Ionic equilibrium is fundamental in understanding many chemical processes encountered in science laboratories, especially those involving solutions and reactions in aqueous media. In the Kenyan context, professionals working in clinical laboratories, water quality testing facilities, and pharmaceutical analysis regularly apply ionic equilibrium concepts to analyze pH, buffer systems, and solubility phenomena. Mastery of ionic equilibrium allows technologists to predict reaction directions, calculate concentrations of ions, and optimize conditions for analytical procedures which are critical for accurate diagnostics and quality control.

1.1 Ionic equilibrium

Ionic equilibrium describes the balance between ions and unionised molecules, or between ions in solution and an undissolved solid, in a system that has reached a steady state. It governs the acidity of water supplies, the action of buffers in blood and soil, the choice of indicator in a titration and the formation of scale or precipitates in pipes and boilers. In this topic we apply the general idea of chemical equilibrium to solutions of electrolytes and learn to calculate pH, ionisation constants, buffer pH, hydrolysis and solubility. All concentrations are in mol/dm³ and all constants are quoted at 25 °C unless stated.

Electrolytes and ionisation

An electrolyte is a substance that conducts electric current in solution or when molten because it produces mobile ions. Strong electrolytes ionise almost completely, while weak electrolytes ionise only partially, so an equilibrium exists between the unionised molecules and their ions. The position of this equilibrium decides how acidic, basic or conducting a solution is. Example: the laboratory technician at a county water office finds that 0.10 M hydrochloric acid conducts far better than 0.10 M ethanoic acid, because the first is fully ionised and the second is only slightly ionised.

Factors that affect the degree of ionisation of a weak electrolyte:

• Dilution: Lowering the concentration shifts the equilibrium towards more ions, so the degree of ionisation increases.
• Temperature: Ionisation of weak electrolytes is often endothermic, so a rise in temperature then increases the degree of ionisation and the equilibrium constant.
• Nature of the solvent: A solvent of high dielectric constant such as water separates ions well and favours ionisation.
• Nature of the electrolyte: Weaker bonds between the ionisable hydrogen and the rest of the molecule give greater ionisation.
• Presence of a common ion: An added ion already present in the equilibrium suppresses ionisation (see the common ion effect below).

Acid and base theories

Several definitions of acids and bases are used, and each is useful in a different setting. The Brønsted-Lowry theory is the most useful for equilibrium work because it treats acid-base reactions as competition for protons, and it leads directly to conjugate pairs.

• Arrhenius theory: An acid produces H⁺ ions in water and a base produces OH⁻ ions in water. It is limited to aqueous solutions.
• Brønsted-Lowry theory: An acid is a proton donor and a base is a proton acceptor. Example: in NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, water is the acid and ammonia is the base.
• Conjugate pairs: When an acid loses a proton it forms its conjugate base, and when a base gains a proton it forms its conjugate acid. The conjugate base of a strong acid is extremely weak, while the conjugate base of a weak acid is weak but noticeably stronger.
• Lewis theory: An acid is an electron-pair acceptor and a base is an electron-pair donor. Example: BF₃ accepts a pair from NH₃ to form BF₃·NH₃.
• Amphoteric species: Some species act as acid or base depending on the partner. Water, HCO₃⁻ and HSO₄⁻ are examples.

Degree of ionisation and Ostwald's dilution law

The degree of ionisation, α, is the fraction of the dissolved electrolyte that has formed ions. For a weak monobasic acid HA of initial concentration C, the equilibrium HA ⇌ H⁺ + A⁻ gives [H⁺] = [A⁻] = Cα and [HA] = C(1 − α). Substituting into the equilibrium expression gives Ostwald's dilution law.

Ka = [H⁺][A⁻] ÷ [HA] = Cα² ÷ (1 − α)

For a weak acid α is very small, so 1 − α ≈ 1 and the law simplifies:

Ka = Cα²

α = √(Ka ÷ C)

[H⁺] = Cα = √(Ka × C)

The law shows that α increases when the solution is diluted, because α is inversely proportional to the square root of C.

Example: find α, [H⁺] and pH of 0.10 M ethanoic acid, Ka = 1.8 × 10⁻⁵.

α = √(Ka ÷ C)

α = √(1.8 × 10⁻⁵ ÷ 0.10)

α = √(1.8 × 10⁻⁴) = 1.34 × 10⁻²

[H⁺] = Cα = 0.10 × 1.34 × 10⁻² = 1.34 × 10⁻³ mol/dm³

pH = −log(1.34 × 10⁻³) = 2.87

The acid is therefore about 1.3 % ionised.

