Introduction to Acids and Bases - AP Chemistry

SECTION 1

Historical Context & Motivation

The classification of substances as acids or bases is one of the oldest organizing schemes in chemistry, rooted in practical observations about taste, corrosivity, and the ability of certain solutions to neutralize one another. Early alchemists noted that acidic substances turned blue litmus red and dissolved metals, while alkaline solutions felt slippery and reversed litmus color changes. These empirical distinctions persisted for centuries before chemists developed molecular-level explanations for why these behaviors occur.

1661 • Robert Boyle's Empirical Definitions

Boyle catalogued observable properties of acids (sour taste, corrosive, turn litmus red) and bases (slippery, turn litmus blue), establishing the first systematic classification.

1884 • Arrhenius Theory

Svante Arrhenius proposed that acids produce H⁺ ions and bases produce OH⁻ ions upon dissolution in water, earning him the 1903 Nobel Prize in Chemistry.

1923 • Brønsted–Lowry Theory

Johannes Brønsted and Thomas Lowry independently defined acids as proton donors and bases as proton acceptors, extending acid–base chemistry beyond aqueous solutions.

1923 • Lewis Theory

Gilbert N. Lewis broadened the definition further: acids accept electron pairs, bases donate them. This framework encompasses reactions with no protons at all, such as BF₃ + NH₃.

The central question driving these successive refinements was: What is the fundamental chemical event that makes a substance acidic or basic? Each model offered a wider lens—from ions in water (Arrhenius) to proton transfer in any solvent (Brønsted–Lowry) to electron-pair transactions in all phases (Lewis). The AP Chemistry curriculum emphasizes the Brønsted–Lowry framework for equilibrium calculations while invoking Lewis theory for coordination chemistry and reaction mechanisms.

SECTION 2

Core Principles & Definitions

Three theoretical frameworks coexist in modern chemistry, each useful in different contexts. The AP exam expects you to move fluently among them, recognizing when to apply each definition and understanding their nested relationship: every Arrhenius acid is also a Brønsted–Lowry acid, and every Brønsted–Lowry acid is also a Lewis acid, but the reverse is not true.

Arrhenius Model

An acid dissociates in water to release H⁺ (or H₃O⁺); a base dissociates to release OH⁻. Applicable only in aqueous solutions. Example: HCl → H⁺ + Cl⁻; NaOH → Na⁺ + OH⁻.

Brønsted–Lowry Model

An acid is a proton (H⁺) donor; a base is a proton acceptor. Every acid–base reaction involves a conjugate acid–base pair. Works in any solvent or gas phase.

Lewis Model

An acid is an electron-pair acceptor; a base is an electron-pair donor. This is the most general framework and includes metal-ion coordination, BF₃ + NH₃, and more.

Conjugate Pairs

When an acid donates a proton it becomes its conjugate base; when a base accepts a proton it becomes its conjugate acid. A strong acid has a very weak conjugate base.

Amphiprotic Species

Substances like water (H₂O) and the hydrogen carbonate ion (HCO₃⁻) can act as either an acid or a base depending on the reaction partner.

SECTION 3

Visual Explanation — Proton Transfer in Action

In this Brønsted–Lowry reaction, HCl donates a proton (H⁺) to water. Two conjugate pairs form: HCl/Cl⁻ (pair 1) and H₂O/H₃O⁺ (pair 2). A strong acid like HCl reacts essentially to completion, producing a very weak conjugate base (Cl⁻).

The diagram above illustrates the essential Brønsted–Lowry mechanism: a proton donor (HCl) transfers H⁺ to a proton acceptor (H₂O). The proton never exists freely in solution; it is immediately captured by water's lone pair to form the hydronium ion (H₃O⁺). Notice that each reactant has a partner on the product side that differs by exactly one proton—these are conjugate pairs. Recognizing conjugate pairs is critical for writing equilibrium expressions and predicting the direction of proton-transfer equilibria.

