Exercises — Acids and bases — Arrhenius, Brønsted-Lowry, Lewis definitions

Arrhenius acid ::: produces ${\text{H}}^{+}$ (really ${\text{H}}_3{\text{O}}^{+}$) in water; base produces ${\text{OH}}^{-}$ in water. Brønsted-Lowry acid ::: a proton (${\text{H}}^{+}$) donor; base a proton acceptor. Lewis acid ::: an electron-pair acceptor; base an electron-pair donor.

What to do: ask "does it release ${\text{H}}^{+}$ in water (acid) or ${\text{OH}}^{-}$ in water (base)?"

• ${\text{HNO}}_3\to {\text{H}}^{+}+{\text{NO}}_3^{-}$ — releases ${\text{H}}^{+}$ → Arrhenius acid.
• $\text{KOH}\to {\text{K}}^{+}+{\text{OH}}^{-}$ — releases ${\text{OH}}^{-}$ → Arrhenius base.
• ${\text{CH}}_4$ — no ionizable ${\text{H}}^{+}$, no ${\text{OH}}^{-}$ → neither.
• ${\text{H}}_2{\text{SO}}_4\to 2{\text{H}}^{+}+{\text{SO}}_4^{2-}$ — releases ${\text{H}}^{+}$ → Arrhenius acid (diprotic).

What to do: find who gives away the proton (acid) and who takes it (base).

• $\text{HCl}$ hands its ${\text{H}}^{+}$ over → Brønsted-Lowry acid.
• ${\text{H}}_2\text{O}$ receives that ${\text{H}}^{+}$ to become ${\text{H}}_3{\text{O}}^{+}$ → Brønsted-Lowry base.

Step 1 — molecular equation. ${\text{H}}_2{\text{SO}}_4$ is diprotic (2 protons), so it needs 2 hydroxides: ${\text{H}}_2{\text{SO}}_4+2\text{KOH}\to {\text{K}}_2{\text{SO}}_4+2{\text{H}}_2\text{O}$ Why the 2: each ${\text{H}}^{+}$ must meet one ${\text{OH}}^{-}$; two protons demand two hydroxides. Stoichiometry $1:2$. Step 2 — strip spectators. ${\text{K}}^{+}$ and ${\text{SO}}_4^{2-}$ stay dissolved and unchanged, so they cancel. What survives is the essence: ${\text{H}}^{+}+{\text{OH}}^{-}\to {\text{H}}_2\text{O}$

What to do: track the proton across the arrow. Two things gain/lose exactly one ${\text{H}}^{+}$.

• ${\text{H}}_2\text{O}$ (acid) loses ${\text{H}}^{+}$ → ${\text{OH}}^{-}$. Pair: ${\text{H}}_2\text{O}/{\text{OH}}^{-}$.
• ${\text{HCO}}_3^{-}$ (base) gains ${\text{H}}^{+}$ → ${\text{H}}_2{\text{CO}}_3$. Pair: ${\text{HCO}}_3^{-}/{\text{H}}_2{\text{CO}}_3$. Here ${\text{HCO}}_3^{-}$ acts as a base. (In Buffer solutions you meet it acting as an acid too — it is amphoteric.)

• Arrhenius: no water, no ${\text{H}}^{+}$, no ${\text{OH}}^{-}$ → does not qualify.
• Brønsted-Lowry: no proton is transferred → does not qualify.
• Lewis: ${\text{BF}}_3$ has an empty orbital (boron: incomplete octet) and accepts an electron pair → Lewis acid; ${\text{NH}}_3$ donates its nitrogen lone pair → Lewis base → qualifies. A dative (coordinate) bond forms. This is the classic example that only Lewis captures.

Key rule: strength of an acid and its conjugate base are inverse. A strong acid gives up its proton eagerly and completely, so its conjugate base has almost no urge to grab the proton back → weak base.

• $\text{HCl}$ strong → ${\text{Cl}}^{-}$ is a very weak base.
• ${\text{CH}}_3\text{COOH}$ weak → ${\text{CH}}_3{\text{COO}}^{-}$ is a stronger base than ${\text{Cl}}^{-}$. Ranking (weaker → stronger): ${\text{Cl}}^{-}<{\text{CH}}_3{\text{COO}}^{-}$. (This inverse relationship is quantified by Acid-base equilibrium constants.)

