Difficulty: Intermediate | Prerequisites: IMF fundamentals, basic equilibrium concepts, logarithms.
Acid-base chemistry is one of the most heavily tested topics in general chemistry. It builds on equilibrium principles and connects molecular structure to solution behaviour. You need to be comfortable with the concept of chemical equilibrium (forward and reverse reactions occurring simultaneously) and with using logarithms before diving in. This material leads directly into buffers and titrations, so mastering it here saves significant time later.
Acids donate protons, bases accept them. Strong acids dissociate completely; weak acids establish an equilibrium described by Ka. The pH scale is a logarithmic measure of hydronium ion concentration. Ka and Kb for a conjugate pair always multiply to give Kw (1.0 × 10⁻¹⁴). Knowing whether a dissolved salt produces an acidic, basic, or neutral solution requires you to trace the ions back to their parent acid and base.
Brønsted-Lowry acid
A species that donates a proton (H⁺) to another species.
In simple terms: it gives away a hydrogen ion.
Brønsted-Lowry base
A species that accepts a proton (H⁺) from another species.
In simple terms: it takes a hydrogen ion.
Conjugate acid
The species formed when a base accepts a proton. It has one more H than the original base.
Think of it as: the base after it has picked up an H⁺.
Conjugate base
The species formed when an acid donates a proton. It has one fewer H than the original acid.
Think of it as: the acid after it has given away an H⁺.
Ka (acid dissociation constant)
The equilibrium constant for the dissociation of a weak acid in water. A larger Ka means a stronger acid (more dissociation).
In simple terms: Ka tells you what fraction of the acid molecules actually let go of their proton in water.
Kb (base dissociation constant)
The equilibrium constant for the reaction of a weak base with water. A larger Kb means a stronger base.
Kw (ion-product constant of water)
Kw = Ka × Kb = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C. This links the strength of a conjugate acid-base pair.
In simple terms: if you know Ka for an acid, you can find Kb for its conjugate base (and vice versa) by dividing Kw by the one you have.
pH
pH = −log[H₃O⁺]. The lower the pH, the more acidic the solution. pH 7 is neutral at 25 °C.
pOH
pOH = −log[OH⁻]. Related to pH by the equation pH + pOH = 14.00 at 25 °C.
pKa
pKa = −log(Ka). A lower pKa means a stronger acid. Useful for comparing acid strengths and for the Henderson-Hasselbalch equation.
Strong acid
An acid that dissociates completely (100%) in water. There are seven to memorise: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₃, HClO₄.
Weak acid
An acid that only partially dissociates in water. An equilibrium exists between the undissociated acid (HA) and the ions (H₃O⁺ and A⁻).
Percent ionisation
The fraction of acid molecules that dissociate, expressed as a percentage: ([H₃O⁺] / [HA]initial) × 100.
In simple terms: out of 100 acid molecules you put in, how many actually let go of their proton.
Equilibrium
The state in which the forward and reverse reactions occur at equal rates. Concentrations remain constant, but the reaction has not stopped.
Think of it as: a tug-of-war where both sides are pulling with equal force, so the rope does not move, but nobody has let go.
For a reaction aA + bB ⇌ cC + dD, the equilibrium expression is:
K = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
Pure liquids and pure solids are excluded (their activity is 1).
Example: for NH₃(g) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)
Kb = [NH₄⁺][OH⁻] / [NH₃]
H₂O(l) is omitted because it is a pure liquid.
In any Brønsted-Lowry acid-base reaction, there are two conjugate pairs. For the reaction above:
NH₃ is the base (it accepts H⁺ from water).
H₂O is the acid (it donates H⁺ to ammonia).
NH₄⁺ is the conjugate acid (NH₃ after gaining H⁺).
OH⁻ is the conjugate base (H₂O after losing H⁺).
To find a conjugate acid: add H⁺ to the species. To find a conjugate base: remove H⁺ from the species.
Examples:
Conjugate acid of HSO₄⁻ → H₂SO₄
Conjugate acid of H₂O → H₃O⁺
Conjugate acid of CH₃COO⁻ → CH₃COOH
Conjugate base of HSO₄⁻ → SO₄²⁻
Conjugate base of H₂O → OH⁻
Conjugate base of HClO₃ → ClO₃⁻
For any conjugate acid-base pair: Ka × Kb = Kw = 1.0 × 10⁻¹⁴
Example: Kb for NH₃ = 1.71 × 10⁻⁵. Therefore Ka for NH₄⁺ = Kw / Kb = (1.0 × 10⁻¹⁴) / (1.71 × 10⁻⁵) = 5.85 × 10⁻¹⁰.
At equilibrium, the reaction has not stopped. Both the forward and reverse reactions continue at equal rates, so concentrations remain constant. "The reaction has stopped" is a false statement about equilibrium.
When interpreting particle diagrams at equilibrium, count the number of X particles and Y particles, then check whether K = [products] / [reactants] (raised to the appropriate powers) matches the given value.