SECTION 4

Mathematical Framework — pH, pOH, and Kw

Quantifying the acidity or basicity of a solution requires a logarithmic scale because hydronium concentrations span many orders of magnitude—from roughly 10 M in concentrated acid to 10⁻¹⁵ M in concentrated base. The pH scale compresses this enormous range into a manageable 0–14 span for typical aqueous systems at 25 °C.

pH DEFINITION

pH = −log[H₃O⁺]

where [H₃O⁺] is the molar concentration of hydronium ion. A low pH (< 7) indicates an acidic solution; a high pH (> 7) indicates a basic solution.

pOH DEFINITION

pOH = −log[OH⁻]

Analogous to pH but measuring hydroxide concentration. At 25 °C, pH + pOH = 14.00.

AUTOIONIZATION OF WATER

Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ (at 25 °C)

K_w is the ion-product constant for water. In pure water at 25 °C, [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ M, giving pH = 7.00. K_w increases with temperature, so neutral pH drops below 7 at elevated temperatures.

Ka AND Kb RELATIONSHIP

Ka × Kb = Kw

For any conjugate acid–base pair, the product of the acid dissociation constant (K_a) and the base dissociation constant (K_b) equals K_w. A stronger acid (larger K_a) necessarily has a weaker conjugate base (smaller K_b).

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AP Exam Tip

SECTION 5

Strong vs. Weak Acids and Bases

The distinction between strong and weak acids (or bases) is one of the most consequential classifications on the AP exam. A strong acid ionizes completely in dilute aqueous solution—there is no equilibrium to consider, and K_a is so large it is not tabulated. A weak acid only partially ionizes, establishing a dynamic equilibrium described by K_a. You must memorize the six common strong acids (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄) and the strong bases (Group 1 hydroxides and Ca(OH)₂, Sr(OH)₂, Ba(OH)₂).

Left: a strong acid (HCl) ionizes completely—no undissociated molecules remain. Right: a weak acid (acetic acid) reaches equilibrium with most molecules still intact. The particle-level representations highlight that concentration of H₃O⁺ is much lower for the weak acid at the same initial concentration.

Property	Strong Acid/Base	Weak Acid/Base
Ionization in water	Complete (→)	Partial (⇌)
Ka or Kb	Very large (not tabulated)	Small (10⁻² to 10⁻¹⁴)
[H₃O⁺] from 0.10 M soln	0.10 M (equals initial conc.)	≪ 0.10 M (requires ICE table)
Conjugate strength	Conjugate is negligibly weak	Conjugate has measurable strength
Common examples (acids)	HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄	CH₃COOH, HF, HCN, H₂CO₃

Comparison of strong and weak acids/bases

SECTION 6

Worked Example — pH of a Weak Acid

Calculate the pH of a 0.25 M acetic acid (CH₃COOH) solution. K_a = 1.8 × 10⁻⁵.

Worked example

1
Step 1 — Write the equilibrium expression

CH₃COOH(aq) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq). The equilibrium expression is K_a = [H₃O⁺][CH₃COO⁻] / [CH₃COOH] = 1.8 × 10⁻⁵.
2
Step 2 — Set up the ICE table

Let x = [H₃O⁺] at equilibrium. Initial: [CH₃COOH] = 0.25, [H₃O⁺] = 0, [CH₃COO⁻] = 0. Change: −x, +x, +x. Equilibrium: (0.25 − x), x, x.
3
Step 3 — Apply the 5% approximation

Check if x ≪ 0.25 by assuming 0.25 − x ≈ 0.25. Then K_a = x² / 0.25 = 1.8 × 10⁻⁵. Solving: x² = 4.5 × 10⁻⁶.

x = 2.12 × 10⁻³ M
4
Step 4 — Validate the approximation

(2.12 × 10⁻³ / 0.25) × 100% = 0.85%, which is well under 5%. The approximation is valid.
5
Step 5 — Calculate pH

pH = −log(2.12 × 10⁻³) = −(−2.674) = 2.67. This is consistent with a weak acid: the pH is below 7 but higher than it would be for a strong acid of the same concentration (which would give pH = 0.60).

