25.1 Acids and Bases
- understand and use the terms conjugate acid and conjugate base
- define conjugate acid–base pairs, identifying such pairs in reactions
- define the terms pH, Ka, pKa and Kw mathematically and use them in calculations (Kb and the equation Kw = Ka × Kb will not be tested)
- calculate [H+(aq)] and pH values for: (a) strong acids (b) strong alkalis (c) weak acids
- (a) define a buffer solution (b) explain how a buffer solution can be made (c) explain how buffer solutions control pH, using chemical equations (d) describe and explain the uses of buffer solutions, including the role of HCO₃⁻ in controlling pH in blood
- calculate relevant concentrations and the pH of buffer solutions, given appropriate data
- understand and use the term solubility product, Ksp
- write an expression for Ksp
- calculate Ksp from concentrations and vice versa
- (a) understand and use the common ion effect to explain the different solubility of a compound in a solution containing a common ion (b) perform calculations using Ksp values and concentration of a common ion
- Write down the definition of a Brønsted–Lowry acid and Brønsted–Lowry base. Compare your definitions with those of another learner.
- Explain the meaning of: equilibrium, equilibrium constant, dissociation, strong acid, strong base, weak acid, weak base, neutralisation.
- Make a list of strong acids and weak acids, and strong bases and weak bases.
- Write an equilibrium expression for a weak acid — the process is the same as for Kc, just written as Ka instead.
- Explain why the equation H⁺ + OH⁻ → H₂O applies to any acid–alkali reaction.
- Describe how you would carry out an experiment to find the enthalpy change of neutralisation of an acid by an alkali.
- Explain the meaning of: solute, solution, salt, sparingly soluble, polarisation, non-polar, enthalpy change, hydration.
1Introduction — Acids and Bases ▾
Forget “acids are sour and dangerous.” The real definition is about one thing only: protons (H⁺ ions).
An acid is a proton donor (gives away H⁺). A base is a proton acceptor (takes H⁺).
🔄 Conjugate Acid–Base Pairs
Every time an acid loses a proton, what’s left behind can accept a proton right back — that makes it a conjugate pair. The two species in a pair differ by exactly one H⁺.
acid₁ base₂ base₁ acid₂
HA and A⁻ are one conjugate pair. B and BH⁺ are the other conjugate pair.
An antacid tablet works exactly like this: it contains a base (e.g. CaCO₃ or Mg(OH)₂) that accepts H⁺ from excess stomach acid (HCl), calming heartburn in minutes.
💪 Strong vs Weak — It’s Not About Concentration!
Common mix-up: “strong” and “concentrated” are not the same thing. Strength is about how completely an acid dissociates into ions — nothing to do with how much of it you dissolved.
A dilute strong acid can still be “strong” (fully dissociated), and a concentrated weak acid is still “weak” (only partially dissociated). Strength = extent of dissociation.
Strong acids and bases — fully dissociated
| Type | Example | Dissociation equation |
|---|---|---|
| Strong acid | Hydrochloric acid | HCl → H⁺ + Cl⁻ |
| Strong acid | Nitric acid | HNO₃ → H⁺ + NO₃⁻ |
| Strong acid (diprotic) | Sulfuric acid | H₂SO₄ → 2H⁺ + SO₄²⁻ |
| Strong base | Sodium hydroxide | NaOH → Na⁺ + OH⁻ |
| Strong base | Potassium hydroxide | KOH → K⁺ + OH⁻ |
Weak acids and bases — only partly dissociated
| Type | Example | Dissociation equation |
|---|---|---|
| Weak acid | Ethanoic acid | CH₃COOH ⇌ H⁺ + CH₃COO⁻ |
| Weak acid | Nitrous acid | HNO₂ ⇌ H⁺ + NO₂⁻ |
| Weak acid | Carbonic acid | H₂CO₃ ⇌ H⁺ + HCO₃⁻ |
| Weak base | Ammonia | NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ |
Notice the arrow: → for complete (strong) dissociation, ⇌ for partial, reversible (weak) dissociation.