Adding NaOH to the NH₃/NH₄⁺/OH⁻ equilibrium increases [OH⁻], which increases pOH... wait, it decreases pOH and therefore increases pH. More directly: adding a strong base raises the pH.
Strong base pH example: 0.075 M Ca(OH)₂
Ca(OH)₂ is a strong base; it dissociates completely.
Each formula unit produces 2 OH⁻ ions: [OH⁻] = 2 × 0.075 = 0.150 M
pOH = −log(0.150) = 0.824
pH = 14.00 − 0.824 = 13.18
Finding [H₃O⁺] from pH: if pH = 3.12, then [H₃O⁺] = 10⁻³·¹² = 7.59 × 10⁻⁴ M
pOH and hydronium concentration: a solution with pOH = 12 has pH = 14 − 12 = 2, meaning [H₃O⁺] = 10⁻² = 0.01 M. A solution with pOH = 2 has pH = 12, so [H₃O⁺] = 10⁻¹² M. The highest hydronium concentration belongs to the solution with the highest pH value subtracted from 14, so pOH = 12 gives the most H₃O⁺.
For a weak acid HA with initial concentration C and dissociation constant Ka:
Set up the ICE table for HA + H₂O ⇌ H₃O⁺ + A⁻.
At equilibrium: [H₃O⁺] = [A⁻] = x, and [HA] = C − x.
Ka = x² / (C − x). If x is small relative to C (the 5% approximation), simplify to Ka ≈ x² / C, so x = √(Ka × C).
pH = −log(x).
Example: 0.60 M HCOOH, Ka = 1.7 × 10⁻⁴
x = √(1.7 × 10⁻⁴ × 0.60) = √(1.02 × 10⁻⁴) = 0.01010 M
pH = −log(0.01010) ≈ 2.00
Work the ICE table backwards. For HNO₂ with Ka = 4.5 × 10⁻⁴ and pH = 2.14:
[H₃O⁺] = 10⁻²·¹⁴ = 7.24 × 10⁻³ M = x
Ka = x² / (C − x), so 4.5 × 10⁻⁴ = (7.24 × 10⁻³)² / (C − 7.24 × 10⁻³)
Solving: C ≈ 0.124 M
Memorize these: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₃, HClO₄.
Everything else is a weak acid unless told otherwise.
Larger Ka = stronger acid = lower pKa = lower pH (at the same concentration).
For the reaction CH₃COOH + ClO₂⁻ ⇌ CH₃COO⁻ + HClO₂:
CH₃COOH: Ka = 1.8 × 10⁻⁵, so pKa = −log(1.8 × 10⁻⁵) ≈ 4.74
HClO₂: Ka = 1.1 × 10⁻², so pKa = −log(1.1 × 10⁻²) ≈ 1.96
HClO₂ is the stronger acid (larger Ka, smaller pKa).
Equilibrium favours the side with the weaker acid and weaker base. Since HClO₂ is stronger than CH₃COOH, equilibrium favours the left (reactant) side, because the products include HClO₂ which would rather donate its proton back.
For three acids HCl (strong), HC₉H₇O₄ (Ka = 3.0 × 10⁻⁴), and H₂CO₃ (Ka = 4.2 × 10⁻⁷):
At equal concentrations, the weakest acid (smallest Ka) produces the fewest H₃O⁺ ions and has the highest pH. That is H₂CO₃.
But when concentrations differ, you must calculate [H₃O⁺] for each. For 0.0020 M HCl: [H₃O⁺] = 0.0020 M, pH = 2.70. For 1.4 M HC₉H₇O₄: x = √(3.0 × 10⁻⁴ × 1.4) ≈ 0.0205, pH ≈ 1.69. For 1.8 M H₂CO₃: x = √(4.2 × 10⁻⁷ × 1.8) ≈ 8.69 × 10⁻⁴, pH ≈ 3.06.
The most acidic (lowest pH) is 1.4 M HC₉H₇O₄.
When an ionic compound dissolves, trace each ion back to its parent acid or base:
Ion from a strong acid or strong base: spectator ion, no effect on pH.
Cation from a weak base (e.g. NH₄⁺): acts as a weak acid in water, lowers pH.
Anion from a weak acid (e.g. CH₃COO⁻, CO₃²⁻, F⁻): acts as a weak base in water, raises pH.
Small, highly charged metal cations (e.g. Cu²⁺, Fe³⁺, Al³⁺): polarise water molecules and release H⁺, lowering pH.
Examples:
CaCO₃ → Ca²⁺ (from strong base Ca(OH)₂, spectator) + CO₃²⁻ (from weak acid H₂CO₃, acts as base). pH increases.
Cu(NO₃)₂ → Cu²⁺ (small, highly charged metal cation, acidic) + NO₃⁻ (from strong acid HNO₃, spectator). pH decreases.
pH = −log[H₃O⁺]
pOH = −log[OH⁻]
pH + pOH = 14.00 (at 25 °C)
[H₃O⁺] = 10⁻ᵖᴴ
[OH⁻] = 10⁻ᵖᴼᴴ
Kw = Ka × Kb = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴
% ionisation = ([H₃O⁺] / [HA]initial) × 100
pKa = −log(Ka)
Ka = x² / (C − x) for weak acid ICE table (x = [H₃O⁺] at equilibrium)
The pH of blood is tightly regulated near 7.4 by buffer systems, and even small deviations can be life-threatening, which is why understanding weak acid equilibria matters in medicine. Antacid tablets work by introducing a weak base (often CaCO₃ or Mg(OH)₂) that reacts with excess stomach acid (HCl), raising the pH.
