Apply Proton Transfer Reactions — Apply Reactivity 3.1—Proton transfer reactions in problem-solving and explanations - IB Chemistry

SECTION 1

Historical Context & Motivation

For centuries, chemists noticed that certain substances tasted sour, turned litmus paper red, or reacted vigorously with metals. Others felt slippery and turned litmus blue. These observations were consistent, but nobody could explain why at the particle level. The quest to understand acid–base behavior ultimately led to one of chemistry's most powerful ideas: the proton transfer reaction. Grasping this concept lets you predict products, explain buffer systems, and solve quantitative problems across the IB Chemistry syllabus.

1884 • Arrhenius Definition

Svante Arrhenius proposed that acids produce H⁺ ions and bases produce OH⁻ ions in water, providing the first particle-level model of acid–base behavior.

1923 • Brønsted–Lowry Theory

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

1923 • Lewis Acid–Base Theory

Gilbert N. Lewis introduced an even broader definition based on electron-pair donation and acceptance, though the Brønsted–Lowry model remains central for proton transfer reactions.

1909 • pH Scale Introduced

Søren Sørensen developed the pH scale as a practical way to express hydrogen-ion concentration, making quantitative acid–base work far more accessible.

Modern IB • Proton Transfer in Problem-Solving

Today, IB Chemistry Reactivity 3.1 asks you to apply the proton transfer model to predict products, identify conjugate pairs, and explain observations in both familiar and unfamiliar contexts.

The key question that ties this history together is straightforward: when an acid meets a base, which particle moves, and where does it go? The Brønsted–Lowry answer—a proton (H⁺) transfers from the acid to the base—gives you a tool to write equations, identify conjugate pairs, and solve problems with confidence.

SECTION 2

Core Principles & Definitions

Before you can apply proton transfer reactions in problem-solving, you need a clear vocabulary. The Brønsted–Lowry model revolves around four connected ideas: what donates, what accepts, and what remains after the transfer. Every proton transfer produces a conjugate acid–base pair, linking the reactant side to the product side of the equation.

Brønsted–Lowry Acid

A species that donates a proton (H⁺) to another species. After donating, it becomes its conjugate base.

Brønsted–Lowry Base

A species that accepts a proton (H⁺) from another species. After accepting, it becomes its conjugate acid.

Conjugate Acid–Base Pair

Two species that differ by exactly one proton. For example, HCl and Cl⁻ form a conjugate pair, as do NH₃ and NH₄⁺.

Amphiprotic / Amphoteric Species

A substance like water (H₂O) that can act as either an acid or a base depending on the reaction partner.

Proton Transfer Direction

Protons transfer from the stronger acid to the stronger base. The reaction favors formation of the weaker acid and weaker base.

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KEY TAKEAWAY

Think of a proton transfer like passing a ball in a game. The acid is the player who throws the ball (H⁺), and the base is the player who catches it. After the throw, the thrower no longer has the ball (conjugate base) and the catcher now holds it (conjugate acid). Every single Brønsted–Lowry reaction is this simple exchange—one particle moves between two partners.

SECTION 3

Visual Explanation — Proton Transfer in Action

The following diagram illustrates the classic proton transfer between hydrochloric acid (HCl) and water (H₂O). Notice how the proton leaves the acid and attaches to the base, creating a conjugate acid–base pair on each side of the equation.

The pink-bordered species (HCl and Cl⁻) form one conjugate pair, while the cyan-bordered species (H₂O and H₃O⁺) form the other. The curved dashed arrow shows the proton traveling from the acid to the base.

In the diagram above, HCl donates its proton to water. Once HCl loses H⁺, it becomes Cl⁻ (the conjugate base of HCl). Water gains that proton and becomes H₃O⁺ (the conjugate acid of H₂O). Every Brønsted–Lowry reaction has exactly two conjugate pairs. Identifying these pairs is often the first step in any IB problem.

SECTION 4

Mathematical Framework — pH, Ka, and Equilibrium

Proton transfer reactions are not just qualitative—many IB problems require you to calculate pH or use the acid dissociation constant (Kₐ). These equations describe how far a proton transfer proceeds at equilibrium.

