Ostwald's dilution law (weak acid) - α = √(Ka - C): Worked examples

The scenario matrix

Every Ostwald problem lands in one of these cells. The rest of the page fills each cell with a worked example.

Cell	Situation	Which tool?	Danger
A	Normal weak acid, $α < 5%$	Simple $α = K_{a} / C$	none
B	Dilution comparison (same acid, two $C$ )	Ratio trick $α \propto 1/ C$	thinking pH falls
C	$α$ turns out large ( $> 5%$ )	Full quadratic $K_{a} = \frac{C α ^{2}}{1 - α}$	using shortcut anyway
D	Reverse problem: given $α$ , find $K_{a}$ or $C$	Rearrange the exact formula	using approximate one
E	Limiting / degenerate: $C \to 0$ , or extremely dilute	Watch $α \to 1$ ; water's own $[H^{+}]$ matters	formula predicting $α > 1$
F	Strong acid mistakenly fed in	Don't use Ostwald; $[H^{+}] = C$	$α > 1$ nonsense
G	Real-world word problem (vinegar)	Translate words → $C, K_{a}$	unit slips
H	Exam twist: percent ionisation given, find pH from scratch	chain $α \to [H^{+}] \to$ pH	forgetting $C$

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

The figure above plots $α$ against $C$ on a log axis. Notice the cyan curve (the exact answer) hugging the white dashed line (the shortcut) at high $C$ , then peeling away toward $α = 1$ as we dilute — that peel-off region is exactly cells C and E.

Cell A — the ordinary weak acid

Worked example A · 0.05 M formic acid,

$K_{a} = 1.8 \times 1 0^{- 4}$ Forecast: guess — will $α$ be above or below $5%$ ? (Formic is ~10× stronger than acetic.)

Step 1. Apply the shortcut $α = K_{a} / C$ . $α = \frac{1.8 \times 1 0 ^{- 4}}{0.05} = 3.6 \times 1 0^{- 3} = 0.0600$ Why this step? The shortcut is the fastest route when we can trust it — we test that next.

Step 2. Check the $5%$ rule. $α = 6.0% > 5%$ . Why this step? The shortcut secretly assumed $1 - α \approx 1$ . At $6%$ that assumption is already fraying, so we should refine — this actually pushes us into cell C territory. Keep this in mind; we redo it properly in Example C.

Verify: plug $α = 0.06$ back into the exact law: $\frac{C α ^{2}}{1 - α} = \frac{0.05 \times 0.0036}{0.94} = 1.91 \times 1 0^{- 4}$ , slightly above the true $1.8 \times 1 0^{- 4}$ — confirming the shortcut over-estimates $α$ here. Lesson: always run the $5%$ test.

Cell B — dilution done right (and the pH trap)

Worked example B · Take the 0.2 M acetic acid above and dilute 100×

Forecast: $α$ goes UP by roughly which factor? And does pH go up or down?

Step 1. New concentration $C = 0.002$ M. Ratio trick: since $α \propto 1/ C$ , diluting $100 \times$ multiplies $α$ by $100 = 10$ . $α_{new} = 9.49 \times 1 0^{- 3} \times 10 = 9.49 \times 1 0^{- 2} = 9.49%$ Why this step? We avoid a fresh square root by exploiting the proportionality — but note $9.49% > 5%$ , so this is now cell C and the answer is an over-estimate.

Step 2. $[H^{+}] = K_{a} C = 1.8 \times 1 0^{- 5} \times 0.002 = 3.6 \times 1 0^{- 8} = 1.90 \times 1 0^{- 4}$ M. Why: to test the pH trap we need the actual ion count, not the percentage.

Step 3. $pH = - lo g (1.90 \times 1 0^{- 4}) = 3.72$ . Why this step: compare with the original pH $= 2.72$ .

The lesson (Mistake-buster): $α$ rose $10 \times$ , yet pH rose from $2.72$ to $3.72$ — the solution became less acidic. Percent-ionised went up; total ions per litre went down.

Verify: $Δ pH = 3.72 - 2.72 = 1.00$ . Since $[H^{+}] \propto C$ and $C$ fell $100 \times$ , $[H^{+}]$ falls $10 \times$ , so pH rises by exactly $lo g 10 = 1.00$ ✓.

Cell C — when the shortcut breaks: the quadratic

Worked example C · 0.001 M HF,

$K_{a} = 6.8 \times 1 0^{- 4}$ (the parent's Example 3, done exactly) Forecast: the shortcut screams $α = 0.68 = 0.825$ . Do you trust $82.5%$ ?

Step 1. Reject the shortcut. $α = 0.825 > 5%$ badly, and $1 - α = 0.175$ is nowhere near $1$ . Why this step? Using the shortcut here would double-count ions that no longer exist as HA.

Step 2. Write the exact law and clear the fraction: $K_{a} = \frac{C α ^{2}}{1 - α} \Rightarrow C α^{2} + K_{a} α - K_{a} = 0.$ Plug numbers: $0.001 α^{2} + 6.8 \times 1 0^{- 4} α - 6.8 \times 1 0^{- 4} = 0$ . Why: a quadratic in $α$ keeps the depletion of HA honest.

