Visual walkthrough — Strong vs weak acids - bases; degree of dissociation α

This page rebuilds the central result of the parent topic one picture at a time. We are chasing this single formula:

$\alpha \approx \sqrt{\frac{K_a}{c}}$

But you are not allowed to trust it yet — we have not even said what the symbols mean. By the end you will have watched it grow out of nothing. Every step tells you WHAT we just did, WHY we did it, and WHAT IT LOOKS LIKE.

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Step 1 — What is a molecule "dissociating"?

WHAT. Picture a single acid molecule. We write it as $\text{HA}$: one hydrogen ($\text{H}$) stuck onto the rest of the molecule ($\text{A}$). "Dissociating" just means the bond snaps and the hydrogen leaves as a charged particle ${\text{H}}^{+}$, while the leftover $\text{A}$ carries the extra electron and becomes ${\text{A}}^{-}$.

WHY. Before any algebra we must see the event we are counting. Everything later is just bookkeeping on how many molecules do this.

PICTURE. On the left, an intact $\text{HA}$. On the right, the snapped pair ${\text{H}}^{+}+{\text{A}}^{-}$. The double arrow means the snap can also un-snap.

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Step 2 — Counting a whole jar, not one molecule

WHAT. Now fill a jar. Let $c$ be the initial concentration — how many acid molecules we poured in per litre, before any snapping. We measure it in mol/L (moles per litre).

WHY. One molecule tells us nothing about "fraction". Fraction needs a population. $c$ is that population size.

PICTURE. A jar of $N$ identical $\text{HA}$ dots. None have snapped yet — this is "time zero".

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Step 3 — Splitting the jar with α

WHAT. Take the jar of concentration $c$ and colour it in. A fraction $\alpha$ snapped, so:

• the snapped hydrogens number $c\alpha$ per litre,
• the leftovers ${\text{A}}^{-}$ also number $c\alpha$ (they come in pairs — one ${\text{H}}^{+}$ per one ${\text{A}}^{-}$),
• the survivors still whole number $c(1-\alpha )$.

WHY. This is the entire trick of the derivation: write every concentration in terms of just two knowns, $c$ and $\alpha$. Once everything is in those two letters, we can solve for α.

PICTURE. The same jar, now three-coloured: white survivors $c(1-\alpha )$, cyan freed ${\text{H}}^{+}$ at $c\alpha$, amber leftovers ${\text{A}}^{-}$ at $c\alpha$.

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Step 4 — The rule that fixes where equilibrium lands: $K_a$

WHAT. The snapping and un-snapping reach a standoff called equilibrium — forward and backward happen at equal rates, so the counts stop changing. Nature obeys one arithmetic rule at that standoff, the equilibrium constant $K_a$:

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

WHY this combination and not, say, a sum? Because equilibrium is a balance of rates, and reaction rates multiply concentrations of the things colliding. Products on top, reactant on the bottom. A big $K_a$ means the top (dissociated stuff) wins → strong acid. A tiny $K_a$ means the bottom wins → weak acid. $K_a$ is a fixed number for a given acid at a given temperature; it does not change with dilution (that is what makes it useful — see Acid-base equilibria and Ka, Kb).

PICTURE. A balance beam: products $[{\text{H}}^{+}][{\text{A}}^{-}]$ on one pan, reactant $[\text{HA}]$ on the other, with $K_a$ as the fixed ratio marked on the fulcrum.

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Step 5 — The clean shortcut: why we drop the $(1-\alpha )$

WHAT. For a weak acid α is tiny (a percent or two). So $1-\alpha$ is almost exactly $1$. We approximate $1-\alpha \approx 1$:

$K_a\approx c{\alpha }^2⟹{\alpha }^2\approx \frac{K_a}{c}⟹\boxed{\alpha \approx \sqrt{\frac{K_a}{c}}}$

WHY. Keeping $(1-\alpha )$ forces a quadratic (Step 7). Dropping it turns the whole thing into one square root. It costs almost nothing when α really is small — and that is precisely the weak-acid case.

PICTURE. A number line from 0 to 1: mark α at ~0.013. The gap from 1 down to $1-\alpha$ is a hair's width — visibly negligible. Alongside, the graph of $\alpha =\sqrt{K_a/c}$ falling as $c$ grows.

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Step 6 — Does the shortcut actually work? A worked check

WHAT. Test it on 0.1 M acetic acid, $K_a=1.8\times 10^{-5}$.