pH = 2.67

SECTION 7

Strengths & Limitations of Each Model

Model	Strengths	Limitations
Arrhenius	Simple, intuitive; directly explains neutralization producing water; easy pH calculations for strong acids/bases in water.	Limited to aqueous solutions; cannot explain why NH₃ is basic (no OH⁻ in formula); cannot address gas-phase or non-aqueous reactions.
Brønsted–Lowry	Works in any solvent; explains conjugate pairs; handles amphiprotic species; directly ties to Ka/Kb equilibria tested on AP exam.	Still requires a transferable proton; cannot classify Lewis acids like BF₃ or metal cations that have no protons to donate.
Lewis	Most general; explains coordination bonds, electrophilic reactions, and why metal cations make solutions acidic; unifies organic and inorganic chemistry.	Does not inherently predict Ka values or pH; can be overly broad—many reactions could technically be classified as Lewis acid–base.

Comparison of acid–base models

SECTION 8

Connection to Advanced Topics

The fundamentals of acid–base theory laid out here form the scaffolding for several advanced AP Chemistry topics. Buffer solutions rely on conjugate acid–base pairs to resist pH changes; titration curves map pH against volume of added titrant, revealing equivalence points where moles of acid equal moles of base. Polyprotic acids like H₂SO₄ and H₃PO₄ undergo stepwise dissociation, each step with its own K_a value. Lewis acid–base theory underpins coordination chemistry, where metal ions act as Lewis acids accepting electron pairs from ligands.

This Lesson Covers	Advanced Extension
pH = −log[H₃O⁺] for monoprotic acids	Henderson–Hasselbalch equation for buffer pH: pH = pKa + log([A⁻]/[HA])
Ka expression and ICE table for one equilibrium	Polyprotic acid equilibria with Ka₁ ≫ Ka₂ ≫ Ka₃; stepwise ICE tables
Strong vs. weak classification	Titration curves: strong/strong, strong/weak, weak/weak; indicator selection
Lewis acid = electron-pair acceptor	Metal aqua-ion hydrolysis: [Fe(H₂O)₆]³⁺ donates H⁺ making solutions acidic

Introductory concepts and their advanced extensions

As you progress through the Acids and Bases unit, every new topic will reference the concepts from this lesson: conjugate pairs, K_a/K_b relationships, the autoionization of water, and the distinction between strong and weak. Mastering these fundamentals now will make buffer and titration problems significantly more approachable.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

In the reaction NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), which species is the Brønsted–Lowry acid?

A. NH₃
B. H₂O
C. NH₄⁺
D. OH⁻

Show answer (B)

B is correct. Water donates a proton to NH₃, making H₂O the Brønsted–Lowry acid. A is wrong because NH₃ accepts the proton, acting as the base. C (NH₄⁺) is the conjugate acid of NH₃ and is a product, not a reactant acting as the acid. D (OH⁻) is the conjugate base of water and is also a product. The conjugate pairs are NH₃/NH₄⁺ and H₂O/OH⁻.

Question 2: BASIC CALCULATION

What is the pH of a 0.0050 M HNO₃ solution at 25 °C?

A. 2.00
B. 2.30
C. 11.70
D. 5.00

Show answer (B)

B is correct. HNO₃ is a strong acid, so [H₃O⁺] = 0.0050 M. pH = −log(0.0050) = −log(5.0 × 10⁻³) = 3 − log 5.0 = 3 − 0.70 = 2.30. A (2.00) would correspond to [H₃O⁺] = 0.010 M, which is double the given concentration. C (11.70) is the pOH, not the pH—a common error of confusing the two scales. D (5.00) would correspond to 10⁻⁵ M H₃O⁺, far lower than expected for this concentration.

Question 3: INTERMEDIATE

A 0.10 M solution of a monoprotic weak acid HA has a pH of 3.00 at 25 °C. What is the Ka of HA?

