Difficulty: Intermediate-Advanced | Prerequisites: Brønsted-Lowry definitions, pKa, the four factors for predicting acid/base strength
Big picture: This is where the theory gets applied. Exam questions will hand you two molecules and ask: what are the products, and which side of the equilibrium is favoured? The challenge is assigning acid and base roles correctly, especially with amphoteric molecules or complex organic structures with multiple types of hydrogen. This section walks through the systematic approach, including what to do when you do not have pKa values to hand.
To solve an acid-base problem: identify the acid and base, draw the products (conjugate base and conjugate acid), then compare pKa values to determine which side is favoured (the side with the weaker acid wins). When pKa values are unavailable, use the "base perspective" to compare the stability of the base and conjugate base using the four structural factors.
Amphoteric
A molecule that can behave as either an acid or a base depending on the reaction partner. Water and ammonia are common examples. When solving problems with amphoteric molecules, you must consider both possible roles and determine which equilibrium is favoured.
Base perspective
A qualitative method for predicting the favoured side of an acid-base equilibrium without pKa values. Compare the base and conjugate base, and choose the more stable of the two as the favoured side. Stability is assessed using elemental, resonance, hybridisation, and inductive effects.
Think of it as: "which lone pair is happier sitting still?" The more stable lone pair belongs to the favoured side.
pKreaction
The difference between the pKa of the acid and the pKa of the conjugate acid: pKreaction = pKa(A-H) - pKa(B-H). A negative value means products are favoured; a positive value means reactants are favoured. The magnitude tells you the order-of-magnitude ratio.
Counter-ion
A spectator ion (such as Na⁺ from NaOH) that does not participate in the acid-base reaction but stays associated with whatever anionic species is present in solution. When drawing acid-base problems, represent metal-nonmetal bonds ionically.
Red flags (problem-solving heuristic)
A set of warning signs to check before finalising your answer: formal charges are not the most important factor; negatively charged atoms almost never act as acids; positively charged atoms almost never act as bases; dianions and dications are almost never needed to describe reactions in solution.
Expand all Lewis structures around the atoms of interest. Represent bonds between main-group elements and metals (Li, Na, K, Mg) ionically.
Identify the base (the species with a lone pair ready to donate electrons). Anions are a strong first guess for the role of base.
Identify the acid (the species with a dissociable bond to hydrogen). When multiple types of H are available, the most acidic H (on the most electronegative atom, or the one producing the most stable conjugate base) reacts first.
Draw the products (conjugate base and conjugate acid).
Determine the favoured side: compare pKa values (lower pKa = stronger acid; equilibrium favours the weaker acid side), or use the base perspective if pKa values are unavailable.
When both molecules can act as either acid or base, you must test both possibilities.
Analysis 1: Water as base, ammonia as acid
Water (base) accepts a proton from ammonia (acid, pKa = 38) to form hydronium (conjugate acid, pKa = -1) and amide (NH₂⁻, conjugate base).
The product side contains the stronger acid (hydronium, pKa = -1), so the equilibrium favours the reactant side (the weaker acid, ammonia, is favoured).
Base perspective: the conjugate base (amide) has its lone pair on nitrogen (less electronegative than oxygen, and the molecule is negatively charged), making it less stable than the original base (water, neutral, lone pair on oxygen). Reactants favoured.
Analysis 2: Water as acid, ammonia as base
Water (acid, pKa = 16) donates a proton to ammonia (base) to form hydroxide (conjugate base) and ammonium (conjugate acid, pKa = 9).
The product side again contains the stronger acid (ammonium, pKa = 9), so the equilibrium again favours reactants (water, pKa = 16, is the weaker acid).
The side with no formal charges is favoured, consistent with the pKa analysis.
Both analyses agree: the reactant side (neutral water + neutral ammonia) is favoured.
First, expand the structure: draw acetic acid with its O-H and C-H bonds visible, and represent NaOH as Na⁺ + HO⁻.
Analysis 1 (most favourable): Hydroxide as base, acetic acid O-H as acid
Hydroxide (base) deprotonates the O-H of acetic acid (pKa = 5) to form acetate (conjugate base) and water (conjugate acid, pKa = 16).
pKreaction = 5 - 16 = -11. Products are favoured by 10¹¹ to 1.
Base perspective (without pKa): Both hydroxide and acetate have their charge on oxygen (same elemental effects). The key difference is that acetate has resonance stabilisation across two oxygens, while hydroxide has none. The more stable conjugate base (acetate) is on the product side. Products favoured.
Analysis 2 (not favoured): Hydroxide as base, acetic acid C-H as acid
The resulting carbon anion conjugate base is far less stable than acetate because carbon is less electronegative than oxygen (elemental effects dominate).
When hydroxide approaches acetic acid, the more acidic O-H proton always reacts first and faster. This pathway is not competitive.
Analysis 3 (not favoured): Acetic acid as base, hydroxide as acid
This would require hydroxide to donate a proton and form oxide (O²⁻), a dianion that is extremely unstable.
Dianions are almost never formed in solution. This pathway is strongly disfavoured.
When pKa values are not available, compare the base and the conjugate base:
Which lone pair is on a more electronegative atom? (Elemental)
Which lone pair is delocalised by resonance? (Resonance)
Which lone pair is in an orbital with more s-character? (Hybridisation)
Which lone pair is near electron-withdrawing groups? (Inductive)
The side of the equilibrium with the more stable lone pair (less reactive base) is favoured.
