted-Lowry theories.
  • ✓By the end of this lesson students will be able to distinguish between strong and weak acids and bases, and understand the concept of the acid dissociation constant (Ka) for Higher Level.
  • ✓By the end of this lesson students will be able to calculate pH values for strong acids and bases.
  • ✓By the end of this lesson students will be able to explain the composition and mechanism of action of buffer solutions for Higher Level.

Key concepts

Arrhenius Theory of Acids and Bases

The Arrhenius theory defines an acid as a substance that dissociates in water to produce hydrogen ions (H⁺). A base is defined as a substance that dissociates in water to produce hydroxide ions (OH⁻). This theory is limited to aqueous solutions.

Acid: HA(aq) → H⁺(aq) + A⁻(aq) Base: BOH(aq) → B⁺(aq) + OH⁻(aq)
Brønsted-Lowry Theory of Acids and Bases

The Brønsted-Lowry theory provides a broader definition. An acid is a proton (H⁺) donor, and a base is a proton (H⁺) acceptor. When an acid donates a proton, it forms its conjugate base. When a base accepts a proton, it forms its conjugate acid. A conjugate acid-base pair consists of two species that differ by a single proton. Substances that can act as both an acid and a base are called amphoteric substances (e.g., water).

Acid: HA + H₂O ⇌ H₃O⁺ + A⁻ Base: B + H₂O ⇌ BH⁺ + OH⁻
Strong vs. Weak Acids and Bases

Strong acids and bases dissociate completely in water. Examples of strong acids include hydrochloric acid (HCl), sulfuric acid (H₂SO₄), and nitric acid (HNO₃). Examples of strong bases include sodium hydroxide (NaOH) and potassium hydroxide (KOH). Weak acids and bases only partially dissociate in water, establishing an equilibrium. Examples of weak acids include ethanoic acid (CH₃COOH) and carbonic acid (H₂CO₃). Examples of weak bases include ammonia (NH₃) and methylamine (CH₃NH₂).

Acid Dissociation Constant (Ka) (Higher Level)

For a weak acid, HA, that partially dissociates in water, an equilibrium is established: HA(aq) + H₂O(l) ⇌ H⁺(aq) + A⁻(aq). The acid dissociation constant, Ka, is the equilibrium constant for this reaction. A larger Ka value indicates a stronger weak acid, meaning it dissociates to a greater extent. The pKa is defined as -log₁₀Ka; a smaller pKa value corresponds to a stronger acid.

Ka = [H⁺][A⁻] / [HA]
The pH Scale

The pH scale is used to express the acidity or alkalinity of a solution. It is defined as the negative logarithm to base 10 of the hydrogen ion concentration, [H⁺]. A pH of 7 is neutral at 25°C, pH < 7 is acidic, and pH > 7 is basic (alkaline). The ion product of water, Kw, is [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ mol² dm⁻⁶ at 25°C. This relationship allows for the calculation of [H⁺] from [OH⁻] and vice versa. Similarly, pOH = -log₁₀[OH⁻], and pH + pOH = 14 at 25°C.

pH = -log₁₀[H⁺] [H⁺] = 10⁻pH Kw = [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ mol² dm⁻⁶ (at 25°C)
Buffer Solutions (Higher Level)

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. Buffer solutions typically consist of a weak acid and its conjugate base (e.g., ethanoic acid and sodium ethanoate) or a weak base and its conjugate acid (e.g., ammonia and ammonium chloride). The weak acid component neutralises added base, while the conjugate base component neutralises added acid, thus maintaining a relatively constant pH.

Key facts to remember

  • 1Arrhenius acids produce H⁺ ions in water; Arrhenius bases produce OH⁻ ions in water.
  • 2Brønsted-Lowry acids are proton (H⁺) donors; Brønsted-Lowry bases are proton (H⁺) acceptors.
  • 3Conjugate acid-base pairs differ by a single proton.
  • 4Strong acids and bases dissociate completely in water; weak acids and bases dissociate partially, establishing an equilibrium.
  • 5The pH scale measures acidity/alkalinity: pH = -log₁₀[H⁺].
  • 6The ion product of water, Kw = [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ mol² dm⁻⁶ at 25°C.
  • 7The acid dissociation constant, Ka, quantifies the strength of a weak acid; a larger Ka indicates a stronger acid (Higher Level).
  • 8Buffer solutions resist changes in pH upon addition of small amounts of acid or base (Higher Level).

