Weak Acid and Base Equilibria

SECTION 1

Historical Context & Motivation

The study of acids and bases stretches back centuries, but the quantitative understanding of weak acid and base equilibria only crystallized in the late nineteenth and early twentieth centuries. Early chemists recognized that certain acids—vinegar, citrus juice—were far less corrosive than hydrochloric acid or sulfuric acid, yet they lacked a theoretical framework to explain why. The breakthrough came when scientists began applying the principles of chemical equilibrium to the ionization of solutes in water, revealing that the degree of dissociation—not merely the identity of the acid or base—determines acidity and basicity in solution.

1884

Arrhenius Theory of Electrolytic Dissociation

Svante Arrhenius proposed that acids produce H⁺ ions and bases produce OH⁻ ions in water, providing the first systematic classification of acid and base strength.

1907

The pH Scale

Søren Sørensen introduced the pH scale at the Carlsberg Laboratory, giving chemists a logarithmic measure of hydrogen ion concentration that made weak acid behavior experimentally tractable.

1909

Ostwald's Dilution Law

Wilhelm Ostwald quantified how the degree of dissociation of a weak electrolyte varies with concentration, linking equilibrium constants to observable conductivity data.

1923

Brønsted–Lowry Definition

Johannes Brønsted and Thomas Lowry independently defined acids as proton donors and bases as proton acceptors, broadening the concept beyond aqueous solutions and enabling analysis of conjugate pairs.

1966

Modern Computational Chemistry

Ab initio calculations began to predict pKₐ values from molecular structure, bridging the gap between empirical equilibrium data and quantum mechanical theory.

The central question this lesson addresses is deceptively simple: if a weak acid does not fully ionize, how do we calculate the equilibrium concentrations of all species in solution? Answering this question requires fluency with equilibrium expressions, ICE tables, and the mathematical approximations that make these calculations manageable on the AP Chemistry exam.

SECTION 2

Core Principles & Definitions

A weak acid is a species that donates a proton to water but does so incompletely—establishing a dynamic equilibrium between the undissociated acid and its ions. Similarly, a weak base accepts a proton from water only partially. The quantitative measure of this tendency is the acid dissociation constant (Kₐ) or base dissociation constant (K_b), which encapsulate the position of the equilibrium at a given temperature.

Partial Ionization

Weak acids and bases ionize only partially in water. At equilibrium, significant concentrations of the molecular (undissociated) form remain, distinguishing them from strong acids and bases that ionize completely.

Conjugate Acid–Base Pairs

Every weak acid HA has a conjugate base A⁻, and every weak base B has a conjugate acid BH⁺. The strengths are inversely related: a stronger weak acid has a weaker conjugate base, and vice versa.

Kₐ and K_b Quantify Strength

Kₐ measures the extent of proton donation; K_b measures the extent of proton acceptance. Larger values mean greater ionization. For a conjugate pair, Kₐ × K_b = K_w = 1.0 × 10⁻¹⁴ at 25 °C.

The ICE Table Framework

Initial–Change–Equilibrium tables organize the algebra needed to solve for unknown concentrations, systematically tracking how the initial concentration shifts as the system reaches equilibrium.

The 5% Approximation

When the ratio of initial concentration to Kₐ (or K_b) exceeds roughly 100, the change x is small enough relative to the initial concentration that we can simplify the equilibrium expression by omitting x from the denominator.

✦ KEY TAKEAWAY

Think of a weak acid in water like a partially opened faucet: only a fraction of the available protons flow into solution. The equilibrium constant Kₐ tells you how far the faucet is opened. A large Kₐ (close to 1) means the faucet is nearly wide open—most molecules ionize. A tiny Kₐ (like 10⁻⁹) means only a trickle escapes. Understanding this "partial release" is the key to every calculation in this unit.

SECTION 3

Visual Explanation — Ionization Equilibrium

This diagram contrasts the complete ionization of a strong acid (HCl) with the partial ionization of a weak acid (acetic acid). In the weak acid solution, the vast majority of molecules remain undissociated, and the equilibrium expression Kₐ governs the relative concentrations of products and reactants.

The diagram above captures the essential distinction on which all weak acid/base calculations rest. On the left, HCl—a strong acid—dissociates completely, so no equilibrium expression is needed; the concentration of H⁺ equals the initial acid concentration. On the right, acetic acid (CH₃COOH) establishes a dynamic equilibrium in which the forward (ionization) and reverse (protonation) reactions proceed at equal rates. The double arrow (⇌) symbolizes this balance. Notice that the colored bars representing ions are far smaller for the weak acid, reflecting the low percent ionization—typically well below 10% for most weak acids at moderate concentrations. This is the visual intuition you should carry into every ICE-table calculation: the change x is small relative to the initial concentration.

