Visual walkthrough — Conjugate acid-base pairs

This page rebuilds the single most important result about conjugate acid-base pairs — that an acid's strength and its conjugate base's strength are locked together by water — from absolute zero, in pictures. Every symbol is earned before it is used. If you have never seen a strength constant before, start at line one and you will be fine: we define each letter the moment it first appears.

Prerequisites we will lean on: Brønsted-Lowry theory (acids give protons, bases take them) and chemical equilibrium (reactions that run both ways). We will pay both of them off inside the walkthrough.

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Step 1 — What is a proton, and what does "transfer" look like?

WHAT. A proton is just a hydrogen atom that lost its single electron, written ${\text{H}}^{+}$. It is a bare positive charge — tiny, and always looking for a molecule to attach to.

WHY start here. Every idea on this page is a story about ONE proton hopping from one molecule to another. If we don't picture the proton clearly, none of the letters later will mean anything.

PICTURE. In the figure, the red dot is the proton. A molecule that is holding the red dot is an acid (it can give it away). The same molecule after it lets go is called its conjugate base — same molecule, one proton lighter.

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Step 2 — The forward story: an acid hands its proton to water

WHAT. Take any acid and call it $\text{HA}$. The letter $\text{A}$ just means "whatever the rest of the molecule is" and the $\text{H}$ in front is the proton it can give away. Drop it into water. Water accepts the proton and becomes ${\text{H}}_3{\text{O}}^{+}$ (hydronium — water wearing an extra proton).

$\underset{\text{acid, holds }{\text{H}}^{+}}{\underbrace{\text{HA}}}+\underset{\text{base, takes it}}{\underbrace{{\text{H}}_2\text{O}}}\rightleftharpoons \underset{\text{water + }{\text{H}}^{+}}{\underbrace{{\text{H}}_3{\text{O}}^{+}}}+\underset{\text{acid − }{\text{H}}^{+}}{\underbrace{{\text{A}}^{-}}}$

WHY water? Because we want a measurable, standard comparison. Every acid gets tested against the same partner — water — so their numbers can be compared fairly.

The $\rightleftharpoons$ symbol (double half-arrows) means the reaction runs both ways at once — this is the equilibrium idea from chemical equilibrium. HA keeps handing protons to water, and ${\text{A}}^{-}$ keeps grabbing them back, until the two rates balance.

PICTURE. The red proton leaves $\text{HA}$ and lands on water. Watch the charges: $\text{HA}$ (neutral here) becomes ${\text{A}}^{-}$ (one proton = one positive charge removed, so it goes down by one), and water becomes ${\text{H}}_3{\text{O}}^{+}$ (gained a positive charge).

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Step 3 — Turning "how far" into a number: the constant $K_a$

WHAT. At balance, we measure how much of each species is floating around. Square brackets $[]$ mean "concentration of" — how crowded that species is in the water. We define the acid strength constant $K_a$ (the "a" is for acid):

$K_a=\frac{[{\text{H}}_3{\text{O}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}$

• $[{\text{H}}_3{\text{O}}^{+}]$ — how much proton got handed off (top, because it's a product)
• $[{\text{A}}^{-}]$ — how much leftover acid-minus-proton there is (top, product)
• $[\text{HA}]$ — how much acid never let go (bottom, reactant)

WHY a fraction of products over reactants? Because we want ONE number that says how far the proton-handoff went. Products on top, starting material on the bottom: if the top is big and the bottom is small, the proton mostly left → big $K_a$ → strong acid. If the top is tiny, almost nothing left → small $K_a$ → weak acid. That single ratio is the whole meaning of "acid strength."

WHY is water not in the fraction? Water is the solvent — it's everywhere, so its concentration barely changes and it's baked into the constant. (This is a rule from equilibrium: pure liquids don't appear.)

PICTURE. A balance beam. Products pile on the top pan, the leftover acid on the bottom. The tilt of the beam is $K_a$: tilted toward products = strong acid.

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Step 4 — Now play the same story backwards: the conjugate base and $K_b$

WHAT. Take the leftover ${\text{A}}^{-}$ from Step 2 and ask the reverse question: can it grab a proton off water? When it does, it pulls a proton off ${\text{H}}_2\text{O}$, rebuilding $\text{HA}$ and leaving behind ${\text{OH}}^{-}$ (hydroxide — water minus a proton).

$\underset{\text{base, wants }{\text{H}}^{+}}{\underbrace{{\text{A}}^{-}}}+{\text{H}}_2\text{O}\rightleftharpoons \underset{\text{rebuilt acid}}{\underbrace{\text{HA}}}+\underset{\text{water − }{\text{H}}^{+}}{\underbrace{{\text{OH}}^{-}}}$

Its strength gets its own number, the base strength constant $K_b$ (the "b" is for base), built exactly like $K_a$ (products over reactants):

$K_b=\frac{[\text{HA}][{\text{OH}}^{-}]}{[{\text{A}}^{-}]}$

• $[\text{HA}]$ — the acid we rebuilt (product, top)
• $[{\text{OH}}^{-}]$ — the hydroxide left behind (product, top)
• $[{\text{A}}^{-}]$ — the base that didn't grab a proton (reactant, bottom)

WHY define $K_b$ at all? Because "strong acid → weak base" is a claim we want to prove, not just assert. To compare acid-strength and base-strength we need both on the same footing: two ratios of the same shape.

WHY is water again missing from the fraction? Same pure-liquid rule as in Step 3: water is the solvent, its concentration is essentially constant, so equilibrium convention absorbs $[{\text{H}}_2\text{O}]$ into the constant and it never appears in $K_b$. Both $K_a$ and $K_b$ drop water for the exact same reason — keep that consistent, it matters in the next step.