Worked examples

Example 1

Calculate the pH of a 0.025 M solution of hydrochloric acid (HCl).

IHCl is a strong acid, so it dissociates completely in water:
IIHCl(aq) → H⁺(aq) + Cl⁻(aq)
IIITherefore, the concentration of H⁺ ions is equal to the initial concentration of HCl.
IV[H⁺] = 0.025 M
VUse the pH formula: pH = -log₁₀[H⁺]
VIpH = -log₁₀(0.025)

Answer

pH = 1.60

Remember that for strong acids, [H⁺] is directly equal to the acid's concentration.

Example 2

Calculate the pH of a 0.010 M solution of sodium hydroxide (NaOH).

INaOH is a strong base, so it dissociates completely in water:
IINaOH(aq) → Na⁺(aq) + OH⁻(aq)
IIITherefore, the concentration of OH⁻ ions is equal to the initial concentration of NaOH.
IV[OH⁻] = 0.010 M
VUse the ion product of water to find [H⁺]: Kw = [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ mol² dm⁻⁶
VI[H⁺] = Kw / [OH⁻] = (1.0 x 10⁻¹⁴) / 0.010
VII[H⁺] = 1.0 x 10⁻¹² M
VIIINow use the pH formula: pH = -log₁₀[H⁺]
9pH = -log₁₀(1.0 x 10⁻¹²)

Answer

pH = 12.00

Alternatively, you could calculate pOH first: pOH = -log₁₀[OH⁻] = -log₁₀(0.010) = 2.00. Then, pH = 14 - pOH = 14 - 2.00 = 12.00.

Example 3

(Higher Level) A 0.10 M solution of a weak acid, HA, has a pH of 2.87. Calculate the acid dissociation constant, Ka, for HA.

IFirst, find the hydrogen ion concentration, [H⁺], from the pH:
II[H⁺] = 10⁻pH = 10⁻²·⁸⁷
III[H⁺] ≈ 1.35 x 10⁻³ M
IVFor a weak acid, HA, the dissociation equilibrium is: HA(aq) ⇌ H⁺(aq) + A⁻(aq)
VAt equilibrium, we can assume that [H⁺] = [A⁻] (since they are produced in a 1:1 ratio from the dissociation of HA).
VIAlso, the concentration of undissociated HA at equilibrium is approximately the initial concentration minus the amount that dissociated:
VII[HA]_equilibrium = [HA]_initial - [H⁺] = 0.10 M - 1.35 x 10⁻³ M
VIII[HA]_equilibrium ≈ 0.09865 M
9Now, write the expression for Ka:
10Ka = [H⁺][A⁻] / [HA]
11Substitute the equilibrium concentrations:
12Ka = (1.35 x 10⁻³)(1.35 x 10⁻³) / (0.09865)

Answer

Ka ≈ 1.85 x 10⁻⁵ mol dm⁻³

For weak acids, the assumption that [HA] at equilibrium is approximately equal to the initial concentration (if [H⁺] is very small compared to [HA]_initial) is often made, but subtracting the dissociated amount gives a more accurate result.

Common mistakes

  • ✗Confusing the Arrhenius and Brønsted-Lowry definitions, especially regarding the role of water.
  • ✗Incorrectly assuming that all acids and bases dissociate completely, particularly when dealing with weak acids/bases.
  • ✗Forgetting to convert [OH⁻] to [H⁺] (or pOH to pH) when calculating the pH of a basic solution.
  • ✗Not showing all steps in pH calculations, especially the formula used.
  • ✗Incorrectly stating units for Ka or pH (pH has no units).

Exam tips

  • ★Always state the definition of an acid or base (Arrhenius or Brønsted-Lowry) clearly when asked.
  • ★For calculations, write down the relevant formula first (e.g., pH = -log₁₀[H⁺]) and then substitute values.
  • ★Pay close attention to significant figures in calculations; pH values are typically given to two decimal places.
  • ★When explaining buffer action (Higher Level), ensure you describe how both the weak acid and its conjugate base (or weak base and its conjugate acid) neutralise added H⁺ and OH⁻ ions, respectively.

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