A. 1.0 × 10⁻³
B. 1.0 × 10⁻⁵
C. 1.0 × 10⁻⁶
D. 1.0 × 10⁻²

Show answer (B)

B is correct. pH = 3.00 means [H₃O⁺] = 1.0 × 10⁻³ M. Since HA ⇌ H₃O⁺ + A⁻, at equilibrium x = 1.0 × 10⁻³. Ka = x²/(0.10 − x) ≈ (1.0 × 10⁻³)²/0.10 = 1.0 × 10⁻⁵. The 5% check: (10⁻³/0.10) × 100% = 1%, valid. A (10⁻³) confuses [H₃O⁺] with Ka. C (10⁻⁶) results from dividing x² by 1.0 instead of 0.10. D (10⁻²) would result from calculating x/0.10 rather than x²/(0.10 − x).

Question 4: APPLIED

Formic acid (HCOOH) has Ka = 1.8 × 10⁻⁴. A student prepares a 0.40 M HCOOH solution. (a) Write the equilibrium expression for the dissociation of formic acid in water. (b) Calculate the equilibrium concentration of H₃O⁺ and the pH of the solution. (c) Calculate the percent ionization of formic acid. (d) The student dilutes the solution to 0.040 M. Without performing a full calculation, predict whether the percent ionization will increase, decrease, or stay the same. Justify your answer.

Show answer

(a) Ka = [H₃O⁺][HCOO⁻]/[HCOOH] = 1.8 × 10⁻⁴. (b) ICE table: let x = [H₃O⁺] = [HCOO⁻] at equilibrium. Ka = x²/(0.40 − x). Assuming x ≪ 0.40: x² = (1.8 × 10⁻⁴)(0.40) = 7.2 × 10⁻⁵, so x = 8.49 × 10⁻³ M. Check: 8.49 × 10⁻³/0.40 = 2.1% < 5%, valid. pH = −log(8.49 × 10⁻³) = 2.07. (c) Percent ionization = ([H₃O⁺]/C₀) × 100% = (8.49 × 10⁻³/0.40) × 100% = 2.1%. (d) Percent ionization increases upon dilution. By Le Châtelier's principle, dilution effectively adds water (increasing volume), decreasing concentrations of all aqueous species, which shifts the equilibrium toward products. Mathematically, Ka = x²/(C₀ − x) is constant, so as C₀ decreases, x/C₀ must increase.

Question 5: CRITICAL THINKING

A student measures the pH of four 0.10 M solutions at 25 °C and records the following data: Solution A: pH = 1.00 Solution B: pH = 2.87 Solution C: pH = 4.74 Solution D: pH = 13.00 (a) Classify each solution as a strong acid, weak acid, weak base, or strong base. Justify each classification. (b) Calculate Ka for Solution C. (c) Identify the conjugate base of Solution C's acid and calculate its Kb. (d) If equal volumes of Solution A and Solution D are mixed, determine the pH of the resulting solution. Assume volumes are additive.

Show answer

(a) Solution A: strong acid. pH = 1.00 corresponds to [H₃O⁺] = 0.10 M, equal to the initial concentration, indicating complete ionization. Solution B: weak acid. pH = 2.87 gives [H₃O⁺] = 1.35 × 10⁻³ M, much less than 0.10 M, indicating partial ionization. Solution C: weak acid. pH = 4.74 gives [H₃O⁺] = 1.82 × 10⁻⁵ M, even less ionization. Solution D: strong base. pOH = 14.00 − 13.00 = 1.00, so [OH⁻] = 0.10 M = initial concentration. (b) For Solution C: Ka = x²/(0.10 − x) where x = 1.82 × 10⁻⁵. Since x ≪ 0.10, Ka ≈ (1.82 × 10⁻⁵)²/0.10 = 3.3 × 10⁻⁹. (c) Kb = Kw/Ka = (1.0 × 10⁻¹⁴)/(3.3 × 10⁻⁹) = 3.0 × 10⁻⁶. (d) Equal volumes of 0.10 M strong acid and 0.10 M strong base: moles H₃O⁺ = moles OH⁻ (both = 0.10 × V). Complete neutralization occurs. The resulting solution contains only water and spectator ions, so pH = 7.00.