Visual walkthrough — Ostwald's dilution law (weak acid) - α = √(Ka - C)

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Step 0 — The one thing we are watching

Before any maths, picture the actual stuff in the beaker.

We have a weak acid. Call it $\text{HA}$. That is just a name for a molecule made of two parts stuck together:

• H — a hydrogen piece that can pop off as a positive particle written ${\text{H}}^{+}$ (this is what makes things "acidic").
• A — the leftover piece, which carries the leftover negative charge, written ${\text{A}}^{-}$.

When $\text{HA}$ splits, we write: $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

The double arrow $\rightleftharpoons$ means: it splits and re-joins, over and over, until the two happen at the same speed. That balance point is called equilibrium.

Everything below is the story of how $\alpha$ depends on $C$.

Look at the picture: left jar is concentrated (crowded, few split), right jar is dilute (roomy, more split). That single observation is the whole law. Now we make it exact.

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Step 1 — Count the particles (the ICE table)

WHAT. We track three quantities as the reaction proceeds: how much $\text{HA}$, ${\text{H}}^{+}$, and ${\text{A}}^{-}$ sit in each litre.

WHY. The constant $K_a$ (coming in Step 2) is built out of these three amounts. We cannot write it until we know each one in terms of $C$ and $\alpha$.

PICTURE. Start with $C$ of $\text{HA}$ and nothing else. A fraction $\alpha$ of them split. So the amount that splits is: $\underset{\text{start amount}}{\underbrace{C}}\times \underset{\text{fraction that go}}{\underbrace{\alpha }}=C\alpha$

Every molecule that leaves $\text{HA}$ becomes one ${\text{H}}^{+}$ and one ${\text{A}}^{-}$ — a one-to-one-to-one deal. So:

	$\text{HA}$	${\text{H}}^{+}$	${\text{A}}^{-}$
Start	$C$	$0$	$0$
Change	$-C\alpha$	$+C\alpha$	$+C\alpha$
End (equilibrium)	$C(1-\alpha )$	$C\alpha$	$C\alpha$

• $C(1-\alpha )$ = what's left un-split (we took $C\alpha$ away from $C$).
• $C\alpha$ appears twice — one for each new ion.

The bars in the figure shrink ($\text{HA}$) and grow (the ions) by exactly the same height $C\alpha$. That equal height is the visual heart of Step 1.

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Step 2 — The rule the beaker must obey ($K_a$)

WHAT. Nature fixes a ratio at equilibrium. For this temperature, that ratio is a constant called $K_a$ (the acid dissociation constant).

WHY. The forward splitting and backward re-joining settle at a specific balance. Chemistry tells us that balance is captured by: $K_a=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}$ where the square brackets $[]$ mean "concentration of, at equilibrium, in mol/L." The products (things made) go on top; the reactant (thing consumed) goes on the bottom.

PICTURE. Now plug in the "End" row from Step 1: $K_a=\frac{\stackrel{[{\text{H}}^{+}]}{\overbrace{(C\alpha )}}\cdot \stackrel{[{\text{A}}^{-}]}{\overbrace{(C\alpha )}}}{\underset{[\text{HA}]}{\underbrace{C(1-\alpha )}}}$

• top: $C\alpha \cdot C\alpha =C^2{\alpha }^2$ (two ions multiplied together).
• bottom: $C(1-\alpha )$ (the survivors).

Cancel one $C$ (top has $C^2$, bottom has $C$): $\boxed{K_a=\frac{C{\alpha }^2}{1-\alpha }}$

This is the exact law — no shortcuts yet. The figure shows $K_a$ as a fixed horizontal chalk line: the beaker is free to change $C$ and $\alpha$, but their combination $\frac{C{\alpha }^2}{1-\alpha }$ must always land back on that line.

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Step 3 — The clever simplification (why $1-\alpha \approx 1$)

WHAT. For a weak acid, very little splits, so $\alpha$ is tiny. We replace $1-\alpha$ with just $1$.

WHY. Weak means shy: a typical $\alpha$ is $0.01$ (1 in 100 splits). Then $1-\alpha =0.99$, which is $1$ to within 1%. Keeping the $-\alpha$ buys us almost nothing but a messier equation.

