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

Before you can read $\alpha =\sqrt{K_a/C}$ you must own every letter in it. This page builds each symbol from nothing — plain words first, then a picture, then the reason the topic can't live without it. Read top to bottom; each block leans on the one above.

1. Concentration, $C$ — "how crowded is the jar?"

A mole is just a counting word — like "dozen" but enormous (about $6\times 10^{23}$ particles). So $C$ answers: how tightly packed are my acid molecules?

Look at the figure below. Left jar: many dots in a small space — high $C$. Right jar: same dots spread into a bigger volume — low $C$. Diluting means adding water = moving from left to right = lowering $C$.

WHY the topic needs it: Ostwald's whole story is "what happens when you change crowding." Without a number for crowding, there is nothing to change.

2. The acid molecule HA and its split — the reaction arrow $\rightleftharpoons$

Read the pieces:

• HA — the whole, un-split acid molecule.
• H⁺ — a hydrogen atom that left its electron behind, so it carries $+1$ charge. This is the "acid part."
• A⁻ — everything else, now carrying $-1$ charge (it kept the electron).
• $\rightleftharpoons$ — the double harpoon. A single arrow $\to$ means "goes all the way." The double harpoon means "goes both ways at once" — some HA splits while some H⁺ and A⁻ rejoin, forever. This is the picture of equilibrium.

WHY the topic needs it: "Weak" acid means the harpoon sits mostly on the HA side. If the arrow went one way ($\to$), nothing would be weak and there'd be no fraction to compute.

3. Degree of ionization, $\alpha$ — "what fraction actually split?"

Read the Greek letter: α = "alpha", the first letter of the Greek alphabet, borrowed by scientists as a name for "this particular fraction."

• $\alpha =0$ → nothing split (not an acid at all here).
• $\alpha =1$ → everything split (a strong acid).
• $\alpha =0.02$ → 2 out of every 100 molecules split (a typical weak acid).

In the figure, out of 100 molecules the yellow ones are the few that split. Count them, divide by 100, and you have $\alpha$. To turn $\alpha$ into a percentage, multiply by 100: $\alpha =0.02$ is $2\%$ ionized.

WHY the topic needs it: $\alpha$ is the quantity Ostwald's law predicts. Everything else is machinery for pinning down this one number.

4. Building the equilibrium amounts: $C\alpha$ and $C(1-\alpha )$

Now combine $C$ (how many total) with $\alpha$ (what fraction split). This is pure arithmetic:

WHAT we did: started with $C$ moles/L. A fraction $\alpha$ splits, so the amount that split is (total) × (fraction) $=C\alpha$. Because HA → H⁺ + A⁻ makes one of each, both $[{\text{H}}^{+}]$ and $[{\text{A}}^{-}]$ equal $C\alpha$. The leftover un-split HA is what's not the fraction: $C-C\alpha =C(1-\alpha )$.

WHAT IT LOOKS LIKE: the parent's ICE table is exactly this — a bookkeeping ledger. "$1-\alpha$" is just "the fraction that did not split," because all fractions in the jar must add to 1.

WHY the topic needs it: these three amounts are the ingredients for the next symbol, Ka.

5. The equilibrium constant, $K_a$ — "the ratio nature holds fixed"

Read it slowly:

• $[{\text{H}}^{+}]$ = concentration of hydrogen ion.
• The subscript $a$ in $K_a$ just labels it "for an acid" (there are $K_b$ for bases, $K_w$ for water).
• $K$ is a constant — a fixed value set only by which acid it is and the temperature. Change the crowding $C$ and $\alpha$ will shift, but this ratio does not budge.

Big vs small $K_a$: a large $K_a$ means the top (ions) dominates → lots of splitting → stronger acid. A tiny $K_a$ (like $10^{-5}$) means the bottom (un-split HA) dominates → weak acid.

WHY the topic needs it: the fixedness of $K_a$ is the lever. Because it can't change, when $C$ drops, $\alpha$ is forced to rise to keep the ratio constant. That forcing is Ostwald's law.

6. Putting the ratio together — where ${\alpha }^2$ appears

Substitute the amounts from §4 into the ratio from §5:

WHAT we did: put $C\alpha$ on top twice (once for H⁺, once for A⁻) and $C(1-\alpha )$ on the bottom.

WHY ${\alpha }^2$? Because two ion concentrations multiply on top, and each carries one factor of $\alpha$. The exponent 2 (written ${\alpha }^2$, read "alpha squared," meaning $\alpha \times \alpha$) is the mathematical fingerprint of "one molecule makes two ions." This squaring is the whole reason a square root shows up later — the square root is simply the operation that undoes squaring.

7. The weak-acid approximation and the square root  ​

The approximation: for a weak acid $\alpha$ is tiny, so $1-\alpha \approx 1$. Drop the denominator: $K_a\approx C{\alpha }^2$

Now solve for $\alpha$ — divide by $C$, then take the square root to undo the squaring: ${\alpha }^2\approx \frac{K_a}{C}⟹\boxed{\alpha =\sqrt{\frac{K_a}{C}}}$

WHAT IT LOOKS LIKE: the figure below plots $\alpha$ against $C$. As $C$ shrinks toward zero (left), $\alpha$ climbs — the curve is the shape of $\sqrt{K_a/C}$. This is the picture behind "dilution increases ionization." Notice the curve eventually leaves the safe green band ($\alpha <0.05$) as $C$ gets small — that is the approximation breaking.

