Visual walkthrough — Acids and bases — Arrhenius, Brønsted-Lowry, Lewis definitions

The parent note gave you three definitions of acids and bases. This page zooms into the heart of all three: what actually moves when an acid meets a base. We will build the single most important reaction in this chapter — a proton being handed from one molecule to another — from absolutely nothing, one picture at a time. By the end you will see why the same drawing explains Arrhenius, Brønsted-Lowry, and Lewis at once.

This deep-dives the parent topic.

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Step 1 — What is a proton, really?

WHAT. Before anything else we must agree on the smallest character in our story: the symbol ${\text{H}}^{+}$. The letter $\text{H}$ is a hydrogen atom — one proton in the centre (the nucleus) with one electron circling it. The little $+$ means "this thing is missing one negative charge." An electron carries one unit of negative charge, so ${\text{H}}^{+}$ is a hydrogen atom that has lost its only electron.

WHY. Every acid-base theory in the parent note revolves around this particle, so we cannot use the symbol ${\text{H}}^{+}$ honestly until we know it is just a bare nucleus — no electron cloud at all. That "nakedness" is the entire reason it is desperate to grab something later.

PICTURE. On the left, a neutral $\text{H}$ atom: a blue nucleus wrapped in a faint electron cloud. Strip the electron away (yellow arrow) and you are left with the tiny red dot on the right — that lone dot is ${\text{H}}^{+}$.

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Step 2 — What a "lone pair" looks like

WHAT. The taker in our story needs something to offer the hungry proton. That something is a lone pair: two electrons sitting on an atom that are not being used to bond to anything. In water, ${\text{H}}_2\text{O}$, the oxygen has two such spare pairs.

WHY. The proton has no electrons; a lone pair is a spare set of two. The whole reaction is possible only because one side is empty-handed and the other has a spare pair to share. If we did not name the lone pair now, the arrows in the next steps would point at nothing.

PICTURE. A water molecule drawn Lewis-style: the red oxygen in the middle, two blue $\text{H}$ atoms bonded to it (green bonds), and — the stars of this step — two yellow dots-in-a-pair sitting on top of the oxygen. Those are the lone pairs.

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Step 3 — The handoff itself (the curved arrow)

WHAT. Now we put the two characters together. The proton ${\text{H}}^{+}$ approaches the lone pair on the water's oxygen. The lone pair reaches out and wraps around the proton, forming a new bond. The product is ${\text{H}}_3{\text{O}}^{+}$, the hydronium ion.

$\underset{\text{electron acceptor}}{\underbrace{{\text{H}}^{+}}}+\underset{\text{electron donor,}\text{the two dots}=\text{lone pair}}{\underbrace{:{\text{OH}}_2}}⟶\underset{\text{new bond formed, }+\text{ charge carried over}}{\underbrace{{\text{H}}_3{\text{O}}^{+}}}$

Read the symbols where they sit: the colon $:$ in $:{\text{OH}}_2$ is literally the lone pair from Step 2, drawn as two dots. The arrow $⟶$ is the handoff. On the right, the $+$ that used to sit on the proton now belongs to the whole ${\text{H}}_3{\text{O}}^{+}$ molecule — charge is never destroyed, it just gets a new home.

WHY. We use a curved arrow (not a straight one) because it shows where the electrons travel: from the lone pair, to the space between oxygen and the incoming proton. Chemistry's arrows always follow the electrons, because electrons are what make bonds. This single curved arrow is the mechanical truth behind every acid-base equation in the parent note.

PICTURE. The proton (red dot) on the left, water on the right. A curved yellow arrow starts at the lone pair and curls to the gap between $\text{O}$ and the proton, showing the new bond forming. The product ${\text{H}}_3{\text{O}}^{+}$ appears with three $\text{H}$'s and a $+$ label.

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Step 4 — Same picture, three names

WHAT. Here is the payoff. The exact drawing from Step 3 is what all three theories describe — they just point at different parts of it.