Ionic product of water, pH and pOH

Pure water ionises very slightly: H₂O ⇌ H⁺ + OH⁻. The product of the two ion concentrations is a constant called the ionic product of water, Kw, which equals 1.0 × 10⁻¹⁴ mol²/dm⁶ at 25 °C. In pure water [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/dm³. Kw increases with temperature because ionisation of water is endothermic, so neutral water at higher temperature has a pH below 7 even though [H⁺] = [OH⁻].

Definitions and relationships:

pH = −log[H⁺]

pOH = −log[OH⁻]

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴

pH + pOH = 14 (at 25 °C)

Worked example (strong acid): 0.0050 M HCl in a school laboratory.

[H⁺] = 5.0 × 10⁻³ mol/dm³ (complete ionisation)

pH = −log(5.0 × 10⁻³) = 2.30

Worked example (strong base): 0.020 M NaOH.

[OH⁻] = 2.0 × 10⁻² mol/dm³

pOH = −log(2.0 × 10⁻²) = 1.70

pH = 14 − 1.70 = 12.30

Points to remember when handling pH:

• Logarithmic scale: A change of one pH unit means a tenfold change in [H⁺].
• Very dilute acids: When [H⁺] from the acid is near 10⁻⁷ M, the H⁺ from water cannot be ignored, so a 10⁻⁸ M HCl solution has a pH just below 7, not 8.
• Diprotic strong acids: For H₂SO₄ the first ionisation is complete, so [H⁺] is at least equal to the acid concentration.
• Temperature: The relation pH + pOH = 14 holds only at 25 °C.
• Measurement: A pH meter calibrated with buffers of known pH gives more reliable readings than indicator paper.

Ionisation constants of weak acids and bases

For a weak acid HA, the acid ionisation constant is Ka = [H⁺][A⁻] ÷ [HA]. For a weak base B, B + H₂O ⇌ BH⁺ + OH⁻, the base ionisation constant is Kb = [BH⁺][OH⁻] ÷ [B]. A larger constant means a stronger acid or base. The constants are often quoted as pKa = −log Ka and pKb = −log Kb, so that a smaller pKa means a stronger acid.

For a conjugate acid-base pair in water:

Ka × Kb = Kw

pKa + pKb = 14

Approximate formulae for weak electrolytes of concentration C, valid when ionisation is small:

[H⁺] = √(Ka × C)

[OH⁻] = √(Kb × C)

Example: pH of 0.10 M aqueous ammonia, Kb = 1.8 × 10⁻⁵.

[OH⁻] = √(Kb × C)

[OH⁻] = √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol/dm³

pOH = −log(1.34 × 10⁻³) = 2.87

pH = 14 − 2.87 = 11.13

Polyprotic acids ionise in steps, each with its own constant. Ka1 is much larger than Ka2, so the pH of a solution of H₂CO₃ or H₃PO₄ is controlled almost entirely by the first step.

Common ion effect

The common ion effect is the suppression of ionisation of a weak electrolyte (or of the solubility of a sparingly soluble salt) by adding a soluble compound that supplies one of the ions already in the equilibrium. By Le Chatelier's principle, the added ion shifts the equilibrium towards the unionised molecules. Example: adding sodium ethanoate to ethanoic acid increases [CH₃COO⁻], so the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ moves left and [H⁺] falls.

Worked example: a school laboratory mixes 0.10 M ethanoic acid with 0.10 M sodium ethanoate (Ka = 1.8 × 10⁻⁵).

Ka = [H⁺][CH₃COO⁻] ÷ [CH₃COOH]

[H⁺] = Ka × [CH₃COOH] ÷ [CH₃COO⁻]

[H⁺] = 1.8 × 10⁻⁵ × 0.10 ÷ 0.10 = 1.8 × 10⁻⁵ mol/dm³

pH = 4.74

Without the salt the same acid had pH 2.87, so the common ion raised the pH by nearly two units.

Uses of the common ion effect:

• Buffer preparation: A weak acid with its salt, or a weak base with its salt, resists changes in pH.
• Qualitative analysis: Adding NH₄Cl before NH₃ lowers [OH⁻] so that only hydroxides of very low solubility, such as Fe(OH)₃ and Al(OH)₃, precipitate.
• Salting out: Adding NaCl to soap solution lowers the solubility of soap, so it separates as a solid in soap making.
• Purification of salts: Passing HCl gas into a saturated NaCl solution precipitates pure NaCl because of the extra Cl⁻.
• Reducing solubility losses: A sparingly soluble precipitate is washed with a dilute solution of a common ion, not pure water, to limit dissolving.