$\alpha \approx \sqrt{\frac{1.8\times 10^{-5}}{0.1}}=\sqrt{1.8\times 10^{-4}}=0.0134=1.34\%$

Then $[{\text{H}}^{+}]=c\alpha =0.1\times 0.0134=1.34\times 10^{-3}\text{M}$, giving $\text{pH}=-\log (1.34\times 10^{-3})=2.87$ (see pH and pOH calculations).

WHY. A formula you cannot check is folklore. Here α came out at 1.34%, so $1-\alpha =0.9866\approx 1$ — the Step 5 assumption held, and the answer is self-consistent.

PICTURE. A ladder of concentrations 0.01 / 0.1 / 1.0 M with α bars 4.24% / 1.34% / 0.42% shrinking as concentration climbs — dilution wins.

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Step 7 — The edge cases the shortcut breaks on

WHAT. The square root is a shortcut, not the law. It fails when α is not small — i.e. when the acid is strong-ish or the solution very dilute. Then you must keep $(1-\alpha )$ and solve the honest quadratic:

$c{\alpha }^2+K_a\alpha -K_a=0⟹\boxed{\alpha =\frac{-K_a+\sqrt{K_a^2+4c{K}_a}}{2c}}$

We keep the + root only, because α cannot be negative (a negative fraction of molecules is meaningless).

WHY / the "rule of 100". Check $c/K_a$. If $c/K_a>100$, α < ~5% and the square root is safe. If $c/K_a\lesssim 100$, the shortcut over-estimates α and you switch to the quadratic.

Degenerate corners:

• Strong acid ($K_a\to \infty$): the balance in Step 4 tips fully right, $\alpha \to 1$, and $[{\text{H}}^{+}]\to c$. The jar is all cyan.
• Zero concentration ($c\to 0$): $\sqrt{K_a/c}\to \infty$ — the shortcut lies (α can't exceed 1). The quadratic correctly caps α at 1, and here water's own ${\text{H}}^{+}$ (autoionization) must be added in too.
• Very dilute weak acid: [H⁺] from the acid drops toward $10^{-7}$ M, where water itself supplies comparable ${\text{H}}^{+}$ — neither formula is complete without water's contribution.

PICTURE. The two curves — shortcut $\sqrt{K_a/c}$ (cyan) and exact quadratic (white) — overlapping on the right (dilute-safe zone) but the cyan one diverging past 1 as $c\to 0$, with the amber "rule of 100" line marking where they part.

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The one-picture summary

Everything above, compressed: jar → split by α → balance rule $K_a$ → drop $(1-\alpha )$ → square root, with the exact quadratic guarding the edges.

Recall Feynman retelling — say it back in plain words

I pour a known amount $c$ of an acid into water. Some molecules snap into an ${\text{H}}^{+}$ and an ${\text{A}}^{-}$; the fraction that snaps is α. So at equilibrium I have $c\alpha$ freed hydrogens, $c\alpha$ leftovers, and $c(1-\alpha )$ survivors — I wrote all three using only $c$ and α. Nature enforces a fixed ratio $K_a=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}$; I substitute my three quantities in and get $K_a=\frac{c{\alpha }^2}{1-\alpha }$. For a weak acid α is tiny, so $1-\alpha \approx 1$, leaving $K_a\approx c{\alpha }^2$, i.e. $\alpha \approx \sqrt{K_a/c}$. That says: stronger acid → more snapping; more crowded jar → less fractional snapping. If α turns out big (dilute or strong-ish, $c/K_a<100$) I stop cheating and solve the full quadratic $c{\alpha }^2+K_a\alpha -K_a=0$, taking the positive root, and at extreme dilution I remember water makes its own ${\text{H}}^{+}$ too.

Recall

What does α physically count? ::: The fraction of the originally-dissolved acid molecules that have dissociated at equilibrium. Why can we replace $[\text{HA}]$ with $c(1-\alpha )$? ::: Because a fraction α of the $c$ molecules snapped, so the fraction that survived whole is $(1-\alpha )$. Why does α increase on dilution while $[{\text{H}}^{+}]$ decreases? ::: Diluting lowers $c$, so $\sqrt{K_a/c}$ grows (bigger slice snaps), but $[{\text{H}}^{+}]=c\alpha \propto \sqrt{c}$ still shrinks because there is less total acid. When must you abandon $\alpha =\sqrt{K_a/c}$ and use the quadratic? ::: When $c/K_a\lesssim 100$ (α not small), so the $1-\alpha \approx 1$ approximation breaks. Why the + root of the quadratic only? ::: α is a fraction of molecules; a negative value is physically impossible.