PICTURE. On a number line from $0$ to $1$, mark $\alpha$ near the far-left wall. The gap $1-\alpha$ is almost the full length $1$. So: $K_a=\frac{C{\alpha }^2}{1-\alpha }\approx \frac{C{\alpha }^2}{1}=C{\alpha }^2$

The rule of thumb: this is safe while $\alpha <0.05$ (under 5% split).

The figure plots the exact curve and the approximate curve on top of each other — they hug tightly on the left (small $\alpha$) and peel apart on the right. We are living on the left.

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Step 4 — Unlock $\alpha$ (where the square root is born)

WHAT. Solve $K_a\approx C{\alpha }^2$ for $\alpha$.

WHY. We want $\alpha$ alone on one side so we can predict it from the two knowns, $K_a$ and $C$.

PICTURE. Undo the algebra one move at a time: $K_a=C{\alpha }^2\stackrel{÷C}{\to }{\alpha }^2=\frac{K_a}{C}\stackrel{\sqrt{}}{\to }\alpha =\sqrt{\frac{K_a}{C}}$

The square root appears because $\alpha$ was squared — and it was squared back in Step 2 because splitting one molecule creates two ions that got multiplied together. That's the physical origin of the ​: it is undoing the "two-ions-multiplied" step.

The curve is $\alpha$ against $C$: it swoops upward as $C\to 0$ (dilute → more splitting) and flattens toward zero as $C$ grows (crowded → less splitting). Trace the red dot sliding leftward as we add water.

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Step 5 — From $\alpha$ to acidity ($[{\text{H}}^{+}]$ and pH)

WHAT. Convert the fraction $\alpha$ into the actual concentration of acid particles, $[{\text{H}}^{+}]$, and then into pH.

WHY. $\alpha$ alone doesn't tell you "how sour." The number that does is $[{\text{H}}^{+}]$ — how many acid particles per litre. From Step 1, $[{\text{H}}^{+}]=C\alpha$.

PICTURE. Substitute the just-found $\alpha$: $[{\text{H}}^{+}]=C\alpha =C\cdot \sqrt{\frac{K_a}{C}}=\sqrt{K_{a}C}$

Watch the $C$: outside is $C^1$, inside the root is $C^{-1}$, so combined they give $C^{+1/2}=\sqrt{C}$. That means $[{\text{H}}^{+}]$ scales with $\sqrt{C}$ — not with $C$. This is the punchline of the whole page:

Now for pH. First two definitions we need:

Take $-{\log }_{10}$ of both sides of $[{\text{H}}^{+}]=\sqrt{K_{a}C}$ (the root turns into a $\frac{1}{2}$ out front, because ${\log }_{10}\sqrt{x}=\frac{1}{2}{\log }_{10}x$): $\text{pH}=-{\log }_{10}\sqrt{K_{a}C}=-\frac{1}{2}{\log }_{10}(K_{a}C)=-\frac{1}{2}({\log }_{10}{K}_a+{\log }_{10}C)=\frac{1}{2}(\text{p}{K}_a-{\log }_{10}C)$

The last step used $-{\log }_{10}{K}_a=\text{p}{K}_a$ (our definition above) and split the product's log into a sum.

Two chalk curves: $\alpha$ rising and $[{\text{H}}^{+}]$ falling as we move toward dilution. They cross the same water-adding arrow going opposite ways — the picture of the paradox.

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Step 6 — The edge case: when the shortcut lies

WHAT. If the acid is not that weak (or extremely dilute), $\alpha$ climbs past $0.05$ and the "$1-\alpha \approx 1$" trick fails.

WHY. When lots of $\text{HA}$ has split, ignoring the missing $\alpha$ in $1-\alpha$ throws away a real chunk. The formula $\sqrt{K_a/C}$ then over-predicts $\alpha$ — it can even spit out a value bigger than $1$, which is impossible (you cannot split more than 100%).