Buffer solutions

A buffer solution resists a change in pH when small amounts of acid or base are added, or when it is diluted. An acidic buffer contains a weak acid and its salt with a strong base (for example CH₃COOH and CH₃COONa), and a basic buffer contains a weak base and its salt with a strong acid (for example NH₃ and NH₄Cl). The weak component neutralises added base or acid, while the conjugate component neutralises the other. Example: the H₂CO₃/HCO₃⁻ system in blood holds blood pH close to 7.4, and a clinical laboratory relies on it when interpreting blood results.

How an acidic buffer works:

• Added H⁺: The conjugate base (A⁻) combines with H⁺ to form HA, so [H⁺] rises very little.
• Added OH⁻: The weak acid (HA) reacts with OH⁻ to form A⁻ and water, so [OH⁻] stays small.
• Large reserves: Both HA and A⁻ are present in comparable amounts, so neither is quickly used up.

Henderson-Hasselbalch equation for an acidic buffer:

pH = pKa + log([A⁻] ÷ [HA])

For a basic buffer:

pOH = pKb + log([BH⁺] ÷ [B])

Worked example: pH of a buffer containing 0.20 M sodium ethanoate and 0.10 M ethanoic acid (pKa = 4.74).

pH = pKa + log([A⁻] ÷ [HA])

pH = 4.74 + log(0.20 ÷ 0.10)

pH = 4.74 + 0.30 = 5.04

Worked example: 0.010 mol of HCl is added to 1.0 dm³ of this buffer (0.10 mol HA and 0.20 mol A⁻).

New moles of A⁻ = 0.20 − 0.010 = 0.19 mol

New moles of HA = 0.10 + 0.010 = 0.11 mol

pH = 4.74 + log(0.19 ÷ 0.11)

pH = 4.74 + 0.24 = 4.98

The pH changed by only 0.06 unit. Adding the same amount of HCl to 1.0 dm³ of pure water would change the pH from 7 to 2.

Steps for preparing an acidic buffer of a chosen pH:

1. Select a weak acid whose pKa is within one unit of the required pH, because buffering is best when pH is close to pKa.
2. Calculate the ratio [A⁻] ÷ [HA] needed by rearranging the Henderson-Hasselbalch equation: ratio = 10^(pH − pKa).
3. Choose a total concentration high enough to give adequate buffer capacity, for example 0.10 M to 0.20 M in total.
4. Weigh the salt and measure the acid (or weigh both solids) in the calculated proportions.
5. Dissolve them in less distilled water than the final volume and mix well.
6. Check the pH with a calibrated pH meter and adjust by adding small amounts of acid or salt solution.
7. Make up to the final volume in a volumetric flask and mix again.

Factors that determine buffer performance:

• Buffer capacity: The amount of acid or base a buffer can absorb before its pH changes significantly, which increases with the total concentration of the components.
• Component ratio: Capacity is greatest when [A⁻] = [HA], that is, when pH = pKa.
• Useful range: A buffer works effectively over about pKa ± 1 pH unit.
• Dilution: Dilution leaves the ratio, and so the pH, nearly unchanged, but lowers the capacity.
• Temperature: Ka depends on temperature, so the pH of a buffer shifts slightly with temperature.

Hydrolysis of salts

Salt hydrolysis is the reaction of the ions of a salt with water that changes the pH of the solution. Only ions that are conjugates of weak acids or weak bases react appreciably, because they take or give protons to water. Example: a tea estate chemist finds that a solution of ammonium sulphate fertiliser is acidic, because NH₄⁺ gives protons to water.

For the anion A⁻ of a weak acid: A⁻ + H₂O ⇌ HA + OH⁻. The hydrolysis constant is:

Kh = Kw ÷ Ka

[OH⁻] = √(Kh × C)

Worked example: pH of 0.10 M sodium ethanoate (Ka of ethanoic acid = 1.8 × 10⁻⁵).

Kh = Kw ÷ Ka

Kh = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰

[OH⁻] = √(5.6 × 10⁻¹⁰ × 0.10) = √(5.6 × 10⁻¹¹) = 7.5 × 10⁻⁶ mol/dm³

pOH = −log(7.5 × 10⁻⁶) = 5.13

pH = 14 − 5.13 = 8.87

For the cation of a weak base, BH⁺ + H₂O ⇌ B + H₃O⁺, Kh = Kw ÷ Kb, and the same method gives [H⁺] = √(Kh × C).

Solubility product and precipitation

A sparingly soluble ionic salt in contact with its saturated solution is in equilibrium, for example AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). The solubility product, Ksp, is the product of the ion concentrations, each raised to the power of its coefficient, in a saturated solution at a given temperature. The solid does not appear in the expression because its concentration is constant.