SECTION 4

Mathematical Framework

Weak Acid Equilibrium Expression

Consider a generic monoprotic weak acid HA dissolving in water. The ionization reaction and its equilibrium expression are the foundation of every pH calculation in this unit.

WEAK ACID IONIZATION

HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)

HA = weak acid, A⁻ = conjugate base, H₃O⁺ = hydronium ion. Water is the solvent and does not appear in the equilibrium expression.

ACID DISSOCIATION CONSTANT

Kₐ = [H₃O⁺][A⁻] / [HA]

All concentrations are equilibrium molar concentrations (mol/L). A larger Kₐ indicates a stronger weak acid—one that ionizes to a greater extent.

Weak Base Equilibrium Expression

WEAK BASE IONIZATION

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

B = weak base (e.g., NH₃), BH⁺ = conjugate acid (e.g., NH₄⁺), OH⁻ = hydroxide ion.

BASE DISSOCIATION CONSTANT

K_b = [BH⁺][OH⁻] / [B]

K_b measures the extent of proton acceptance from water. For a conjugate pair, Kₐ × K_b = K_w = 1.0 × 10⁻¹⁴ at 25 °C.

The ICE Table Method

To solve for the equilibrium concentrations, we construct an ICE table (Initial, Change, Equilibrium). For a weak acid HA with initial concentration C₀ and assuming no initial H₃O⁺ or A⁻ beyond the autoionization of water (which is negligible), we let x represent the molar concentration of HA that ionizes. At equilibrium: [HA] = C₀ − x, [H₃O⁺] = x, and [A⁻] = x. Substituting into the Kₐ expression yields Kₐ = x² / (C₀ − x). If C₀ / Kₐ ≥ 100, the approximation C₀ − x ≈ C₀ is valid, simplifying to x = √(Kₐ × C₀). Always verify afterward that x / C₀ < 0.05 (the 5% rule); if not, use the quadratic formula.

SIMPLIFIED WEAK ACID FORMULA

[H₃O⁺] = x ≈ √(Kₐ × C₀)

Valid when C₀ / Kₐ ≥ 100. Then pH = −log[H₃O⁺] = −log(x).

⚠️ When the 5% Rule Fails

If the percent ionization exceeds 5%, you must solve the full quadratic: x² + Kₐx − KₐC₀ = 0, giving x = (−Kₐ + √(Kₐ² + 4KₐC₀)) / 2. On the AP exam, the College Board expects you to recognize when the approximation is valid and when the quadratic is required.

SECTION 5

Percent Ionization & Concentration Effects

The percent ionization of a weak acid is defined as ([H₃O⁺]_eq / C₀) × 100%. A crucial trend tested on the AP exam is that percent ionization increases as the initial concentration decreases. This counterintuitive result follows directly from Le Châtelier's principle: diluting the solution shifts the equilibrium toward greater ionization to partially restore equilibrium concentrations. While the absolute [H₃O⁺] decreases upon dilution, the fraction of acid that has ionized increases.

The amber curve shows that percent ionization rises as C₀ decreases. At 0.001 M acetic acid, about 4.2% of molecules ionize, whereas at 1.00 M only ~0.13% ionize. The cyan curve shows that despite higher percent ionization at low concentrations, the absolute [H₃O⁺] still decreases upon dilution.

This inverse relationship between concentration and percent ionization is a direct consequence of the equilibrium law. When you dilute a weak acid, the reaction quotient Q temporarily drops below Kₐ because all concentrations decrease, but the denominator (containing the single [HA] term) decreases more slowly than the numerator (containing the product [H₃O⁺][A⁻]). The system responds by shifting right, producing more ions until Q = Kₐ again. On the AP exam, you should be prepared to explain this trend qualitatively using Le Châtelier's principle and quantitatively using the simplified formula or the quadratic.

Acetic acid (Kₐ = 1.8 × 10⁻⁵) at various concentrations at 25 °C

C₀ (M)	[H₃O⁺] (M)	pH	% Ionization
1.00	4.2 × 10⁻³	2.37	0.42%
0.10	1.3 × 10⁻³	2.87	1.3%
0.010	4.2 × 10⁻⁴	3.38	4.2%
0.0010	1.3 × 10⁻⁴	3.89	13%

SECTION 6

Worked Example — Finding pH of a Weak Acid

Let us work through a complete calculation to find the pH of a 0.25 M solution of hydrofluoric acid (HF), given that Kₐ = 6.8 × 10⁻⁴ at 25 °C.

pH of 0.25 M HF

Step 1 — Write the Ionization EquationHF(aq) + H₂O(l) ⇌ H₃O⁺(aq) + F⁻(aq). The equilibrium expression is Kₐ = [H₃O⁺][F⁻] / [HF] = 6.8 × 10⁻⁴.