PH DEFINITION

pH = −log₁₀[H⁺]

[H⁺] is the hydrogen-ion concentration in mol dm⁻³ (M). A lower pH means a higher concentration of H⁺ ions, indicating a more acidic solution.

ACID DISSOCIATION CONSTANT

Kₐ = [H⁺][A⁻] / [HA]

HA represents the undissociated acid, A⁻ is the conjugate base, and [H⁺] is the hydronium-ion concentration. A large Kₐ means the acid is strong and donates protons readily.

IONIC PRODUCT OF WATER

Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 25 °C

This constant links the proton transfer self-ionisation of water: H₂O + H₂O ⇌ H₃O⁺ + OH⁻. At 25 °C, pH + pOH = 14.

pKa RELATIONSHIP

pKₐ = −log₁₀Kₐ

A smaller pKₐ corresponds to a stronger acid. This value is commonly found in the IB Data Booklet and is useful for comparing acid strengths.

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IB Exam Tip

On the IB exam, you will often be told to assume that the concentration of a strong acid equals [H⁺] directly. For weak acids, you must set up an ICE table or use the Kₐ expression. Always check whether the acid is strong or weak before choosing your approach.

SECTION 5

Classifying Acid–Base Strength and Predicting Reaction Direction

A critical skill in IB Chemistry is predicting which direction a proton transfer will favour. The rule is elegant: protons move from the stronger acid to the stronger base, producing the weaker conjugate pair. Equilibrium always lies on the side of the weaker acid and weaker base.

The left column lists acids from strongest (top) to weakest (bottom). Their conjugate bases appear in the right column in inverse order: the strongest acid produces the weakest conjugate base. Proton transfer proceeds from the stronger acid to the stronger base, favouring products when the stronger acid and stronger base appear on the same side of the equation.

Category	Examples	Key Characteristic
Strong Acid	HCl, HNO₃, H₂SO₄ (first proton)	Ionises completely in water; Kₐ is very large
Weak Acid	CH₃COOH, H₂CO₃, HF	Ionises partially; Kₐ < 1; equilibrium favours HA
Strong Base	NaOH, KOH, Ba(OH)₂	Dissociates completely; provides OH⁻ which accepts H⁺
Weak Base	NH₃, CH₃NH₂	Accepts protons partially; Kb < 1; equilibrium favours reactants

Classification of common acids and bases for IB Chemistry

SECTION 6

Worked Example — Proton Transfer Problem

Let's walk through a typical IB-style problem that asks you to identify conjugate pairs, write the proton transfer equation, and calculate the pH.

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Problem Statement

A 0.10 mol dm⁻³ solution of ethanoic acid (CH₃COOH) has a Kₐ of 1.8 × 10⁻⁵ at 25 °C. (a) Write the equation for the proton transfer reaction with water and identify both conjugate pairs. (b) Calculate the pH of the solution.

Proton Transfer & pH of Ethanoic Acid

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Step 1 — Write the proton transfer equation

Ethanoic acid donates a proton to water: CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq). Because ethanoic acid is a weak acid, we use the equilibrium arrow (⇌) rather than a single arrow.

CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
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Step 2 — Identify conjugate pairs

Conjugate pair 1: CH₃COOH (acid) and CH₃COO⁻ (conjugate base) — they differ by one H⁺. Conjugate pair 2: H₂O (base) and H₃O⁺ (conjugate acid) — they also differ by one H⁺.

Pair 1: CH₃COOH / CH₃COO⁻ | Pair 2: H₂O / H₃O⁺
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Step 3 — Set up the Kₐ expression

Kₐ = [H₃O⁺][CH₃COO⁻] / [CH₃COOH]. Let x = [H₃O⁺] = [CH₃COO⁻] at equilibrium. Because the acid is weak and Kₐ is small, we can approximate [CH₃COOH] ≈ 0.10 − x ≈ 0.10.