Vinegar is only about 5% ethanoic acid, yet it’s a weak acid even at higher concentrations — that’s why you can safely put it on chips, while battery acid (dilute sulfuric acid) is dangerously corrosive at a similar concentration because it’s strong.
2Ionic Product of Water and pH ▾
Ionic Product of Water
Pure water conducts electricity slightly, so it must be ionised to a small extent:
Because [H₂O(l)] is a constant, we can define a new constant called the ionic product of water, Kw:
Because [H₃O⁺] = [H⁺(aq)] (hydrogen ions in water are all hydrated), this expression is often simplified to Kw = [H⁺][OH⁻].
The ionic product of water, Kw is the equilibrium constant for the dissociation of water at 298 K. Kw = [H⁺][OH⁻]
📏 Calculation of pH
- Acids have a pH below 7.0
- Pure water is neutral with a pH of 7.0
- Bases and alkalis have a pH above 7.0
Neutral means [H⁺] = [OH⁻]. At 25 °C, this happens at pH 7.00 because Kw = 1.00 × 10⁻¹⁴. At other temperatures, Kw changes, so the pH of neutral water may not be 7.00.
pH of strong acids
They ionise completely. Therefore, for a strong monoprotic acid: [H⁺] ≈ concentration of acid
Find the pH of 0.010 mol dm⁻³ HCl.
pH = −log₁₀(0.010)
Answer: pH = 2.00
pH of strong alkalis
They ionise completely to produce OH⁻. Use [H⁺][OH⁻] = Kw, then calculate pH using pH = −log[H⁺].
Find the pH of 0.0050 mol dm⁻³ NaOH.
[H⁺] = Kw ÷ [OH⁻] = (1.0 × 10⁻¹⁴) ÷ 0.0050 = 2.0 × 10⁻¹² mol dm⁻³
pH = −log₁₀(2.0 × 10⁻¹²)
Answer: pH = 11.70
pH of weak acids
Acid dissociation constant, Ka: the equilibrium constant for the dissociation of a weak acid.
pKa values: the values of Ka expressed as a logarithm of base 10. A small “p” in front of a symbol related to acids and bases means the negative logarithm (to base 10) — so pH is −log₁₀[H⁺] and pKa = −log₁₀Ka.
Ethanoic acid, Ka = 1.7 × 10⁻⁵ mol dm⁻³, concentration = 0.10 mol dm⁻³. Find the pH.
[H⁺] = √(Ka × [HA]) = √(1.7 × 10⁻⁵ × 0.10) = 1.30 × 10⁻³ mol dm⁻³
pH = −log₁₀(1.30 × 10⁻³)
Answer: pH = 2.89
Percentage ionisation
- The greater the Ka value, the more strongly acidic the acid is
- The greater the pKa value, the less strongly acidic the acid is
3Buffer Solutions ▾
Buffer solution: a solution in which the pH does not change significantly when small amounts of acids and bases are added — that is, a solution that minimises changes in pH when small amounts of acid or base are added.
How is a buffer made?
- a weak acid – conjugate base (e.g. CH₃COOH + CH₃COO⁻ from CH₃COONa)
- a weak base – conjugate acid (e.g. NH₃ + NH₄⁺)
Method 1 — mix a weak acid with its salt
Method 2 — partially neutralise a weak acid with a strong alkali
How does a buffer work?
Added acid: CH₃COO⁻ + H⁺ → CH₃COOH
- added H⁺ is removed
- [H⁺] does not increase much
- pH changes only slightly
Added alkali: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
- OH⁻ is removed
- [OH⁻] does not increase much
- pH changes only slightly
🩸 How your blood stays at pH 7.35–7.45
Your cells constantly produce CO₂. It reacts with water in blood plasma to form the HCO₃⁻ / CO₂ buffer pair:
- If H⁺ rises → equilibrium shifts LEFT, mopping up the excess H⁺
- If H⁺ falls → equilibrium shifts RIGHT, releasing more H⁺
If this buffer fails, blood pH drops and acidosis can result — severe cases can cause organ malfunction and even coma.
