Acids and bases — Arrhenius, Brønsted-Lowry, Lewis definitions: Worked examples

This is the drill page for the Acids and Bases topic. The parent note gave you the three definitions; here we hunt down every kind of question an exam can build from them and solve one of each — slowly, with a forecast you make first and a verification at the end.

Before we touch a single example, let us lay out the whole battlefield.

The scenario matrix

Think of every acid–base question as living in one cell of a grid. The two big axes are:

Which definition governs? (Arrhenius / Brønsted-Lowry / Lewis)
What kind of species or twist is thrown at you? (simple, polyprotic, amphoteric, no-proton, ion, degenerate/limiting)

Here is the grid. Each cell names the trap it hides, and the example number that defuses it.

Cell	Case class	The trap it hides	Example
A1	Arrhenius, simple strong acid + base	writing full eq vs net ionic	Ex 1
A2	Arrhenius, polyprotic (many H⁺)	getting the 1:2 stoichiometry wrong	Ex 2
A3	Arrhenius limitation (base with no OH⁻)	why the definition strains	Ex 3
B1	Brønsted-Lowry conjugate pairs	pairing the wrong two species	Ex 4
B2	Brønsted-Lowry amphoteric ion	one species, two roles	Ex 5
B3	Brønsted-Lowry strength / direction	which side equilibrium favours	Ex 6
L1	Lewis, no proton at all	still calling it acid–base	Ex 7
L2	Lewis metal–ligand (real world)	seeing complexation as acid–base	Ex 8
D	Degenerate input: bare $\ce H +$	the ultimate Lewis acid	Ex 9
X	Exam twist: classify one reaction under all three	knowing which lens even applies	Ex 10

The rule: by the end, every cell above must be struck through in your mind. Let's go.

Ex 1 — Cell A1: simple strong acid meets strong base

Forecast: guess before reading — will you need more, less, or the same volume of base? (Same concentration, and $\ce H C l$ gives one $\ce H +$ per molecule…)

Write the molecular equation. $\ce H C l (a q) + N a O H (a q) - > N a C l (a q) + H 2 O (l)$ Why this step? We must see what actually forms before stripping anything away.
Cancel spectator ions. $\ce N a +$ and $\ce C l -$ float free on both sides. $\ce H + (a q) + O H - (a q) - > H 2 O (l)$ Why this step? The essence of Arrhenius neutralisation is always $\ce H +$ meeting $\ce O H -$ ; spectators only balance charge.
Match moles. Moles of $\ce H +$ $= 0.030 L \times 0.10 mol L^{- 1} = 0.0030 mol$ . Since ratio is $1 : 1$ , we need $0.0030 mol$ of $\ce O H -$ . Why this step? Step 2 told us the ratio is one-to-one, so equal moles neutralise.
Solve for volume. $V = \frac{0.0030}{0.10} = 0.030 L = 30 mL$ .

Verify: $30 mL$ — same as the acid. Matches the forecast (equal conc, 1:1 acid). Moles $\ce H +$ = moles $\ce O H -$ = $0.0030$ . ✓

Ex 2 — Cell A2: polyprotic acid (the stoichiometry trap)

Forecast: $\ce H 2 S O 4$ is diprotic — will you need $0.25$ , $0.50$ , or $0.125$ mol of base?

Count the protons per acid molecule. Each $\ce H 2 S O 4$ can release two $\ce H +$ : $\ce H 2 S O 4 (a q) - > 2 H + (a q) + S O 4^{2 -} (a q)$ Why this step? Polyprotic acids donate more than one proton; miss this and every downstream number halves.
Balance the neutralisation. Two $\ce O H -$ are needed per acid: $\ce H 2 S O 4 (a q) + 2 N a O H (a q) - > N a 2 S O 4 (a q) + 2 H 2 O (l)$ Why this step? Each $\ce O H -$ mops up exactly one $\ce H +$ ; two protons demand two hydroxides.
Apply the 1:2 ratio. $0.25 mol \ce H 2 S O 4 \times 2 = 0.50 mol \ce N a O H$ . Why this step? Stoichiometry directly scales the mole count.

Verify: $0.50 mol$ . Protons available $= 0.25 \times 2 = 0.50$ ; hydroxides $= 0.50$ . Balanced. ✓ Forecast $0.50$ confirmed.

Ex 3 — Cell A3: an Arrhenius base with no OH⁻ in it

Forecast: where does the $\ce O H -$ come from if the molecule has none?

Let ammonia react with the solvent. $\ce N H 3 (a q) + H 2 O (l) <=> N H 4 + (a q) + O H - (a q)$ Why this step? Arrhenius requires $\ce O H -$ to appear in water — so we must find a reaction that produces it.
Identify the source of $\ce O H -$ . Ammonia pulls an $\ce H +$ off water, leaving $\ce O H -$ behind. Why this step? The hydroxide is generated indirectly, not donated by $\ce N H 3$ itself.
State the strain. Arrhenius can technically call $\ce N H 3$ a base ("it produced $\ce O H -$ "), but the real action is a proton grab — which Arrhenius can't name. Why this step? This is exactly the gap Brønsted-Lowry fills (see Ex 4).