1.8 × 10⁻⁵ = x² / 0.10
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Step 4 — Solve for x

x² = (1.8 × 10⁻⁵)(0.10) = 1.8 × 10⁻⁶. Taking the square root: x = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³. This is [H₃O⁺].

[H₃O⁺] = 1.34 × 10⁻³ mol dm⁻³
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Step 5 — Calculate pH

pH = −log₁₀(1.34 × 10⁻³) = −(−2.87) = 2.87. We can verify the approximation: x / 0.10 = 1.3%, which is well below 5%, so the approximation is valid.

pH ≈ 2.87

SECTION 7

Strengths & Limitations of the Brønsted–Lowry Model

The Brønsted–Lowry model is incredibly useful, but like all models in chemistry, it has boundaries. Understanding where it excels and where it falls short will help you choose the right approach on IB exam questions.

Strengths	Limitations
Applies to any solvent, not just water (unlike Arrhenius)	Cannot explain acid–base reactions with no proton, e.g., BF₃ + NH₃
Clearly identifies conjugate pairs and direction of proton transfer	Does not account for the role of electron pairs directly (Lewis theory needed)
Easily combined with Kₐ/Kb expressions for quantitative work	For polyprotic acids, each proton transfer must be considered separately, adding complexity
Explains buffer systems, hydrolysis, and neutralisation reactions	Cannot explain why some species (like metal cations) act as acids in solution

Comparison of the Brønsted–Lowry model's strengths and limitations

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KEY TAKEAWAY

Think of the Brønsted–Lowry model as a reliable GPS that works perfectly in most cities (proton transfer reactions) but cannot navigate off-road terrain (non-proton acid–base reactions). For IB Chemistry, this GPS is your primary tool—you will use it in the vast majority of acid–base questions, and the Lewis model only becomes essential when protons are not involved.

SECTION 8

Connection to Lewis Theory and Advanced Applications

The Brønsted–Lowry framework is a subset of the broader Lewis acid–base theory. Every Brønsted–Lowry base is also a Lewis base (it has a lone pair to accept a proton), but not every Lewis acid involves a proton. The table below compares the two models, which is a common exam question at Higher Level.

Feature	Brønsted–Lowry	Lewis
Acid definition	Proton (H⁺) donor	Electron-pair acceptor
Base definition	Proton (H⁺) acceptor	Electron-pair donor
Key particle	H⁺ (proton)	Electron pair
Scope	Reactions involving a proton transfer only	All Brønsted–Lowry reactions plus coordination, metal-ion hydrolysis, etc.
IB relevance	Reactivity 3.1 — primary model for acid–base calculations	Reactivity 3.1 (HL) — extends understanding to non-proton transfers

Brønsted–Lowry vs. Lewis acid–base models

As you move into HL content and university chemistry, you will encounter situations where the proton transfer model is insufficient. For instance, the reaction between BF₃ and NH₃ forms a coordinate (dative) bond without any proton being exchanged. In such cases, the Lewis model is essential. However, for the vast majority of IB Reactivity 3.1 questions, the Brønsted–Lowry proton transfer framework will give you everything you need.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

In the reaction NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), identify the Brønsted–Lowry acid, the Brønsted–Lowry base, and both conjugate pairs.

Show answer

Water (H₂O) acts as the Brønsted–Lowry acid because it donates a proton to ammonia. Ammonia (NH₃) is the Brønsted–Lowry base because it accepts the proton. Conjugate pair 1 is H₂O / OH⁻ (acid and its conjugate base). Conjugate pair 2 is NH₃ / NH₄⁺ (base and its conjugate acid). This also demonstrates that water is amphiprotic—it acts as an acid here, whereas in the HCl reaction it acted as a base.

Question 2: BASIC CALCULATION

Calculate the pH of a 0.020 mol dm⁻³ solution of hydrochloric acid (HCl). Assume complete dissociation.

Show answer

Since HCl is a strong acid, it ionises completely: HCl → H⁺ + Cl⁻. Therefore [H⁺] = 0.020 mol dm⁻³. Using pH = −log₁₀[H⁺] = −log₁₀(0.020) = −(−1.70) = 1.70. The pH of the solution is 1.70. This straightforward calculation works because no equilibrium considerations are needed for strong acids.