Step 2 — Construct the ICE TableInitial: [HF] = 0.25 M, [H₃O⁺] = 0, [F⁻] = 0. Change: [HF] decreases by x, [H₃O⁺] increases by x, [F⁻] increases by x. Equilibrium: [HF] = 0.25 − x, [H₃O⁺] = x, [F⁻] = x.

Step 3 — Check the 5% ApproximationC₀ / Kₐ = 0.25 / (6.8 × 10⁻⁴) = 368. Since 368 > 100, the approximation 0.25 − x ≈ 0.25 is likely valid. We proceed with the simplified equation: Kₐ = x² / 0.25.

Step 4 — Solve for xx² = Kₐ × C₀ = (6.8 × 10⁻⁴)(0.25) = 1.70 × 10⁻⁴. Taking the square root: x = √(1.70 × 10⁻⁴) = 0.0130 M. Therefore [H₃O⁺] = 0.0130 M.

[H₃O⁺] = 0.0130 M

Step 5 — Verify the ApproximationPercent ionization = (0.0130 / 0.25) × 100% = 5.2%. This is slightly above 5%, so for maximum accuracy you would use the quadratic. However, at 5.2% the approximation introduces only modest error, and many AP scoring guidelines accept this approach when the margin is small. Using the quadratic formula yields x = 0.0127 M—a difference of about 2%.

Step 6 — Calculate pHpH = −log(0.0130) = −(−1.89) = 1.89. Using the more precise quadratic result, pH = −log(0.0127) = 1.90. Either value would receive full credit on most AP free-response questions when work is shown.

pH ≈ 1.89

💡 AP Exam Tip

Always show your ICE table and state whether you are using the approximation or the quadratic. Even if your final numerical answer has a minor rounding error, you earn most of the rubric points for correct setup and logical reasoning.

SECTION 7

Comparing Weak Acids, Weak Bases, and Their Conjugates

One of the most powerful relationships in this unit is the conjugate pair relationship: for any weak acid HA and its conjugate base A⁻, the product Kₐ × K_b = K_w = 1.0 × 10⁻¹⁴ at 25 °C. This means knowing either Kₐ or K_b immediately gives you the other. The table below compares several common weak acids and bases with their conjugate partners, illustrating the inverse relationship between strength of an acid and strength of its conjugate base.

Selected conjugate acid–base pairs at 25 °C. Note Kₐ × K_b = 1.0 × 10⁻¹⁴ for each pair.

Weak Acid	Kₐ	Conjugate Base	K_b	Relative Strength
HF	6.8 × 10⁻⁴	F⁻	1.5 × 10⁻¹¹	Stronger acid → weaker conjugate base
CH₃COOH	1.8 × 10⁻⁵	CH₃COO⁻	5.6 × 10⁻¹⁰	Moderate acid → moderate conjugate base
HCN	6.2 × 10⁻¹⁰	CN⁻	1.6 × 10⁻⁵	Weaker acid → stronger conjugate base
NH₄⁺	5.6 × 10⁻¹⁰	NH₃	1.8 × 10⁻⁵	Weak conjugate acid of a well-known weak base

✦ KEY TAKEAWAY

The Kₐ–K_b relationship is like a seesaw: when one side (the acid) is strong, the other side (the conjugate base) must be weak, because their product is fixed at K_w. In engineering terms, this is a constraint equation—one degree of freedom is removed, so specifying Kₐ completely determines K_b. This relationship is essential for predicting whether salt solutions will be acidic, basic, or neutral.

SECTION 8

Connection to Buffer Systems and Polyprotic Acids

The weak acid/base equilibrium framework you have learned here is the foundation for two advanced topics that appear prominently on the AP Chemistry exam: buffer solutions and polyprotic acid equilibria. A buffer is simply a solution containing significant concentrations of both a weak acid and its conjugate base (or a weak base and its conjugate acid). The Henderson–Hasselbalch equation, pH = pKₐ + log([A⁻]/[HA]), is a direct algebraic rearrangement of the Kₐ expression. Polyprotic acids like H₂SO₃ or H₃PO₄ have multiple ionization steps, each with its own Kₐ, and the same ICE-table logic applies to each successive equilibrium.