AgCl: Ksp = [Ag⁺][Cl⁻]

PbI₂: Ksp = [Pb²⁺][I⁻]²

Relation between Ksp and molar solubility, s (mol/dm³):

• Salts of type AB (for example AgCl, BaSO₄): Ksp = s²
• Salts of type AB₂ or A₂B (for example PbI₂, Ag₂CrO₄): Ksp = 4s³
• Salts of type AB₃ or A₃B (for example Fe(OH)₃): Ksp = 27s⁴

Worked example (AB type): solubility of AgCl, Ksp = 1.8 × 10⁻¹⁰.

s = √Ksp

s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/dm³

Worked example (AB₂ type): a salt with Ksp = 3.2 × 10⁻¹¹.

Ksp = 4s³

s³ = 3.2 × 10⁻¹¹ ÷ 4 = 8.0 × 10⁻¹²

s = 2.0 × 10⁻⁴ mol/dm³

Common ion effect on solubility: in 0.10 M NaCl, [Cl⁻] = 0.10 M, so the solubility of AgCl falls to:

s = Ksp ÷ [Cl⁻]

s = 1.8 × 10⁻¹⁰ ÷ 0.10 = 1.8 × 10⁻⁹ mol/dm³

Predicting precipitation: compare the ionic product, Qsp, of the mixed solution with Ksp.

• If Qsp < Ksp, the solution is unsaturated and no precipitate forms.
• If Qsp = Ksp, the solution is just saturated.
• If Qsp > Ksp, the solution is supersaturated and a precipitate forms until Qsp = Ksp.

Worked example: 50 cm³ of 1.0 × 10⁻³ M AgNO₃ is mixed with 50 cm³ of 1.0 × 10⁻³ M NaCl. Will AgCl precipitate?

Total volume doubles, so each concentration halves: [Ag⁺] = [Cl⁻] = 5.0 × 10⁻⁴ mol/dm³

Qsp = [Ag⁺][Cl⁻] = 5.0 × 10⁻⁴ × 5.0 × 10⁻⁴ = 2.5 × 10⁻⁷

Qsp (2.5 × 10⁻⁷) > Ksp (1.8 × 10⁻¹⁰), so AgCl precipitates.

Applications of solubility equilibria:

• Water softening: Calcium and magnesium ions are removed as insoluble carbonates or hydroxides. Example: a hotel boiler develops CaCO₃ scale where hard water is heated.
• Gravimetric analysis: An ion is quantitatively precipitated as a sparingly soluble salt, such as sulphate as BaSO₄, and weighed.
• Selective precipitation: Ions are separated by adding a reagent that precipitates the least soluble salt first.
• Effect of pH: Salts of weak acids, such as carbonates and hydroxides, dissolve in acid because the anion is removed as the weak acid, which is why limescale is removed with dilute acid.
• Medical uses: Barium sulphate is safe as an X-ray contrast meal because its very low solubility keeps [Ba²⁺] in the body extremely small.

Acid-base indicators and titration curves

An acid-base indicator is a weak acid or base whose unionised and ionised forms have different colours. For an indicator HIn ⇌ H⁺ + In⁻, the colour seen depends on the ratio [In⁻] ÷ [HIn], which is set by the pH of the solution. The colour changes over a range of about two pH units centred on pKin. Example: a farm laboratory titrating soil extract uses phenolphthalein, which is colourless in acid and pink in alkali.

The pH at the equivalence point depends on the strengths of the acid and base, and the indicator must change colour within the steep part of the titration curve.

• Strong acid and strong base: Equivalence point is at pH 7, and either methyl orange or phenolphthalein is suitable because the vertical section is very steep.
• Weak acid and strong base: Equivalence point is above 7 because of hydrolysis of the salt, so phenolphthalein is suitable and methyl orange is not.
• Strong acid and weak base: Equivalence point is below 7, so methyl orange is suitable and phenolphthalein is not.
• Weak acid and weak base: The pH change near the equivalence point is gradual, so no simple indicator gives a sharp end point.
• Half-equivalence point: For a weak acid titrated with a strong base, the pH at half the volume needed for neutralisation equals pKa, because [HA] = [A⁻].

Chapter Summary

This chapter explored the fundamental concepts of physical chemistry essential for understanding chemical behavior and reactions. It began with ionic equilibrium, explaining how the balance between ions in solution influences properties such as pH and solubility. The discussion then moved to electrochemistry principles, detailing the mechanisms of electron transfer in redox reactions and the generation of electrical energy in galvanic cells. Chemical kinetics was examined next, focusing on the factors that affect reaction rates and the mathematical models used to describe these rates. Finally, the chapter covered chemical thermodynamics, emphasizing the laws that govern energy changes and the spontaneity of chemical processes. Together, these topics provide a comprehensive framework for analyzing chemical systems in both theoretical and practical contexts.