• Arrhenius looks at the water on the left of a base reaction and asks "did we make ${\text{OH}}^{-}$?"
• Brønsted-Lowry watches the proton move and calls the giver an acid, the taker a base.
• Lewis watches the electron pair move and calls the giver (of electrons) a base, the taker an acid.

WHY. Notice they describe the same arrow from opposite ends. Brønsted follows the proton going one way; Lewis follows the electron pair going the other way. They can never disagree about a proton transfer — they are two labels on one event. That is why the parent note calls the theories "nested."

PICTURE. One central handoff drawing with three coloured brackets: a blue bracket labelling the proton's journey (Brønsted view), a yellow bracket labelling the electron pair's journey (Lewis view), and a green note pointing at the ${\text{OH}}^{-}$/${\text{H}}_3{\text{O}}^{+}$ produced (Arrhenius view).

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Step 5 — Conjugate pairs: the handoff runs backwards

WHAT. Every handoff can reverse. If $\text{HA}$ gives a proton to $\text{B}$, then the products ${\text{A}}^{-}$ and ${\text{HB}}^{+}$ can hand the proton back. So each side of the equation contains a giver and a taker.

$\underset{\text{acid: has }\text{H to give}}{\underbrace{\text{HA}}}+\underset{\text{base: has lone pair}}{\underbrace{\text{B}}}\rightleftharpoons \underset{\text{conjugate base: }{\text{HA minus H}}^{+}}{\underbrace{{\text{A}}^{-}}}+\underset{\text{conjugate acid: }{\text{B plus H}}^{+}}{\underbrace{{\text{HB}}^{+}}}$

Read the charges: $\text{HA}$ loses a positive ${\text{H}}^{+}$, so what remains, ${\text{A}}^{-}$, is one unit more negative — that is where the minus comes from. $\text{B}$ gains a positive ${\text{H}}^{+}$, so ${\text{HB}}^{+}$ is one unit more positive. The half-arrows $\rightleftharpoons$ mean the handoff happens both ways at once.

WHY. We need the double arrow because a proton is small and mobile; nothing stops it going back. A pair related by "differ by exactly one ${\text{H}}^{+}$" is called a conjugate pair, and spotting them is how you check any Brønsted equation is balanced.

PICTURE. The forward handoff on top (blue arrow, proton leaving $\text{HA}$) and the reverse handoff below (red arrow, proton returning). Two brackets link $\text{HA}\leftrightarrow {\text{A}}^{-}$ and $\text{B}\leftrightarrow {\text{HB}}^{+}$ as conjugate partners.

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Step 6 — Edge case: what if there is no proton at all?

WHAT. Now the degenerate-but-crucial case. What happens to the handoff when the "acid" is not ${\text{H}}^{+}$ but something else that is also electron-hungry — like boron in ${\text{BF}}_3$? Boron here has only 6 electrons around it (an incomplete octet), so it has an empty orbital, exactly like the proton had empty space.

$\underset{\text{empty orbital = the "hole"}}{\underbrace{{\text{BF}}_3}}+\underset{\text{lone pair on N}}{\underbrace{:{\text{NH}}_3}}⟶\underset{\text{dative bond, both electrons from N}}{\underbrace{{\text{F}}_3\text{B}-{\text{NH}}_3}}$

The lone pair on nitrogen ($:{\text{NH}}_3$) does exactly what water's pair did in Step 3 — it flows into the empty spot. The new bond $\text{B}-\text{N}$ is a dative (coordinate) bond: both shared electrons came from one atom.

WHY. This is the case Brønsted-Lowry cannot describe (no proton moved), yet the picture is identical to Step 3 — an empty acceptor, a lone-pair donor, one curved arrow. This is precisely why Lewis is the most general: it drops the requirement that the acceptor be ${\text{H}}^{+}$ and keeps only the arrow.