WHY the  ​ and not, say, dividing? Because the unknown $\alpha$ appears squared. Only the square root peels off that exponent. If ionization made one ion instead of two, there'd be no square and no root — the two-ion split is what earns the root.

8. From $\alpha$ to acidity: deriving $[{\text{H}}^{+}]=\sqrt{K_{a}C}$ here

We never need the parent page for this — it follows in one line from what we already have.

WHAT we know: from §4, the concentration of hydrogen ion is $[{\text{H}}^{+}]=C\alpha$. From §7, $\alpha =\sqrt{K_a/C}$. Substitute the second into the first: $[{\text{H}}^{+}]=C\alpha =C\sqrt{\frac{K_a}{C}}.$

Now simplify the algebra step by step. Write $C=\sqrt{C}\cdot \sqrt{C}$ (since a number is its own square root times itself). Then $C\sqrt{\frac{K_a}{C}}=\sqrt{C}\cdot \sqrt{C}\cdot \frac{\sqrt{K_a}}{\sqrt{C}}=\sqrt{C}\cdot \sqrt{K_a}=\sqrt{K_{a}C}.$ One $\sqrt{C}$ on top cancels the $\sqrt{C}$ hidden in the denominator, leaving: $\boxed{[{\text{H}}^{+}]=\sqrt{K_{a}C}}$

WHAT IT MEANS: the ion concentration grows like $\sqrt{C}$, not like $C$. So halving $C$ does not halve the acidity — it cuts it only by a factor $1/\sqrt{2}\approx 0.71$. This is the mathematical root of the famous "dilution makes a weak acid less acidic, but by less than you'd guess."

9. Logarithms and pH — reading tiny numbers comfortably

Concentrations like $[{\text{H}}^{+}]=0.00134$ are awkward. The base-10 logarithm rescales them.

We also name $\text{p}{K}_a=-{\log }_{10}{K}_a$ — the same trick applied to $K_a$.

Now derive the pH formula from §8. Start from $[{\text{H}}^{+}]=\sqrt{K_{a}C}$ and take $-{\log }_{10}$ of both sides: $\text{pH}=-{\log }_{10}\sqrt{K_{a}C}.$

Step A — a square root is a power of $\frac{1}{2}$. By definition $\sqrt{x}=x^{1/2}$, and the log of a power pulls the exponent out front (${\log }_{10}(x^n)=n{\log }_{10}x$). So $\text{pH}=-{\log }_{10}((K_{a}C{)}^{1/2})=-\frac{1}{2}{\log }_{10}(K_{a}C).$ This is where the $\frac{1}{2}$ comes from — it is the exponent $1/2$ of the square root, dragged out by the power rule.

Step B — log of a product is a sum. The core log rule ${\log }_{10}(xy)={\log }_{10}x+{\log }_{10}y$ splits the product $K_{a}C$: $\text{pH}=-\frac{1}{2}({\log }_{10}{K}_a+{\log }_{10}C).$

Step C — distribute the $-\frac{1}{2}$ and recognise $\text{p}{K}_a$. Multiply through: $\text{pH}=-\frac{1}{2}{\log }_{10}{K}_a-\frac{1}{2}{\log }_{10}C=\underset{=\text{p}{K}_a}{\underbrace{(-{\log }_{10}{K}_a)}}\cdot \frac{1}{2}-\frac{1}{2}{\log }_{10}C.$ Since $-{\log }_{10}{K}_a=\text{p}{K}_a$, the first piece is $\frac{1}{2}\text{p}{K}_a$, giving the finished formula: $\boxed{\text{pH}=\frac{1}{2}(\text{p}{K}_a-{\log }_{10}C)}$

WHY the topic needs it: the whole point of computing $\alpha$ is to report acidity, and chemists report acidity as pH. Logs are the translator, and this formula is the fast shortcut for a weak acid.

How the foundations feed the topic — in plain words

Read this as a story, top to bottom — each row is used by the next:

1. Concentration $C$ and the fraction $\alpha$ are our two raw numbers: how much acid, and how much of it split.
2. Multiply them to get the equilibrium amounts $C\alpha$ (the ions) and $C(1-\alpha )$ (the leftover HA).
3. Feed those amounts into the fixed ratio $K_a$, and you get the exact relation $K_a=C{\alpha }^2/(1-\alpha )$.
4. For a weak acid, the approximation $1-\alpha \approx 1$ simplifies it, and the square root undoes the squaring — out pops Ostwald's law $\alpha =\sqrt{K_a/C}$.
5. Put $\alpha$ back into $[{\text{H}}^{+}]=C\alpha$ to get $[{\text{H}}^{+}]=\sqrt{K_{a}C}$, then apply base-10 logs to report the answer as pH.

The same steps as a quick map:

See the parent note Ostwald's dilution law for where this machine is put to work. The rising-$\alpha$ story connects to Le Chatelier's Principle; the pH-reporting step feeds pH Calculation Methods and Henderson-Hasselbalch Equation; the fixed-ratio idea underlies Common Ion Effect and Buffer Capacity; multi-step acids appear in Polyprotic Acids; and $\alpha$'s effect on ion count links to Conductivity of Solutions.

Equipment checklist

Cover the right side and test yourself — you're ready for the derivation when every line is automatic.