From monoprotic weak acid equilibria to buffers and polyprotic systems

Concept	This Lesson (Monoprotic Weak Acids/Bases)	Advanced Extension
Species in solution	HA, A⁻, H₃O⁺ at equilibrium	Buffer: both HA and A⁻ at significant concentrations before equilibrium
Governing equation	Kₐ = x² / (C₀ − x)	Henderson–Hasselbalch: pH = pKₐ + log([A⁻]/[HA])
Number of equilibria	One ionization step	Polyprotic: 2–3 sequential steps (Kₐ₁ >> Kₐ₂ >> Kₐ₃)
pH response to dilution	pH increases (less acidic) upon dilution	Buffer: pH is resistant to dilution—ratio [A⁻]/[HA] is preserved
Typical AP FRQ	Calculate pH of a weak acid solution	Calculate pH after adding strong acid/base to a buffer; titration curve analysis

Mastering the single-equilibrium ICE-table calculation is non-negotiable before moving to these more complex systems. Every buffer problem begins by identifying the weak acid/conjugate base pair and their respective concentrations, and every polyprotic problem reduces, step by step, to repeated application of the Kₐ expression. The mathematical and conceptual tools you have developed in this lesson will serve as your scaffold for the rest of the acids-and-bases unit.

SECTION 9

Practice Problems

A student dissolves equal molar amounts of acetic acid (Kₐ = 1.8 × 10⁻⁵) and hydrofluoric acid (Kₐ = 6.8 × 10⁻⁴) in separate beakers of water at the same concentration and temperature. Which of the following correctly compares the two solutions at equilibrium?

What is the pH of a 0.50 M solution of formic acid (HCOOH, Kₐ = 1.8 × 10⁻⁴) at 25 °C?

A 0.020 M solution of a monoprotic weak acid HA has a pH of 3.40 at 25 °C. Which of the following is the best estimate of the Kₐ of this acid?

PROBLEM 4 — APPLIED

Ammonia (NH₃) is a weak base with K_b = 1.8 × 10⁻⁵ at 25 °C. A chemist prepares a 0.15 M ammonia solution for use in a cleaning product. (a) Write the equilibrium expression for the ionization of ammonia in water. (b) Using an ICE table, calculate the equilibrium concentration of OH⁻ in the solution. Clearly state and justify any approximations. (c) Calculate the pH of the solution. (d) The chemist dilutes the solution to 0.015 M. Without performing a full calculation, predict whether the percent ionization of ammonia will increase, decrease, or remain the same. Justify your answer using Le Châtelier's principle.

PROBLEM 5 — CRITICAL THINKING

A research team studies two unknown monoprotic weak acids, Acid X and Acid Y. They prepare 0.10 M solutions of each and measure the pH. They then dilute each solution to 0.010 M and measure the pH again. The data are shown in the table below. | Acid | pH at 0.10 M | pH at 0.010 M | |------|-------------|---------------| | X | 2.88 | 3.39 | | Y | 4.87 | 5.38 | (a) Calculate the Kₐ of Acid X using the 0.10 M data. Show all work. (b) Calculate the Kₐ of Acid Y using the 0.10 M data. Show all work. (c) For each acid, calculate the percent ionization at both concentrations. Organize your results in a table. (d) A student claims that for any weak acid, decreasing the concentration by a factor of 10 should increase the percent ionization by a factor of 10. Using the data from part (c), evaluate this claim. Provide a mathematical argument for why the student's claim is or is not generally valid.

SUMMARY

Lesson Summary

Weak acids and weak bases ionize only partially in water, establishing a dynamic equilibrium between the molecular form and its ions. The extent of ionization is quantified by Kₐ (for acids) and K_b (for bases), which are related by Kₐ × K_b = K_w = 1.0 × 10⁻¹⁴ at 25 °C for any conjugate pair. The ICE table is the systematic tool for calculating equilibrium concentrations: define x as the amount that ionizes, substitute into the equilibrium expression, and solve—using the 5% approximation (valid when C₀/Kₐ ≥ 100) or the quadratic formula when necessary.

Key trends to remember: percent ionization increases as initial concentration decreases (a consequence of Le Châtelier's principle), and a stronger weak acid has a weaker conjugate base. These principles underpin buffer chemistry, titration curve analysis, and polyprotic acid calculations—all of which build directly on the monoprotic equilibrium framework mastered in this lesson. On the AP exam, always show your ICE table, state your approximation, verify it with the 5% rule, and report pH to two decimal places.