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

Before you start, pin down every symbol so no trap can hide behind notation.

Two pictures anchor the whole page. The first shows the single fact every trap twists — that $α$ climbs while $[H^{+}]$ falls as you dilute. The second shows exactly where the square-root shortcut breaks away from the honest quadratic curve.

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

True or false — justify

A weak acid diluted with water becomes more strongly ionized, so its solution becomes more acidic.

False. The fraction ionized (

α

) rises, but

[H^{+}] = K_{a} C

falls with

C

, so pH goes up — less acidic. More splitting per molecule is not more ions per litre (see the crossing curves in figure s01).

Since

α = K_{a} / C

, doubling

K_{a}

doubles

α

False.

α

scales with

K_{a}

, so doubling

K_{a}

multiplies

α

by only

2 \approx 1.41

, not by

2

. The square root softens every change.

K_{a}

and

α

are two names for the same idea of "acid strength."

False.

K_{a}

is a fixed constant (temperature only);

α

is a variable that slides with concentration. A single acid has one

K_{a}

but infinitely many

α

values.

As you keep diluting toward

C \to 0

α

predicted by

K_{a} / C

grows without bound.

True as written, and that is exactly the warning. The shortcut does blow past

1

, which is physically impossible — signalling the approximation has died and you must return to

K_{a} = C α^{2} / (1 - α)

, whose

α

caps at

1

Ostwald's law works fine for

0.1 M

HCl if you just plug in HCl's

K_{a}

False. Strong acids have

α \approx 1

, violating the

α ≪ 1

assumption behind the derivation; the formula returns absurd

α > 1

. For strong acids simply use

[H^{+}] = C

At the same concentration, a weaker acid has a smaller

α

True. Smaller

K_{a}

means less eagerness to split, and

α = K_{a} / C

shrinks with

K_{a}

. Weaker acid, fewer molecules broken apart.

Doubling the concentration halves

α

False.

α \propto 1/ C

, so doubling

C

multiplies

α

1/ 2 \approx 0.71

, not

0.5

. The square root again blunts the effect.

The relation

pH = \frac{1}{2} (p K_{a} - lo g C)

holds for every concentration of a weak acid.

False. It inherits the

α ≪ 1

approximation, so it fails once ionization exceeds roughly

5%

(dilute weak acids, or moderately strong ones like HF).

Because

K_{a}

is written from concentrations, it is a truly universal constant at fixed temperature, valid at any concentration.

False. The thermodynamic

K_{a}

is built from activities; the concentration version we use only matches it in the ideal-dilute limit, drifting as the activity coefficient

γ

falls below

1

in ionically crowded solutions.

Spot the error

"Adding water shifts

K_{a}

upward, which is why more of the acid ionizes."

Error:

K_{a}

never changes with dilution — only temperature moves it. Dilution raises

α

by changing

C

K_{a} = C α^{2} / (1 - α)

; the constant stays put while

α

compensates.

α = K_{a} / C

, therefore

[H^{+}] = K_{a} / C

Error:

[H^{+}] = C α

, not

α

. Substitute step by step:

[H^{+}] = C \cdot K_{a} / C = C^{2} \cdot K_{a} / C = K_{a} C

— pulling the

C

inside the root turns

C^{2} / C

into a single

C

, so

[H^{+}]

grows with

C

, the opposite

C

-dependence from

α

"For

0.001 M

HF (

K_{a} = 6.8 \times 1 0^{- 4}

α = 6.8 \times 1 0^{- 4} /0.001 = 0.68 = 0.825

, so 82.5% ionizes."

Error:

α = 0.825

is far above

5%

, so

(1 - α) \approx 1

is illegal here. Solve the honest quadratic

0.001 α^{2} + (6.8 \times 1 0^{- 4}) α - 6.8 \times 1 0^{- 4} = 0

; the quadratic formula gives one positive root

α \approx 0.54

and one negative root (rejected — a fraction cannot be below

0

), so

α \approx 0.54

, not

0.825

"Because

[H^{+}] = K_{a} C

, a very dilute weak acid can reach

pH > 7

Error: The formula ignores water's own

[H^{+}] = 1 0^{- 7} M

. Below about

1 0^{- 6} M

acid, water autoionization dominates and pH approaches

7

from below — it never crosses to basic.

"The

\frac{1}{2}

pH = \frac{1}{2} (p K_{a} - lo g C)

comes from the acid being diprotic."

Error: The

\frac{1}{2}

comes from

lo g K_{a} C = \frac{1}{2} lo g (K_{a} C)

— the square root becoming a factor of

\frac{1}{2}

when you take the log. It has nothing to do with the number of protons.

"Since dissociation makes ions, conductivity of a weak acid must rise steadily as you dilute."

Error: Conductance

\propto C α

. Even though

α

rises, the product

C α = K_{a} C

falls with dilution — fewer ions per litre carry less current (see Conductivity of Solutions).

Why questions

Why does

(1 - α) \approx 1

make the algebra collapse from a quadratic to a square root?

Keeping

(1 - α)

leaves

C α^{2} + K_{a} α - K_{a} = 0

(quadratic in

α

); setting the denominator to

1

gives

K_{a} = C α^{2}

, which solves in one step as

α = K_{a} / C

. Dropping the depletion term removes the linear term.