Verify: Charge balance — left: $0$ ; right: $(+ 1) + (- 1) = 0$ . ✓ The equilibrium arrow ( $⇌$ ) is correct because ammonia is a weak base (mostly undissociated).

Ex 4 — Cell B1: naming both conjugate pairs

Forecast: which species differ by exactly one $\ce H +$ ? Circle them mentally.

Find the proton donor and its remnant. $\ce H F$ gives away $\ce H +$ and becomes $\ce F -$ . $acid \ce H F ⟶ conjugate base \ce F -$ Why this step? A Brønsted acid loses one proton; its leftover is its conjugate base.
Find the proton acceptor and its product. $\ce H 2 O$ accepts $\ce H +$ and becomes $\ce H 3 O +$ . $base \ce H 2 O ⟶ conjugate acid \ce H 3 O +$ Why this step? A base gains a proton; its product is its conjugate acid.
Confirm the "differ by one H⁺" rule for each pair.

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

Look at the figure: the red curved arrow is the single proton hopping from $\ce H F$ to $\ce H 2 O$ . Each vertical pair is linked by that one hop.

Verify: $\ce H F / \ce F -$ differ by one $\ce H$ and one $+$ of charge (H removed, charge drops by 1). $\ce H 2 O / \ce H 3 O +$ likewise. ✓

Ex 5 — Cell B2: an amphoteric ion (one species, two masks)

Forecast: does $\ce H C O 3 -$ have both a donatable proton and a lone pair to accept one? (Look at its formula — there's an $\ce H$ , and oxygens with lone pairs.)

As an acid — donate the proton to a stronger base like $\ce O H -$ : $\ce H C O 3 - + O H - - > C O 3^{2 -} + H 2 O$ Why this step? $\ce H C O 3 -$ has a removable $\ce H$ , so it can play donor.
As a base — accept a proton from an acid like $\ce H 3 O +$ : $\ce H C O 3 - + H + - > H 2 C O 3$ Why this step? Its oxygen lone pairs can grab an incoming proton, so it can play acceptor.
Conclude: amphoteric. Same ion, opposite roles, decided purely by its partner. Why this step? This is the exact definition of amphoteric — and why bicarbonate is nature's buffer.

Verify: Reaction 1 charge — left $(- 1) + (- 1) = - 2$ ; right $(- 2) + 0 = - 2$ . ✓ Reaction 2 charge — left $(- 1) + (+ 1) = 0$ ; right $0$ . ✓ Both balanced, so both roles are legitimate. (This is the chemistry behind Buffer solutions.)

Ex 6 — Cell B3: strength decides direction

Forecast: strong acids have weak conjugate bases. Which acid "wants" to keep its proton more?

Rank the two acids. $\ce H C l$ (strong) holds its proton loosely; $\ce C H 3 C O O H$ (weak) holds it tightly. Why this step? Strength = willingness to give up $\ce H +$ ; this decides who donates.
Proton flows to form the weaker acid. $\ce H C l$ donates → the strongly-holding $\ce C H 3 C O O H$ forms. Why this step? Equilibrium always favours the side with the weaker acid and weaker base (the more "content" species).
State the direction. Reaction lies far to the right (products favoured).

The figure shows a see-saw: the red weight (weaker acid $\ce C H 3 C O O H$ ) sits low on the product side — that's the stable side equilibrium settles toward.

Verify: Rule check — strong acid $\ce H C l$ ⇒ its conjugate base $\ce C l -$ is very weak ⇒ $\ce C l -$ won't grab the proton back ⇒ forward reaction dominant. ✓ Consistent with Acid-base equilibrium constants ( $K_{a}$ of $\ce H C l ≫ K_{a}$ of $\ce C H 3 C O O H$ ).

Ex 7 — Cell L1: acid–base with no proton anywhere

Forecast: if no proton moves, what does move? (Hint: count boron's valence electrons.)

Spot the electron-deficient centre. Boron in $\ce B F 3$ has only 6 valence electrons — an empty orbital, hungry for a pair. Why this step? A Lewis acid is defined as an electron-pair acceptor; the empty orbital is the acceptor.
Spot the electron donor. Nitrogen in $\ce N H 3$ carries a lone pair. Why this step? A Lewis base is an electron-pair donor; that lone pair is the gift.
Form the dative (coordinate) bond. The lone pair fills boron's empty orbital, making $\ce F 3 B - N H 3$ . Why this step? One shared pair both atoms provided from one side = a coordinate covalent bond = the Lewis adduct.