PICTURE. The red proton now travels the OTHER way — off water, onto ${\text{A}}^{-}$. Notice the same pair $(\text{HA},{\text{A}}^{-})$ appears in both Step 3 and Step 4, just swapped between top and bottom. Hold that thought — it is the whole trick.

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Step 5 — Multiply the two numbers and watch the cancellation

WHAT. Multiply $K_a$ (Step 3) by $K_b$ (Step 4):

$K_a\times K_b=\frac{[{\text{H}}_3{\text{O}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}\times \frac{[\text{HA}][{\text{OH}}^{-}]}{[{\text{A}}^{-}]}$

WHY multiply? Because the acid ($K_a$) and its conjugate base ($K_b$) are two views of the same pair. Multiplying is how we ask "what do these two facts, combined, force to be true?"

Now the magic: $[{\text{A}}^{-}]$ is on the top of the first fraction and the bottom of the second — it cancels. $[\text{HA}]$ is on the bottom of the first and the top of the second — it cancels too. Everything about the specific acid disappears:

$K_a\times K_b=[{\text{H}}_3{\text{O}}^{+}][{\text{OH}}^{-}]$

PICTURE. The two fractions side by side. Cross out $[{\text{A}}^{-}]$ against $[{\text{A}}^{-}]$, and $[\text{HA}]$ against $[\text{HA}]$ — the red strokes show what evaporates. Only water's own pieces survive.

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Step 6 — What survived is just water talking to itself: $K_w$

WHAT. Water quietly hands protons to itself all day long:

${\text{H}}_2\text{O}+{\text{H}}_2\text{O}\rightleftharpoons {\text{H}}_3{\text{O}}^{+}+{\text{OH}}^{-}$

The number for how far this goes has its own name, the water strength constant $K_w$ (the "w" is for water):

$K_w=[{\text{H}}_3{\text{O}}^{+}][{\text{OH}}^{-}]=1.0\times 10^{-14}\text{(at }25^{\circ }\text{C)}$

But that product $[{\text{H}}_3{\text{O}}^{+}][{\text{OH}}^{-}]$ is exactly what survived our cancellation in Step 5. So:

$\boxed{K_a\times K_b=K_w=1.0\times 10^{-14}}$

WHY this is beautiful. The specific acid never mattered. Whatever $\text{HA}$ you picked, its $K_a$ and its conjugate's $K_b$ must multiply to the same fixed water number. That's why strength is inverse: if $K_a$ goes up, $K_b$ must go down to keep the product fixed.

PICTURE. Two water molecules, one red proton hopping between them, producing exactly the ${\text{H}}_3{\text{O}}^{+}$ and ${\text{OH}}^{-}$ that our leftover product was made of.

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Step 7 — Edge and degenerate cases (never leave the reader stranded)

Case A — the very strong acid ($K_a\gg 1$, e.g. HCl). For a practically complete reaction the products must vastly outweigh the reactant, so $K_a$ must be far greater than 1 — not merely $\ge 1$. (At $K_a=1$ the top and bottom of the fraction are equal, meaning products and reactants sit at the same concentration — that is only "half-dissociated," nowhere near complete.) When $K_a\gg 1$, then $K_b=K_w/K_a$ is minuscule. ${\text{Cl}}^{-}$ has essentially zero desire to grab a proton back — it is a negligible base, a spectator. This is the Brønsted meaning of "the reaction goes to completion."

Case B — the very weak acid ($K_a\ll 1$). Then $K_b=K_w/K_a$ becomes large: a weak acid has a genuinely strong conjugate base.

Case C — water itself (the degenerate case). Set the acid equal to water: $\text{HA}={\text{H}}_2\text{O}$ and ${\text{A}}^{-}={\text{OH}}^{-}$. Then $K_a$ of water $\times$ $K_b$ of ${\text{OH}}^{-}$ still equals $K_w$. Water is its own example — it is amphoteric, sitting exactly at the balance point. This is why pure water has equal $[{\text{H}}_3{\text{O}}^{+}]$ and $[{\text{OH}}^{-}]$.

Case D — temperature is not $25^{\circ }$C. $K_w$ is only $10^{-14}$ at $25^{\circ }$C. Heat water and $K_w$ grows (more self-ionization). The relationship $K_a{K}_b=K_w$ still holds — just plug in the new $K_w$.

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The one-picture summary

This single figure is the whole derivation: the acid loop (top), the base loop (bottom), the shared pair cancelling in the middle, and water surviving as $K_w$.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a red ball (a proton) that molecules toss around. An acid is a molecule holding the ball; once it throws it, we call the ball-less version its conjugate base. We drop the acid in water and count how eagerly it throws — that count is $K_a$. Then we take the ball-less leftover and count how eagerly it catches a ball back off water — that count is $K_b$. When we multiply the two counts, every mention of our specific molecule cancels out, top against bottom, and all that's left is water tossing balls to itself — which has its own fixed count, $K_w=10^{-14}$. So no matter which acid you chose, its throwing-count times its catching-count is always the same water number. That's the punchline: good throwers are bad catchers. A strong acid must have a weak conjugate base, because their product can never change.

Recall Quick self-test

Why does the specific acid cancel out? ::: Because $[\text{HA}]$ and $[{\text{A}}^{-}]$ each appear once on a numerator and once on a denominator when $K_a$ and $K_b$ are multiplied. If $K_a=1.0\times 10^{-3}$, what is $K_b$ of the conjugate at 25°C? ::: $K_b=10^{-14}/10^{-3}=1.0\times 10^{-11}$. What does a large $K_a$ mean physically? ::: The proton mostly left the acid — products dominate — so it is a strong acid, hence a very weak conjugate base.