Formal charges are useful but not the most important factor. Do not let a formal charge override elemental or resonance analysis.
Negatively charged atoms almost never play the role of acid (they would need to lose H to form a dianion).
Positively charged atoms almost never play the role of base (they would need to gain H to form a dication).
Dianions and dications are so disfavoured that they are almost never needed to describe reactions in solution.
Students often assume the equilibrium favours the side with the stronger acid. The opposite is true: the stronger acid dissociates and pushes the equilibrium toward the products, so the product side (with the weaker conjugate acid) is favoured.
Students often pick the wrong proton to deprotonate. When a molecule has multiple types of hydrogen (e.g. both O-H and C-H in acetic acid), the most acidic proton reacts preferentially. Always consider which conjugate base would be most stable.
Students often forget to consider both roles when working with amphoteric molecules. If both molecules can act as acid or base, you must draw both possible reactions and determine which is favoured.
Students often propose dianion or dication products. These are almost never formed in solution. If your proposed products include a species with a 2- or 2+ charge, reconsider your assignment of acid and base roles.
⚠️ "Predict the products and determine which side is favoured" is one of the most common exam question formats. Practise the full method: expand structures, assign roles, draw products, compare pKa or use the base perspective.
⚠️ Questions involving amphoteric molecules (water, ammonia, alcohols, amines) are popular because they test whether you can consider multiple possible reactions.
⚠️ The base perspective is essential for questions that do not provide pKa values. You must be able to compare conjugate base stability using the four structural factors.
⚠️ Know the red flags. If your answer involves a dianion, a positively charged base, or a negatively charged acid, something has gone wrong.
True or false: When water reacts with ammonia, the side with no formal charges is favoured. (True)
Fill in the blank: pKreaction = pKa() - pKa(). (Acid, conjugate acid)
True or false: A negatively charged atom is a good candidate to act as the acid in an equilibrium. (False, it would form a dianion)
Fill in the blank: When pKa values are unavailable, use the "____" to compare conjugate base stability. (Base perspective)
True or false: When acetic acid reacts with NaOH, the C-H bond is deprotonated preferentially over the O-H bond. (False, the more acidic O-H reacts first)
Q: Predict the products of the reaction between phenol (C₆H₅OH, pKa ≈ 10) and sodium hydroxide (NaOH). Which side is favoured?
A: Hydroxide (base) deprotonates the O-H of phenol to form phenoxide (C₆H₅O⁻) and water (pKa = 16). pKreaction = 10 - 16 = -6. Products are favoured by 10⁶ to 1. The phenoxide conjugate base is stabilised by resonance with the aromatic ring, making it more stable than hydroxide.
Q: Water and ammonia are both amphoteric. For both possible reactions (water as acid, water as base), determine which side is favoured. Do the two analyses agree?
A: Analysis 1 (water as base, ammonia as acid): products are hydronium (pKa = -1) and amide (NH₂⁻). Product side has the stronger acid, so reactants are favoured. Analysis 2 (water as acid, ammonia as base): products are hydroxide and ammonium (pKa = 9). Product side has the stronger acid (ammonium, 9 < 16), so reactants are favoured. Both analyses agree: the neutral reactant side is favoured.
Q: Without pKa values, use the base perspective to determine whether the reaction of ethanol with sodium amide (NaNH₂) favours products or reactants.
A: Sodium amide provides the amide ion (NH₂⁻) as the base. Ethanol's O-H is deprotonated to form ethoxide (CH₃CH₂O⁻) and ammonia. Comparing the two bases: amide has its lone pair on nitrogen (less electronegative), while ethoxide has its lone pair on oxygen (more electronegative). The more stable base is ethoxide (elemental effects: oxygen holds the charge better). The more stable base is on the product side, so products are favoured.
Q: Why is the reaction of hydroxide with the C-H bond of acetic acid not competitive with the O-H deprotonation, even though both pathways are possible?
A: The carbon anion conjugate base formed by C-H deprotonation is far less stable than the acetate formed by O-H deprotonation. Carbon is less electronegative than oxygen (elemental effects), and acetate has resonance stabilisation across two oxygens. The O-H proton is more acidic and always reacts preferentially.
Q: A student proposes that hydroxide (HO⁻) acts as an acid when mixed with acetic acid, losing its proton to form oxide (O²⁻). Explain why this is wrong.
A: Oxide (O²⁻) is a dianion, which is extremely unstable and almost never formed in solution. Anions are typically very poor acids because deprotonation would require forming a dianion. This violates one of the key red flags for acid-base problem-solving.
The problem-solving framework here applies directly to predicting the outcomes of reactions involving nucleophiles and electrophiles later in the course. A strong base is often (but not always) a strong nucleophile, and the same stability arguments apply to leaving group quality.
Extraction and separation techniques (the broader topic of Chapter 7) rely on acid-base chemistry to convert organic compounds between their neutral and ionic forms, which determines whether they dissolve in aqueous or organic layers.
acid-base equilibrium, predicting products, favoured side, pKreaction, base perspective, conjugate base stability, amphoteric, water as acid, water as base, acetic acid NaOH reaction, red flags acid-base, dianion, formal charge, proton transfer, deprotonation, organic chemistry problem solving, extraction, separation, Chapter 7.4