PICTURE. Take $0.001\text{M}$ HF, $K_a=6.8\times 10^{-4}$. The shortcut says: ${\alpha }_{\text{shortcut}}=\sqrt{\frac{6.8\times 10^{-4}}{0.001}}=\sqrt{0.68}=0.825(82.5\%)$ That's way over the 5% line — untrustworthy. Go back to the exact Step-2 equation and keep $1-\alpha$. Here is the rearrangement, one honest move at a time:

\;\xrightarrow{\;\times(1-\alpha)\;}\; K_a(1-\alpha) = C\alpha^2 \;\xrightarrow{\;\text{expand}\;}\; K_a - K_a\alpha = C\alpha^2 \;\xrightarrow{\;\text{all to one side}\;}\; C\alpha^2 + K_a\alpha - K_a = 0$$ - We first cleared the fraction by multiplying both sides by $(1-\alpha)$. - Then we expanded $K_a(1-\alpha) = K_a - K_a\alpha$. - Finally we moved everything to the left, landing on a **quadratic** — an equation with an $\alpha^2$ term, an $\alpha$ term, and a constant. Plug the HF numbers in and solve: $0.001\,\alpha^2 + (6.8\times10^{-4})\alpha - 6.8\times10^{-4} = 0$, giving $\alpha \approx 0.54$ — a full **54\%**, not the phantom 82.5\%. ![[deepdives/dd-chemistry-2.6.12-d2-s07.png]] The figure overlays the shortcut curve (which shoots off the top past $\alpha=1$) and the honest quadratic curve (which politely bends and never exceeds $1$). Where they split apart is exactly the danger zone. > [!mistake] The trap > Blindly using $\alpha=\sqrt{K_a/C}$ for a not-so-weak acid gives $\alpha>1$ — physically nonsense. **Check the 5\% rule first.** If it fails, solve the quadratic $C\alpha^2 + K_a\alpha - K_a = 0$. --- ## The one-picture summary ![[deepdives/dd-chemistry-2.6.12-d2-s08.png]] Everything on one board: **left**, crowded jar → small $\alpha$, high $[\text{H}^+]$; **right**, roomy jar → big $\alpha$, low $[\text{H}^+]$; **middle**, the chain that connects them — $$\text{HA}\rightleftharpoons\text{H}^++\text{A}^- \;\to\; K_a=\frac{C\alpha^2}{1-\alpha} \;\xrightarrow{\alpha\ll1}\; \alpha=\sqrt{\tfrac{K_a}{C}} \;\to\; [\text{H}^+]=\sqrt{K_a C}.$$ > [!recall]- Feynman retelling of the whole walkthrough > You've got a jar of shy "acid robots," each able to snap into an $\text{H}^+$ piece and an $\text{A}^-$ piece. We started by *counting*: if a fraction $\alpha$ of the $C$ robots snap, then $C\alpha$ snapped, leaving $C(1-\alpha)$ whole — and each snap makes **two** pieces. > > Nature insists on a fixed balance number $K_a$: pieces-made times pieces-made, over robots-left. Plugging the counts in gives $K_a = C\alpha^2/(1-\alpha)$. Because the robots are shy, hardly any snapped, so "$1-\alpha$" is basically $1$ — drop it. That leaves $K_a = C\alpha^2$. Un-squaring (that's where the square root comes from — it undoes the "two pieces multiplied") gives $\alpha = \sqrt{K_a/C}$. > > Add water: $C$ shrinks, and since $C$ sits under the root, $\alpha$ **grows** — more robots feel brave. But the sourness $[\text{H}^+] = \sqrt{K_a C}$ still **drops**, because you spread the same robots over more water. Finally, if the robots aren't shy after all ($\alpha$ big), the "drop the $1-\alpha$" trick breaks and you must solve the honest quadratic — otherwise the formula claims more than 100\% snapped, which no jar can do. > [!recall]- Quick self-test > Why is there a square root in Ostwald's law? > Because splitting one molecule makes two ions that get **multiplied** in $K_a$, giving $\alpha^2$; solving for $\alpha$ un-squares it. > > Diluting increases $\alpha$ but does the acid get more sour? > No — $[\text{H}^+]=\sqrt{K_aC}$ **decreases** with dilution; pH rises. > > When does $\alpha=\sqrt{K_a/C}$ fail? > When $\alpha>0.05$; then $1-\alpha\ne1$ and you solve $C\alpha^2+K_a\alpha-K_a=0$. **Connections:** [[Le Chatelier's Principle]] (why dilution favours splitting) · [[pH Calculation Methods]] (the $[\text{H}^+]=\sqrt{K_aC}$ shortcut) · [[Common Ion Effect]] (adding $\text{A}^-$ pushes $\alpha$ down) · [[Conductivity of Solutions]] (conductance $\propto C\alpha$) · [[Henderson-Hasselbalch Equation]] · [[Buffer Capacity]] · [[Polyprotic Acids]].