THE CONCEPT
Proton Transfer Always Runs Toward the Weaker Acid
Every organic acid-base reaction is fundamentally a proton transfer: an acid donates H⁺ to a base, generating a new acid (the original base, now holding that proton) and a new base (the original acid, now missing that proton) on the product side. Which direction this equilibrium favors — reactants or products — comes down to a single comparison: which side has the weaker acid, since a stable, weak acid is more comfortable holding onto its proton than an unstable, strong acid is.
The general rule is direct: proton transfer runs from a stronger acid to a stronger base, producing a weaker acid and a weaker base — and since weaker acids correspond to higher pKa values, this translates into a simple numeric comparison: if the product-side acid has a HIGHER pKa than the reactant-side acid, the reaction is thermodynamically favorable (Keq > 1) and proceeds substantially toward products.
💡 Memory Trick
The hub's trick states the rule directly: acid-base reaction direction is predicted by pKa — equilibrium favors the side with the weaker acid. The hub's own worked example is genuinely instructive, including its self-correction: comparing water (pKa ≈ 15.7) reacting with sodium hydride, the naive comparison might look like it's roughly balanced — but the hub catches its own arithmetic and clarifies that the REAL comparison is against H₂'s pKa (~35, since NaH's conjugate acid is H₂, not NaOH) — meaning NaH deprotonates water essentially completely, since forming the much weaker acid H₂ (pKa 35) from the much stronger acid H₂O (pKa 15.7) is enormously favorable.
APPLYING THE PKA COMPARISON SYSTEMATICALLY
A Four-Step Check for Any Acid-Base Pair
1
Identify the acid and base on the reactant side
Determine which species is donating a proton (the acid) and which is accepting it (the base) as drawn in the starting materials.
2
Identify the conjugate acid and conjugate base formed
Once the proton transfers, determine the new acid (the original base, now protonated) and new base (the original acid, now deprotonated) that appear on the product side.
3
Look up or estimate the pKa of both acids
Find the pKa of the reactant-side acid and the pKa of the newly formed product-side acid, using known reference values or comparative reasoning (electronegativity, resonance, induction) if exact values aren't memorized.
4
Compare the two pKa values to predict direction
If the product-side acid has a higher pKa (is weaker) than the reactant-side acid, the reaction is favorable as written; if the product-side acid has a lower pKa (is stronger), the reaction actually favors the reverse direction instead.
🧪 Lab Application
You're asked whether acetic acid (pKa ≈ 4.76) will react favorably with sodium ethoxide, whose conjugate acid, ethanol, has a pKa of about 16.
1
Identify the reactant-side acid. Acetic acid (pKa ≈ 4.76) is the proton donor in this reaction.
2
Identify the product-side acid. Once acetic acid donates its proton to ethoxide, ethanol (the conjugate acid of ethoxide) forms as the new acid on the product side, with a pKa of about 16.
3
Compare the two pKa values. The product-side acid (ethanol, pKa 16) has a much HIGHER pKa than the reactant-side acid (acetic acid, pKa 4.76) — meaning the product side has the weaker acid.
4
Predict the reaction direction. Since the equilibrium favors the weaker acid, and that's the product side here, this reaction is strongly favorable as written (Keq >> 1) — acetic acid will readily protonate ethoxide.
📌 Exam Application
Acid-base direction questions are a common and efficient way to test pKa fluency — always explicitly identify both the reactant-side and product-side acids before comparing their pKa values, rather than trying to judge favorability from intuition alone.
⚠️ Most Common Predicting Acid-Base Reactions Mistakes
The most common mistake is comparing the wrong pair of species — remember to compare the ACID on each side (not the base), and make sure you've correctly identified each conjugate acid rather than mixing up which species is which after the proton transfer. The other frequent trap, exactly as the hub's own worked example highlights, is forgetting to trace all the way to the correct conjugate acid (like H₂, not NaOH, as the conjugate acid of NaH) rather than stopping at an intermediate or incorrect comparison point.
✓ Quick Self-Test
1) What general rule predicts the direction of an acid-base equilibrium? 2) If the product-side acid has a higher pKa than the reactant-side acid, is the reaction favorable or unfavorable as written? 3) What is the conjugate acid of hydride (H⁻)? 4) Why is NaH able to deprotonate water essentially completely? 5) What is the first step in systematically predicting the direction of any organic acid-base reaction?
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