Question 3: INTERMEDIATE

A 0.25 mol dm⁻³ solution of methanoic acid (HCOOH) has Kₐ = 1.8 × 10⁻⁴ at 25 °C. Calculate the pH and the percentage ionisation of the acid.

Show answer

Set up Kₐ = x² / 0.25 where x = [H⁺] = [HCOO⁻]. Solving: x² = (1.8 × 10⁻⁴)(0.25) = 4.5 × 10⁻⁵, so x = √(4.5 × 10⁻⁵) = 6.71 × 10⁻³ mol dm⁻³. pH = −log₁₀(6.71 × 10⁻³) = 2.17. Percentage ionisation = (6.71 × 10⁻³ / 0.25) × 100% = 2.68%. Since this is well below 5%, the approximation [HA] ≈ 0.25 is valid. If it exceeded 5%, you would need to use the quadratic formula.

Question 4: APPLIED

Vinegar contains approximately 0.87 mol dm⁻³ ethanoic acid (Kₐ = 1.8 × 10⁻⁵). A student adds a small amount of sodium ethanoate (CH₃COONa) to create a buffer solution. Explain, using proton transfer equations, how this buffer resists changes in pH when a small amount of HCl is added.

Show answer

The buffer contains both CH₃COOH (weak acid) and CH₃COO⁻ (its conjugate base, from the dissolved salt). When HCl is added, the extra H⁺ ions are consumed by the conjugate base via the proton transfer: H⁺(aq) + CH₃COO⁻(aq) → CH₃COOH(aq). This converts some conjugate base into the weak acid, but because neither species is fully depleted, the ratio [CH₃COOH]/[CH₃COO⁻] changes only slightly. The pH therefore remains nearly constant. The key insight is that the proton from HCl is transferred to CH₃COO⁻ rather than remaining free in solution, which is why the pH does not drop significantly.

Question 5: CRITICAL THINKING

Hydrogen fluoride (HF, Kₐ = 6.8 × 10⁻⁴) is classified as a weak acid, yet it is considered extremely dangerous. Ethanoic acid (CH₃COOH, Kₐ = 1.8 × 10⁻⁵) is present in household vinegar. Compare the extent of proton transfer in 0.10 mol dm⁻³ solutions of each acid, calculate the pH of both, and discuss why Kₐ alone does not fully determine how hazardous an acid is.

Show answer

For HF: x² / 0.10 = 6.8 × 10⁻⁴, so x = √(6.8 × 10⁻⁵) = 8.25 × 10⁻³ mol dm⁻³, giving pH = 2.08 and 8.25% ionisation. For CH₃COOH: x² / 0.10 = 1.8 × 10⁻⁵, so x = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³, giving pH = 2.87 and 1.34% ionisation. HF undergoes a greater extent of proton transfer (higher Kₐ), producing a lower pH. However, the danger of HF is not primarily about pH—it lies in the fluoride ion's ability to penetrate skin and bind calcium and magnesium ions in the body, disrupting critical biological processes. This illustrates that acidity (proton transfer extent) and toxicity are separate concepts; Kₐ measures the thermodynamic tendency to donate a proton but does not capture the chemical reactivity of the conjugate base.

SUMMARY

Lesson Summary

Summary

In this lesson you learned that a proton transfer reaction is the defining event in Brønsted–Lowry acid–base chemistry: a Brønsted–Lowry acid donates H⁺ to a Brønsted–Lowry base, producing two conjugate acid–base pairs. You can predict the direction of proton transfer by comparing acid strengths: the equilibrium always favours formation of the weaker acid and weaker base.

Quantitatively, you applied pH = −log₁₀[H⁺] for strong acids and used the Kₐ expression with an approximation for weak acids to calculate [H⁺] and pH. You also explored how buffer solutions use conjugate pairs to resist pH changes, and you compared the Brønsted–Lowry model with the broader Lewis acid–base theory. These tools will help you tackle proton transfer questions throughout the IB Chemistry course and beyond.