Organic Chemistry · Polar Covalent Bonds; Acids and Bases

Predicting Acid–Base Reactions from pKa Values

9 min read

Reference pKa values (acetic acid 4.76, water 15.7, ammonia 38, acetylene ~25, aspirin ~3.5) are standard textbook values; verify against current primary sources before formal citation.

On this page 9 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Every acid–base reaction is an equilibrium, and the single most useful number for predicting where that equilibrium lies is the — a logarithmic measure of acid strength defined as pK_a = -logK_a. The lower the pKa, the stronger the acid; the higher the pKa, the weaker the acid and the more reluctant the molecule is to surrender a proton. Because pKa values of thousands of compounds are tabulated, you can predict the outcome of almost any proton-transfer reaction without running it: the equilibrium always favors the side with the (and the weaker base). In quantitative form, the equilibrium constant is K_eq = 10^{(pK_a^{acid on product side} - pK_a^{acid on reactant side})}. A pKa gap of about 10 units corresponds to an equilibrium constant of 10¹⁰, which means the reaction is essentially complete. This topic turns memorized acid strength lists into a predictive tool you can apply to any new molecule.

Why this matters

Reagent selection in the lab: choosing a base strong enough to deprotonate a given substrate is a routine synthetic decision. For example, terminal alkynes (pK_a ≈ 25) require sodium amide, NaNH₂, not aqueous sodium hydroxide, because water (pK_a = 15.7) is too weak an acid to be pushed to completion by hydroxide.
Drug behavior in the body: whether a drug is ionized or neutral at physiological pH controls absorption, distribution, and excretion. The pKa of aspirin ( ≈ 3.5) explains why it crosses the stomach lining in its neutral form.
Mechanistic reasoning: most organic mechanisms contain proton-transfer steps, and knowing which side a proton transfer favors is often the deciding step between two plausible pathways.
Exams: pKa prediction problems appear in almost every organic chemistry assessment, and the "weaker acid wins" rule is the fastest route to the answer.

The college version

Core Concepts

pKa measures acid strength on a logarithmic scale

Acid strength is quantified by the equilibrium constant for dissociation in water:

$HA + H 2 O ⇌ H 3 O + + A - K a = [H 3 O +][A -] [HA]$

Because K_a values range over many orders of magnitude, chemists compress the scale: pK_a = -logK_a. A difference of one pKa unit means a tenfold difference in K_a — and therefore a tenfold difference in the equilibrium constant of a proton-transfer reaction. Useful reference points: hydronium ion, pK_a = -1.74; acetic acid, 4.76; phenol, about 10; water, 15.7; ethanol, about 16; terminal alkyne, about 25; ammonia, about 38; alkane, about 50.

The fundamental rule: the weaker acid (and weaker base) wins

Consider a general proton transfer, where HB⁺ is the conjugate acid of the base B:

$HA + B ⇌ A - + HB +$

The equilibrium constant is the ratio of the two acidity constants:

$K eq = K a (HA) K a (HB +) = 10 p K a (HB +) - p K a (HA)$

If pK_a(HB⁺) > pK_a(HA), then K_eq > 1 and products are favored — meaning the acid on the product side (HB⁺) is the weaker acid. The rule in words: proton transfer proceeds from the and stronger base toward the weaker acid and weaker base. The stronger acid and the stronger base always sit on the same side of the equation.

Rules of thumb for pKa gaps

ΔpK_a ≥ 10: essentially complete (better than 99.99% conversion; K_eq ≥ 10¹⁰).
ΔpK_a of 3–9: strongly favorable, usually usable for synthesis.
ΔpK_a of 0–2: a real mixture of both sides; the reaction is not a clean way to make products.
Negative ΔpK_a: the reaction runs in the reverse direction; the "products" written are actually the reactants.

From pKa to ionization state: the Henderson–Hasselbalch equation

For a weak acid HA in a solution of known pH:

$pH = p K a + log [A -] [HA]$

At pH = pK_a, the acid is 50% ionized. At pH one unit above the pKa, about 91% is ionized; one unit below, about 91% is neutral. This relationship lets you predict whether a molecule exists as a neutral species or a charged ion at any biological or laboratory pH — the key to solubility and membrane-crossing behavior.