Why does the exponent on

α

end up as a square (hence the square root)?

Each dissociation makes two ions,

H^{+}

and

A^{-}

, both equal to

C α

; their product in

K_{a}

multiplies

α \times α = α^{2}

. Solving for

α

then undoes the square with a root.

Why must we take

- lo g

rather than just report

[H^{+}]

directly?

Because

[H^{+}]

spans many powers of ten across acids;

- lo g

compresses that range into a small readable number (pH) and — via

lo g (x y) = lo g x + lo g y

— turns the product

K_{a} C

into a tidy sum of

p K_{a}

and

lo g C

terms.

Why does dilution favour ionization, in Le Chatelier terms?

One HA molecule becomes two particles (

H^{+} + A^{-}

); adding water dilutes both sides, but the equilibrium relieves this by shifting toward the side with more particles, splitting more HA. See Le Chatelier's Principle.

Why is

K_{a}

allowed to stay constant while

α

[H^{+}]

and

[HA]

all change?

K_{a}

is a ratio arranged so that its value is fixed by temperature alone; the concentrations rearrange themselves to keep that ratio true. It's the constraint, not a participant.

Why can't Ostwald's law describe a strong acid even though the algebra "works"?

The derivation assumed

α ≪ 1

to drop

(1 - α)

. A strong acid has

α \approx 1

, so that step was never valid — the formula outputs nonsense like

α > 1

because you fed it a case it was built to exclude.

Why does the

5%

rule use

α

and not concentration directly?

The approximation error is exactly the size of the term thrown away,

(1 - α)

1

; when

α < 0.05

that gap is under

5%

. The threshold is about how much dissociated, which is

α

itself, not

C

Why does the quadratic

K_{a} = C α^{2} / (1 - α)

always have exactly one usable root?

It is a genuine quadratic with two algebraic solutions, but

α

is a fraction of molecules, so it must satisfy

0 \leq α \leq 1

; the second root falls outside that window (typically negative) and is discarded as unphysical.

Why does buffer action need

α

near

0.5

rather than the tiny

α

typical of a weak acid?

A buffer needs comparable amounts of HA and

A^{-}

; that balance sits around

α \approx 0.5

, reached by adding conjugate base, not by the acid alone. See Buffer Capacity and Henderson-Hasselbalch Equation.

Edge cases

What happens to

α

from the simple formula exactly at the point where

C = K_{a}

You get

α = K_{a} / K_{a} = 1

, i.e. "100% ionized" — but

α = 1

already breaks

α ≪ 1

, so this is the boundary where you must switch to the quadratic. The simple law is only trustworthy well before here.

What does adding a common ion (extra

A^{-}

) do to

α

, and does Ostwald's simple form still apply?

Extra

A^{-}

pushes the equilibrium left (fewer HA split), so

α

drops below

K_{a} / C

; the simple law no longer holds because it assumed

[H^{+}] = [A^{-}]

, which the added ion breaks. Use the Common Ion Effect setup instead.

For a diprotic acid, can you use

α = K_{a} / C

on both dissociation steps?

Approximately yes, step by step, because

K_{a 2} ≪ K_{a 1}

means the second step barely proceeds and hardly disturbs the first. Each step gets its own

K_{a}

; see Polyprotic Acids.

At extreme dilution the simple

α

exceeds

1

— what actually happens physically to the number of ions?

Physically

α

saturates at

1

(all molecules split), but the total ion count

C α = K_{a} C \to 0

: you approach complete ionization of almost nothing. Fraction maxes out; absolute amount vanishes.

What is

[H^{+}]

C \to 0

, and why doesn't pH become basic?

K_{a} C \to 0

suggests

pH \to \infty

, but real water supplies

\approx 1 0^{- 7} M H^{+}

, so pH levels off just under

7

. The acid formula must be combined with water's autoionization near this limit.

If temperature rises, which of

K_{a}

α

C

can change and which is fixed?

C

(moles per litre) is essentially fixed;

K_{a}

changes because it depends on temperature; and

α

then adjusts through

α = K_{a} / C

to track the new

K_{a}

. Temperature is the one lever that moves the "constant."

Two acids share the same

K_{a}

but you measure different

α

— what must differ?

Their concentrations

C

differ: same

K_{a}

but

α = K_{a} / C

means the more dilute one shows the larger

α

. Equal strength, unequal crowding.

Why does using concentrations instead of activities eventually mislead you?

In a crowded ionic solution each ion's effective concentration is scaled by an activity coefficient

γ < 1

, so the true

K_{a}

(built from activities) no longer equals

C α^{2} / (1 - α)

exactly. Ostwald's law is quantitatively reliable only in the ideal-dilute limit where

γ \to 1

Recall One-line summary of every trap

The recurring illusion is confusing the fraction $α$ with the amount $C α$ : dilution raises $α$ but lowers $[H^{+}] = K_{a} C$ . The recurring boundary is $α \geq 0.05$ , where the square-root shortcut dies and the quadratic $K_{a} = C α^{2} / (1 - α)$ must return — remembering that only its physical ( $0 \leq α \leq 1$ ) root counts, and that the whole law is an ideal-dilute (activity $\to$ concentration) approximation.