Step 3. Quadratic formula, taking the positive root (a fraction can't be negative): $α = \frac{- 6.8 \times 1 0 ^{- 4} + ( 6.8 \times 1 0 ^{- 4} ) ^{2} + 4 ( 0.001 ) ( 6.8 \times 1 0 ^{- 4} )}{2 ( 0.001 )} = 0.552$ Why the positive root? $α$ lives in $[0, 1]$ ; the negative root is physically meaningless.

Verify: put $α = 0.552$ back: $\frac{0.001 \times 0.55 2 ^{2}}{1 - 0.552} = \frac{3.05 \times 1 0 ^{- 4}}{0.448} = 6.80 \times 1 0^{- 4} = K_{a}$ ✓. The true $α$ is $\approx 55%$ , far below the fake $82.5%$ .

Cell D — the reverse problem

Cell E — limiting behaviour and the degenerate case

Worked example E · Push acetic acid toward infinite dilution

Forecast: the formula says $α = K_{a} / C \to \infty$ as $C \to 0$ . But $α$ can't exceed $1$ . What actually happens?

Step 1. Find the $C$ at which the shortcut would (falsely) reach $α = 1$ : set $K_{a} / C = 1 \Rightarrow C = K_{a} = 1.8 \times 1 0^{- 5}$ M. Why this step? This marks where the shortcut breaks down completely — below it the shortcut is nonsense.

Step 2. For $C ≪ K_{a}$ the acid is essentially fully ionised: $α \to 1$ , so $[H^{+}] \to C$ . Why: physically, with so few acid molecules and so much water, every molecule finds room to split.

Step 3 (the real degeneracy). Take $C = 1 0^{- 8}$ M. Even fully ionised that gives only $1 0^{- 8}$ M of $H^{+}$ from the acid — less than water's own $1 0^{- 7}$ M. So the pH is set by water, not the acid: pH $\approx 7$ (slightly below), NOT $8$ . Why: Ostwald's law ignores water's autoionisation; at extreme dilution you must add it back.

Verify: a full water-inclusive calc for $1 0^{- 8}$ M strong-ish acid gives $[H^{+}] \approx 1.05 \times 1 0^{- 7}$ , pH $\approx 6.98$ — acidic, never basic ✓. The naive $pH = \frac{1}{2} (p K_{a} - lo g C)$ would give a wrong pH $> 7$ .

Cell F — the strong-acid trap

Cell G — real-world word problem

Worked example G · Household vinegar is

$5.0%$ acetic acid by mass; density $\approx 1.0$ g/mL. Find its pH. ( $K_{a} = 1.8 \times 1 0^{- 5}$ , molar mass $60$ g/mol.) Forecast: vinegar tastes sharply sour — guess pH between $2$ and $3$ .

Step 1. Convert " $5%$ by mass" to molarity. In $1$ L $= 1000$ g solution, mass of acid $= 50$ g. Why this step? Ostwald needs $C$ in mol/L; the label gives mass. $C = \frac{50 g}{60 g/mol} \div 1 L = 0.833 M$

Step 2. $α = K_{a} / C = 1.8 \times 1 0^{- 5} /0.833 = 2.16 \times 1 0^{- 5} = 4.65 \times 1 0^{- 3}$ . Why: $0.47% ≪ 5%$ , shortcut valid.

Step 3. $[H^{+}] = K_{a} C = 1.8 \times 1 0^{- 5} \times 0.833 = 1.50 \times 1 0^{- 5} = 3.87 \times 1 0^{- 3}$ M. $pH = - lo g (3.87 \times 1 0^{- 3}) = 2.41$ Why this step? The question asked for pH, the everyday measure of sourness; $- lo g [H^{+}]$ is the bridge from ion concentration to that number. Verify: $[H^{+}] = C α = 0.833 \times 4.65 \times 1 0^{- 3} = 3.87 \times 1 0^{- 3}$ ✓, and the pH $2.41$ sits in the forecast band $2$ – $3$ ✓. Real vinegar measures pH $\approx 2.4$ – $2.9$ — spot on.

Cell H — exam twist: percent ionisation → pH

Recall

Recall What triggers the quadratic instead of the shortcut?

When $α = K_{a} / C$ exceeds $0.05$ (i.e. $K_{a} / C > 2.5 \times 1 0^{- 3}$ ). Then use $C α^{2} + K_{a} α - K_{a} = 0$ .

When you dilute a weak acid

100 \times

, by what factor does

α

rise?

100 = 10 \times

(since

α \propto 1/ C

Vinegar 5% by mass, density 1 g/mL — what is its molarity?

50 g /60 g mol^{- 1} = 0.833

Why does the Ostwald formula give

α > 1

for strong acids?

Its derivation assumed

α ≪ 1

so that

1 - α \approx 1

; strong acids break that premise, so the result is meaningless — use

[H^{+}] = C

C = 1 0^{- 8}

M, why isn't the pH above 7?

The acid contributes fewer

H^{+}

than water's own

1 0^{- 7}

M, so water sets the pH near (just below) 7 — the acid can never make it basic.

Connections:

Common Ion Effect: cell D/H reverse-solve if $[A^{-}]$ is spiked.
pH Calculation Methods: cells C and E are exactly where the shortcut is replaced by full methods.
Le Chatelier's Principle: cell B's dilution shift explained.
Buffer Capacity and Henderson-Hasselbalch Equation: what happens once $α$ nears $0.5$ .