Common Confusions

Do Not Confuse	With	Difference
pKa	pH	pKa is a property of the acid (how easily it donates a proton); pH is a property of the solution (its current proton concentration).
Stronger acid = larger pKa	Stronger acid = smaller pKa	The scale is inverted: lower pKa means a stronger acid.
Acid on the product side	Acid on the reactant side	Equilibrium favors the weaker acid, which is the one with the higher pKa — always check which side has the higher pKa.
K_eq	K_a	K_a describes one acid's dissociation; K_eq describes the whole proton-transfer reaction and is a ratio of two K_a values.
Using the pKa of the base	Using the pKa of the base's conjugate acid	When a base like NH₂^- is a reactant, substitute the pKa of its conjugate acid (NH₃, 38) — not the pKa of NH₄⁺ (9.24).
ΔpK_a = 1 means slightly favorable	ΔpK_a = 1 gives K_eq = 10	One pKa unit = factor of 10 in K_eq, which is a meaningful but not overwhelming drive; reactions with ΔpK_a near zero are mixtures.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine every acid has a number that says how badly it wants to give away a hydrogen atom: small number = really wants to give it away, big number = holding on tight. When two acids face off, the one that wants the hydrogen less (the bigger number) ends up with it. So the "winner" of an acid–base fight is the weaker acid — the one that's happiest to hold the hydrogen — and the stronger acid is forced to give its hydrogen away.

Worked example

Example 1: Is sodium hydroxide enough to deprotonate acetic acid?

Problem: Predict whether CH₃COOH + NaOH goes to completion in water.

Plan and formula: identify the acid on each side. The reactant acid is acetic acid, pK_a = 4.76. The base is hydroxide; its conjugate acid is water, pK_a = 15.7. Then:

$K eq = 10 p K a (water) - p K a (acetic acid) = 10 15.7 - 4.76 = 10 10.94 \approx 8.7 \times 1010$

Interpretation: with K_eq ≈ 10¹¹, the deprotonation of acetic acid by hydroxide is essentially complete — vinegar is fully neutralized by a stoichiometric amount of NaOH, which is why CH₃COO^- Na⁺ (sodium acetate) is easy to isolate.

Example 2: Which base can deprotonate a terminal alkyne — hydroxide or amide?

Problem: Acetylene, HC ≡ CH, has pK_a ≈ 25. Which reagent generates the acetylide ion: aqueous NaOH (pK_a of water = 15.7) or NaNH₂ (pK_a of ammonia = 38)?

Plan and formula: compute K_eq for each candidate base using the same formula, substituting the conjugate acid of each base:

$K eq (OH -) = 10 15.7 - 25 = 10 -9.3 \approx 5 \times 10 -10 (reactants favored; negligible reaction)$

$K eq (NH 2 -) = 10 38 - 25 = 10 13 (products favored; essentially complete)$

Interpretation: hydroxide in water cannot deprotonate a terminal alkyne — the equilibrium constant is 10^-9.3, meaning almost no acetylide forms. Sodium amide, however, drives the reaction to completion because ammonia (pK_a 38) is a much weaker acid than the alkyne. This is exactly why NaNH₂ in liquid ammonia is the standard reagent for making acetylide anions for carbon–carbon bond formation.

Example 3: Why is aspirin absorbed in the stomach?

Problem: Aspirin (acetylsalicylic acid) has pK_a ≈ 3.5. Compare its ionization in the stomach (pH ≈ 2) and in the blood (pH ≈ 7.4).

Plan and formula: use the Henderson–Hasselbalch equation to find the ratio of ionized to neutral forms at each pH:

$log [A -] [HA] = pH - p K a$

Substitution (stomach, pH 2):

$log [A -] [HA] = 2 - 3.5 = -1.5 \Rightarrow [A -] [HA] = 10 -1.5 \approx 0.032$

Only about 3% is ionized — 97% of aspirin is the neutral, lipid-soluble form that crosses the stomach lining. Substitution (blood, pH 7.4):

$log [A -] [HA] = 7.4 - 3.5 = 3.9 \Rightarrow [A -] [HA] = 10 3.9 \approx 8000$

Interpretation: in blood the drug is essentially fully ionized (charged carboxylate), which keeps it trapped on the blood side of membranes — a classic example of "ion trapping" driven by pKa and pH.