As an acid (donates its proton to a base like ${\text{OH}}^{-}$): $\text{ce}HCO3-+OH-->CO{3}^2-+H2O$ As a base (accepts a proton from an acid like ${\text{H}}_3{\text{O}}^{+}$): $\text{ce}HCO3-+H+->H2CO3$ Why both: ${\text{HCO}}_3^{-}$ carries both a removable $\text{H}$ (so it can donate) and oxygen lone pairs (so it can accept). Having both tools is exactly what "amphoteric" means. See Amphoteric oxides for the same idea in oxides.

The claim uses the Brønsted-Lowry ruler on a reaction that lives in the Lewis ring.

• ${\text{Ag}}^{+}$ has empty orbitals and a positive charge → it accepts electron pairs → Lewis acid.
• ${\text{NH}}_3$ donates its nitrogen lone pair → Lewis base. Two dative bonds form, giving the complex ion. It is an acid–base reaction — just under the broadest definition. All metal–ligand Coordination compounds are Lewis acid–base chemistry.

Brønsted-Lowry (proton view): ${\text{H}}_2\text{O}$ donates ${\text{H}}^{+}$ (acid) → ${\text{OH}}^{-}$; ${\text{NH}}_3$ accepts ${\text{H}}^{+}$ (base) → ${\text{NH}}_4^{+}$. Conjugate pairs: ${\text{H}}_2\text{O}/{\text{OH}}^{-}$ and ${\text{NH}}_3/{\text{NH}}_4^{+}$. Arrhenius (water-outcome view): the reaction produces ${\text{OH}}^{-}$ in water, so ${\text{NH}}_3$ counts as an Arrhenius base — but only indirectly, which is why the definition felt forced. Lewis (electron view): nitrogen's lone pair (Lewis base) attacks the proton of water; that proton, being electron-deficient, is the Lewis acid. The lone pair forms the new N–H bond. Same reaction, three lenses: one physical event — nitrogen's lone pair capturing a proton — read as (a) ${\text{OH}}^{-}$ appearing, (b) a proton transferring, or (c) an electron pair donating. Each definition zooms out from the last.

Product: solid ammonium chloride: $\text{ce}NH3(g)+HCl(g)->NH4Cl(s)$ Brønsted-Lowry: $\text{HCl}$ donates ${\text{H}}^{+}$ (acid), ${\text{NH}}_3$ accepts it (base), giving ${\text{NH}}_4^{+}$ and ${\text{Cl}}^{-}$ that pair as the white solid. Why Arrhenius fails: there is no water and no solution — Arrhenius is defined only for aqueous ${\text{H}}^{+}/{\text{OH}}^{-}$ release. With no ${\text{OH}}^{-}$ produced and no solvent at all, Arrhenius has nothing to say. This is exactly the gap that motivated Brønsted-Lowry.

Step 1 — Lewis capture. ${\text{Al}}^{3+}$ is small and highly charged → a strong Lewis acid. It grabs the oxygen lone pairs of water molecules, forming the hydrated ion $\text{ce}[Al(H2O)6{]}^3+$. Each water is a Lewis base donating a lone pair into aluminium's empty orbitals. Step 2 — proton loosening (Brønsted). The $3+$ charge pulls electron density away from the O–H bonds of those coordinated waters, weakening them. One coordinated water then donates a proton to a free water: $\text{ce}[Al(H2O)6{]}^3++H2O<=>[Al(H2O)5(OH){]}^2++H3O+$ Here the hydrated ion acts as a Brønsted acid. Step 3 — Arrhenius outcome. The result is extra ${\text{H}}_3{\text{O}}^{+}$ in water → the solution is acidic, which is exactly what Arrhenius measures (though it can't explain the mechanism). Name: this is cationic hydrolysis (salt hydrolysis). One physical cause — a Lewis-acidic cation — read out as lower pH.

Step 1 — what got added. ${\text{NH}}_4\text{Cl}$ dissolves fully into ${\text{NH}}_4^{+}+{\text{Cl}}^{-}$. The ${\text{NH}}_4^{+}$ is a product of the equilibrium and is the conjugate acid of ${\text{NH}}_3$. Step 2 — Le Chatelier. Adding a product pushes the equilibrium left (the system relieves the stress by consuming the added ${\text{NH}}_4^{+}$). Step 3 — consequences. Shifting left consumes ${\text{OH}}^{-}$, so $[{\text{OH}}^{-}]$ drops; therefore the solution becomes less basic (pH moves down toward 7). Conjugate-pair tie-in: we suppressed the base ${\text{NH}}_3$'s ionization by flooding in its conjugate acid — the common-ion effect that underlies every buffer. A mixture of ${\text{NH}}_3$ and ${\text{NH}}_4\text{Cl}$ resists pH change for exactly this reason.