The red arrow in the figure is the lone pair moving from N into B's empty box — that single arrow is the acid–base event.

Verify: After bonding, boron reaches 8 electrons ( $6 + 2$ ) — a complete octet, explaining why the adduct is stable. ✓ No protons needed, yet the neutralisation logic (acid + base → stable product) survives.

Ex 8 — Cell L2: metal–ligand complex (real-world)

Forecast: how many ammonia molecules bind, and how many electron pairs does silver take in?

Identify the Lewis acid. $\ce A g +$ has empty orbitals and a $+$ charge pulling on electrons. Why this step? Positive, orbital-empty ⇒ electron-pair acceptor.
Identify the Lewis base(s). Each $\ce N H 3$ donates its nitrogen lone pair. Why this step? Lone-pair donor ⇒ Lewis base; two of them bind here.
Count the donations. $2 \ce N H 3$ ⇒ 2 lone pairs enter silver's orbitals, giving the linear complex $\ce [A g (N H 3) 2] +$ . Why this step? Coordination number here is 2; this is straight Coordination compounds chemistry.

Verify: Charge — left $(+ 1) + 2 (0) = + 1$ ; right $= + 1$ . ✓ Two Ag–N dative bonds ⇒ 2 electron pairs. This is precisely why $\ce A g C l$ redissolves in qualitative analysis. ✓

Ex 9 — Cell D: the degenerate input, a bare proton

Forecast: a proton with no electrons — can it possibly be a Brønsted acid (which must donate a proton)?

Arrhenius view. $\ce H +$ is the ion Arrhenius acids produce in water — the very definition. So under Arrhenius, $\ce H +$ is an acid by production. Why this step? Tests the narrowest lens on the barest input.
Brønsted view. Here $\ce H +$ is the proton being transferred; $\ce H 2 O$ is the base that accepts it. So water is the Brønsted base, and $\ce H +$ isn't a donor here — it's the cargo. Why this step? Shows the same species plays a different structural role depending on the lens.
Lewis view — the degenerate extreme. A bare $\ce H +$ has no electrons at all, so it can only accept a pair. Water's oxygen lone pair fills its need. $\ce H + + : O H 2 - > H 3 O +$ Why this step? Zero-electron input is the most electron-hungry species possible ⇒ the ultimate Lewis acid.

Verify: Charge — left $(+ 1) + 0 = + 1$ ; right $+ 1$ . ✓ Electron count: $\ce H +$ gains one shared pair (2 e⁻) to sit in $\ce H 3 O +$ — the degenerate zero-electron case resolves cleanly. Every proton transfer is thus a Lewis event, proving Brønsted ⊂ Lewis. ✓

Ex 10 — Cell X: exam twist, classify one reaction under all three lenses

Forecast: will all three labels stick to ammonia, or will one refuse?

Arrhenius. $\ce O H -$ is produced in water ⇒ $\ce N H 3$ counts as an Arrhenius base (indirectly, as in Ex 3). Why this step? Check the narrowest theory first; it just barely applies.
Brønsted. $\ce N H 3$ accepts a proton from water ⇒ Brønsted base; water is the Brønsted acid. Why this step? The cleanest description — proton transfer named directly.
Lewis. Nitrogen's lone pair is donated toward the proton it captures ⇒ Lewis base. Why this step? The broadest lens; ammonia's lone pair is the electron-pair donor.

Answer: all three apply; $\ce N H 3$ is a base in every framework — but for three different stated reasons (produces $\ce O H -$ / accepts $\ce H +$ / donates an electron pair).

Verify: Charge balance (from Ex 3) already checked: $0 = 0$ . ✓ Consistency: a species basic under Lewis (lone-pair donor) must be able to accept a proton, hence basic under Brønsted — the nested-frameworks claim of the parent note holds. ✓ Related: Hydrolysis of salts, where $\ce N H 4 +$ (the conjugate acid) reacts back with water.

Recall Did you clear every cell?

Which example handled the polyprotic 1:2 stoichiometry? ::: Ex 2 ( $\ce H 2 S O 4$ + $2 \ce N a O H$ ) Which two examples involve no proton transfer at all? ::: Ex 7 ( $\ce B F 3 + \ce N H 3$ ) and Ex 8 ( $\ce A g + + \ce N H 3$ ) What makes bare $\ce H +$ the ultimate Lewis acid? ::: It has zero electrons, so it can only accept an electron pair (Ex 9) In Ex 6, why does equilibrium lie to the right? ::: The proton flows to form the weaker acid ( $\ce C H 3 C O O H$ ); strong $\ce H C l$ readily donates One ion, both acid and base — which example? ::: Ex 5, bicarbonate $\ce H C O 3 -$ (amphoteric)

See also: Le Chatelier's Principle (why equilibria in Ex 6 shift), pH and pOH, Amphoteric oxides.