Key takeaways

Lower pKa = stronger acid; each pKa unit = 10× in equilibrium constant.
The reaction always favors the weaker acid and weaker base; stronger acid and stronger base are on the same side.
K_eq = 10^{pK_a(product acid) - pK_a(reactant acid)} — show the formula, then substitute.
ΔpK_a ≥ 10 → essentially complete; 3–9 → favorable; 0–2 → mixture; negative → reverse direction.
Common reference pKas: H₃O⁺ −1.74; acetic acid 4.76; phenol ~10; water 15.7; ethanol ~16; terminal alkyne ~25; ammonia ~38.
Henderson–Hasselbalch: pH = pK_a + log([A^-]/[HA]); at pH = pKa the acid is half-ionized.
Conjugate acid–base pairs: a strong acid has a weak conjugate base and vice versa.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

A reaction has a reactant acid with pK_a = 5 and a base whose conjugate acid has pK_a = 20. Does the reaction favor products or reactants? Compute K_eq.
Show answer
Favor products. K_eq = 10^{20 - 5} = 10¹⁵ — the weaker acid on the product side (conjugate acid of the base, pKa 20) wins overwhelmingly.
Explain in one sentence why a strong acid has a weak conjugate base.
Show answer
A strong acid ionizes almost completely, so its conjugate base has essentially no tendency to grab a proton back — a weak base.
At what pH is a weak acid with pK_a = 5.0 exactly 50% ionized? What fraction is ionized at pH = 6.0?
Show answer
50% ionized at pH = pK_a = 5.0. At pH 6.0: log([A^-]/[HA]) = 6.0 - 5.0 = 1.0, so [A^-]/[HA] = 10, meaning 10/11 ≈ 91% ionized.
Can NaHCO₃ (pK_a of H₂CO₃ ≈ 6.4) deprotonate a phenol (pK_a ≈ 10)? Show the K_eq calculation.
Show answer
K_eq = 10^{6.4 - 10} = 10^-3.6 ≈ 2.5 × 10^-4 — reactants favored; bicarbonate is far too weak a base to deprotonate phenol appreciably.
Why does NaNH₂ deprotonate terminal alkynes while aqueous NaOH does not?
Show answer
The equilibrium constant scales with ΔpK_a: amide gives 10¹³ (complete), hydroxide gives 10^-9.3 (negligible), because water (15.7) is a much stronger acid than ammonia (38), so hydroxide's conjugate acid is too strong for the equilibrium to favor acetylide.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

pKa: -logK_a; a number that measures how easily an acid donates a proton (lower = stronger acid)
Ka: Equilibrium constant for acid dissociation in water
Equilibrium constant (Keq): Ratio of product to reactant concentrations at equilibrium
Stronger acid: Species with the lower pKa; donates protons readily
Weaker acid: Species with the higher pKa; holds protons tightly
Conjugate acid–base pair: Two species that differ by one proton (e.g., CH₃COOH / CH₃COO^-)
Henderson–Hasselbalch equation: pH = pK_a + log([A^-]/[HA])

Sources & references

openstax.org — Organic Chemistry

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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In 30 seconds

Why this matters

The college version

Core Concepts

pKa measures acid strength on a logarithmic scale

The fundamental rule: the weaker acid (and weaker base) wins

Rules of thumb for pKa gaps

From pKa to ionization state: the Henderson–Hasselbalch equation

Common Confusions

The same idea, in plain words

Worked example

Example 1: Is sodium hydroxide enough to deprotonate acetic acid?

Example 2: Which base can deprotonate a terminal alkyne — hydroxide or amide?

Example 3: Why is aspirin absorbed in the stomach?

Key takeaways

Check yourself

Keep learning

Key vocabulary

Related lessons

Sources & references