Students often believe that at equilibrium the reaction has stopped. It has not. Both forward and reverse reactions continue; their rates are simply equal.
A common error is forgetting to exclude pure liquids and solids from equilibrium expressions. H₂O(l) does not appear in Ka or Kb expressions.
Students sometimes assume that a weak acid with a large concentration must produce more H₃O⁺ than a strong acid at low concentration. This is not always true; you must calculate [H₃O⁺] for each.
Confusing pKa direction: a lower pKa means a stronger acid. Students often reverse this.
⚠️ Identifying conjugate pairs in a reaction is a high-frequency exam question. Practice adding and removing H⁺ systematically.
⚠️ Strong base pH calculations require you to account for stoichiometry first (e.g. Ca(OH)₂ gives 2 OH⁻ per formula unit).
⚠️ ICE table problems appear in multiple forms: given Ka and concentration, find pH; given pH and Ka, find concentration. Know both directions.
⚠️ Salt-solution pH questions test whether you can trace each ion back to its parent acid or base and predict whether the solution is acidic, basic, or neutral.
⚠️ The seven strong acids must be memorised. If it is not on the list, treat it as weak.
True or false: a Brønsted-Lowry base donates protons. (False; it accepts protons)
Fill in the blank: Ka × Kb = ___ for a conjugate pair. (Kw = 1.0 × 10⁻¹⁴)
True or false: at equilibrium, the forward and reverse reactions have both stopped. (False; both reactions continue at equal rates)
Fill in the blank: the conjugate base of H₂O is ___. (OH⁻)
True or false: a lower pKa indicates a weaker acid. (False; lower pKa = stronger acid)
Q: Write the equilibrium expression for NH₃(g) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq).
A: Kb = [NH₄⁺][OH⁻] / [NH₃]. Water is excluded as a pure liquid.
Q: In the reaction above, identify the acid, base, conjugate acid, and conjugate base.
A: Acid = H₂O (donates H⁺). Base = NH₃ (accepts H⁺). Conjugate acid = NH₄⁺. Conjugate base = OH⁻.
Q: If Kb for NH₃ is 1.71 × 10⁻⁵, what is Ka for NH₄⁺?
A: Ka = Kw / Kb = (1.0 × 10⁻¹⁴) / (1.71 × 10⁻⁵) = 5.85 × 10⁻¹⁰.
Q: Calculate the pH of a 0.075 M Ca(OH)₂ solution.
A: [OH⁻] = 2 × 0.075 = 0.150 M. pOH = −log(0.150) = 0.824. pH = 14.00 − 0.824 = 13.18.
Q: What is [H₃O⁺] in a solution with pH 3.12?
A: [H₃O⁺] = 10⁻³·¹² = 7.59 × 10⁻⁴ M.
Q: Three solutions have pOH values of 2, 7, and 12. Which has the highest [H₃O⁺]?
A: pOH = 12 → pH = 2 → [H₃O⁺] = 0.01 M. This is the highest.
Q: Find the pH of 0.60 M formic acid (HCOOH, Ka = 1.7 × 10⁻⁴).
A: x = √(1.7 × 10⁻⁴ × 0.60) ≈ 0.0101 M. pH ≈ 2.00.
Q: Which is the stronger acid, CH₃COOH (Ka = 1.8 × 10⁻⁵) or HClO₂ (Ka = 1.1 × 10⁻²)?
A: HClO₂. It has the larger Ka.
Q: What happens to pH when CaCO₃ dissolves in water?
A: pH increases. CO₃²⁻ is the conjugate base of the weak acid H₂CO₃, so it acts as a base.
Q: What happens to pH when Cu(NO₃)₂ dissolves in water?
A: pH decreases. Cu²⁺ is a small, highly charged metal cation that polarises water and releases H⁺. NO₃⁻ is a spectator ion (from strong acid HNO₃).
The Ka and Kb relationship feeds directly into buffer calculations (Henderson-Hasselbalch uses pKa). Equilibrium concepts return in solubility product (Ksp) problems later in the course. Understanding which ions are acidic or basic in water is essential for predicting the pH at the equivalence point of a titration.
Brønsted-Lowry acid, Brønsted-Lowry base, proton donor, proton acceptor, conjugate acid, conjugate base, Ka, Kb, Kw, ion-product constant, pH, pOH, pKa, strong acid, weak acid, percent ionisation, percent ionization, ICE table, equilibrium expression, equilibrium constant, seven strong acids, salt hydrolysis, acidic salt, basic salt, neutral salt, hydronium, hydroxide, CHM 11200, general chemistry, Purdue