Acids and Bases Practice Questions

Use ARIO. Resonance is the deciding factor here: the conjugate base of acetic acid (acetate) delocalizes its negative charge across two oxygens, while the conjugate base of ethanol (ethoxide) has the charge stuck on one oxygen. A resonance-stabilized conjugate base means a much stronger acid — acetic acid's pKa (~4.8) is far lower than ethanol's (~16).

Use ARIO. Both conjugate bases have the same resonance stabilization (both are carboxylates), so resonance is a tie. The deciding factor is induction: fluorine is highly electronegative and pulls electron density through the σ bonds, further stabilizing the negative charge on the conjugate base.

Use ARIO. Resonance and induction are not relevant here, as both conjugate bases lack resonance and have no extra electronegative substituents. The deciding factor is Atom: sulfur is larger and more polarizable than oxygen. Even though oxygen is more electronegative, sulfur's larger size spreads out the negative charge on the conjugate base more comfortably, making the thiol more acidic despite the electronegativity trend.

Use ARIO. Resonance is the deciding factor: phenol's conjugate base (phenoxide) can delocalize its negative charge into the aromatic ring through resonance, spreading it across multiple ring carbons. Cyclohexanol's conjugate base has no adjacent π system, so the charge stays localized entirely on oxygen.

Use ARIO. With no resonance or induction differences relevant here, the deciding factor is Orbital hybridization. The terminal alkyne carbon is sp hybridized (50% s-character), while the alkane carbon is sp3 (25% s-character). More s-character means electrons sit closer to the nucleus, stabilizing the negative charge on the conjugate base. This is why terminal alkynes (pKa ≈ 25) are dramatically more acidic than alkanes (pKa ≈ 50).

Equilibrium favors the side with the weaker acid — the acid with the higher pKa. Water (pKa ≈ 15.7) is a much weaker acid than acetic acid (pKa ≈ 4.8), so the reaction strongly favors the product side, where water now holds the proton.

Compare the pKa values of the acid on each side. On the reactant side, the alcohol has pKa ≈ 16. On the product side, the acid formed is acetic acid with pKa ≈ 4.8. Since the reactant side has the weaker acid (higher pKa), equilibrium favors staying on the reactant side — acetate is not a strong enough base to deprotonate the alcohol significantly.

The acid on the product side (the amine, pKa ≈ 38) is far weaker than water (pKa ≈ 15.7) on the reactant side. Equilibrium favors the weaker acid, so this strongly favors products — this is exactly why strong bases like NaNH₂ are destroyed instantly by water or any protic solvent.

The acid donates a proton and becomes the conjugate base. The base accepts a proton and becomes the conjugate acid. Track which species gained or lost the H+ to identify each role correctly.

The amine's nitrogen lone pair accepts the proton from the carboxylic acid's O-H, making the amine the base and the resulting ammonium the conjugate acid.

Even though the ketone doesn't look like a typical acid, any molecule donating a proton in the reaction is acting as the acid. The enolate that remains after proton loss is the conjugate base — and it's resonance stabilized, which is why this proton is acidic enough to remove at all.

Acetic acid's conjugate base delocalizes charge across two oxygens (strongest resonance stabilization). Phenol's conjugate base delocalizes into an aromatic ring (moderate resonance stabilization, but spread across carbon, which is less electronegative than oxygen). Cyclohexanol's conjugate base has no resonance at all, leaving charge stuck on one oxygen — making it the weakest acid of the three.

All three share identical resonance stabilization in their conjugate bases (all are carboxylates). The deciding factor is induction: each additional electronegative chlorine atom withdraws more electron density, further stabilizing the negative charge. More inductive withdrawal means a stronger acid.

Alkyl groups donate electron density inductively, making the nitrogen lone pair more available and the amine more basic than ammonia. Aniline is the weakest base because its nitrogen lone pair is delocalized into the aromatic ring through resonance, making it less available to accept a proton — the same resonance effect that makes phenol more acidic makes aniline less basic.

A base can only deprotonate an acid if its own conjugate acid has a higher pKa than the acid being removed. Ammonia's pKa (≈ 38) is higher than the alkyne's pKa (≈ 25), so NaNH₂ can deprotonate it. Water's pKa (≈ 15.7) is lower than the alkyne's pKa, so NaOH cannot.

NaH's conjugate acid (H₂, pKa ≈ 36) is far higher than the alcohol's pKa (≈ 16), making this reaction essentially irreversible and complete. NaHCO₃'s conjugate acid (carbonic acid, pKa ≈ 6.3) is much lower than the alcohol's pKa, so it cannot deprotonate the alcohol at all.