Calculating buffer pH
A buffer contains 0.20 mol dm⁻³ CH₃COOH and 0.10 mol dm⁻³ CH₃COONa. Ka = 1.7 × 10⁻⁵ mol dm⁻³.
pH = −log₁₀(3.4 × 10⁻⁵)
Answer: pH = 4.47
4Solubility Product and Common Ion Effect ▾
In a saturated solution, an equilibrium exists between the undissolved solid and its dissolved ions. At equilibrium, the rate at which ions leave the solid equals the rate at which ions return to the solid. For example:
Solubility product, Ksp: the product of the concentrations of each ion in a saturated solution of a sparingly soluble salt at 298 K, raised to the power of their relative concentrations.
For example, for PbCl₂ ⇌ Pb²⁺ + 2Cl⁻:
The solubility product is only relevant for salts which are sparingly (very slightly) soluble. It cannot be used for soluble salts such as sodium chloride. Soluble salts include salts of Group I elements, all nitrates and ammonium salts, and many sulfates. Halides are generally soluble except for lead(II) halides and silver halides.
For a general salt C(s) ⇌ aAx⁺(aq) + bBy⁻(aq):
- Smaller Ksp generally means a less soluble salt
- Solubility tells us how much salt dissolves
- Ksp tells us about the equilibrium between the solid and its ions
The common ion effect
The common ion effect: the reduction of the solubility of a dissolved salt by adding a compound that has an ion in common with the dissolved salt — e.g. adding sodium chloride to a solution of very slightly soluble lead(II) chloride.
Example: AgCl(s) ⇌ Ag⁺ + Cl⁻
Adding KCl increases [Cl⁻], shifting the equilibrium left and causing AgCl to precipitate.
- Precipitate forms if ion product > Ksp
- If ion product < Ksp → no precipitate
The shell of this nautilus is composed mainly of calcium carbonate. The nautilus adjusts conditions so that shell material forms only when the concentration of calcium ions and carbonate ions in seawater is high enough to precipitate calcium carbonate.
Kidney stones often form from calcium oxalate once its Ksp is exceeded in urine.
Calculations involving titrations also draw on these solubility ideas — make sure you can combine Ksp expressions with concentration data from a titration.
🔑 Key Formulas at a Glance
| Quantity | Formula |
|---|---|
| Ionic product of water | Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ (25 °C) |
| pH | pH = −log₁₀[H⁺] [H⁺] = 10⁻ᵖᴴ |
| Acid dissociation constant | Ka = [H⁺][A⁻] ⁄ [HA] |
| pKa | pKa = −log₁₀ Ka |
| [H⁺] of a weak acid | [H⁺] = √(Ka × [HA]) |
| Buffer [H⁺] | [H⁺] = Ka × ([acid] ⁄ [salt]) |
| Buffer pH | pH = pKa + log₁₀([salt] ⁄ [acid]) |
| Solubility product | Ksp = product of ionic concentrations, each to the power of its relative amount |
- Strong/weak = extent of dissociation, NOT concentration
- → means complete dissociation; ⇌ means partial, reversible dissociation
- Buffer explanations need equations AND equilibrium-shift reasoning
- Ksp only applies to sparingly soluble salts
- Kw only valid at 298 K unless stated otherwise
- Strong acids/alkalis: fully dissociated (→ not ⇌)
- Weak acids: use Ka, NEVER assume full dissociation
- pKa and Ka have an inverse relationship (lower pKa = stronger acid)
- Ksp only for sparingly soluble salts (not Group 1 salts, nitrates, ammonium salts)
- Common ion effect: equilibrium shifts left when a common ion is added
- Buffer pH formula: learn both forms ([H⁺] = Ka × [acid]/[salt] AND pH = pKa + log([salt]/[acid]))
- For buffer calculations: [salt] ≈ initial salt concentration, [acid] ≈ initial acid concentration