PICTURE. Left: ${\text{BF}}_3$ with a labelled empty orbital (a dashed red hole). Right: ammonia with its yellow lone pair. A curved yellow arrow flows from the nitrogen pair into boron's hole, forming the green $\text{B}-\text{N}$ dative bond.

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Step 7 — Edge case: the amphoteric molecule (giver and taker)

WHAT. One more degenerate case: a molecule that can play either role. Water is the champion. Depending on its partner, water either gives a proton or takes one.

${\text{H}}_2\text{O}+\text{HCl}⟶{\text{H}}_3{\text{O}}^{+}+{\text{Cl}}^{-}(\text{water is the base — takes }{\text{H}}^{+})$ ${\text{H}}_2\text{O}+{\text{NH}}_3⟶{\text{OH}}^{-}+{\text{NH}}_4^{+}(\text{water is the acid — gives }{\text{H}}^{+})$

In the top line water uses its lone pair (Step 2) to grab a proton. In the bottom line water hands one of its own $\text{H}$'s away, becoming ${\text{OH}}^{-}$.

WHY. Water can do both because it owns both tools at once: two lone pairs to accept, and two O–H bonds to donate. A substance with both tools is amphoteric. This is the same molecule bending to whichever role its partner leaves open.

PICTURE. Water in the centre. To the left, facing $\text{HCl}$, its lone pair grabs a proton (arrow pointing in). To the right, facing ${\text{NH}}_3$, one O–H bond releases a proton (arrow pointing out). Same molecule, two directions.

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The one-picture summary

Everything above is one drawing: an empty acceptor, a lone-pair donor, and one curved arrow between them. Change the acceptor and you switch theories; reverse the arrow and you switch acid/base labels.

Recall Feynman retelling — say it in plain words

Imagine a naked positive dot with no electrons — that is a proton, and it is starving for electrons. Now imagine a water molecule carrying two spare pairs of electrons it isn't using. When they meet, one spare pair reaches out and wraps around the naked dot, gluing them together — that new glued bond makes hydronium. That reach-out is the only thing happening. Brønsted-Lowry stands on one side and says "look, a proton just moved — the giver is the acid, the taker is the base." Lewis stands on the other side and says "no, look, an electron pair just moved — the one offering electrons is the base, the one grabbing them is the acid." They are describing the same handshake from opposite chairs, so they never disagree. And if instead of a naked proton you bring something else with an empty slot — like boron in ${\text{BF}}_3$ — the very same electron pair still reaches out and glues on. No proton needed. That is why Lewis is the biggest umbrella: it only cares about the electron pair and the empty slot, never about what the empty slot happens to be. Finally, water is special because it carries both tools — spare pairs to grab a proton and its own hydrogens to give one away — so it plays acid or base depending on who shows up. Same picture, every time.

Recall Quick self-check

A proton has how many electrons of its own? ::: Zero — it is a bare nucleus. In the handoff, which particle physically travels in the curved arrow? ::: The electron pair (from the donor's lone pair). Brønsted watches the ___ move; Lewis watches the ___ move. ::: proton; electron pair Why can ${\text{BF}}_3+{\text{NH}}_3$ be an acid-base reaction with no proton? ::: Boron has an empty orbital (acceptor) and nitrogen donates a lone pair — a Lewis handoff. What makes water amphoteric? ::: It has lone pairs (can accept ${\text{H}}^{+}$) and O–H bonds (can donate ${\text{H}}^{+}$).

Where to go next: Acid-base equilibrium constants quantifies how far the reverse arrow of Step 5 runs; pH and pOH counts the ${\text{H}}_3{\text{O}}^{+}$ from Step 3; Buffer solutions and Hydrolysis of salts use conjugate pairs directly; Coordination compounds are Step 6 repeated; Amphoteric oxides extend Step 7; Le Chatelier's Principle governs which way the $\rightleftharpoons$ leans.