Strong vs weak acids - bases; degree of dissociation α: Question bank

This is a rapid-fire concept bank for the parent topic on α. No heavy arithmetic here — every item targets a misconception or a boundary case that trips people up. Reveal each answer only after you have committed to a reason of your own.

Before we start, one word we lean on constantly:

Keep this distinction burning in your mind the whole way down: $K_{a}$ is a property of the acid; $α$ is a property of the acid in this particular beaker (it depends on concentration).

True or false — justify

True or false: a strong acid always produces more

H^{+}

than a weak acid.

False. Only true at equal concentration. A very concentrated weak acid can out-produce an extremely dilute strong acid, because

[H^{+}]

depends on concentration too, not just strength.

True or false: for a strong acid

α = 1

regardless of concentration.

Essentially true in ordinary conditions — a strong acid ionises completely, so

α \approx 1

at any normal concentration. (In extremely concentrated solutions non-ideal effects blur this, but at the level of this chapter, treat it as exactly 1.)

True or false: diluting a weak acid increases its degree of dissociation

α

True. Le Chatelier: adding water reduces

[HA]

; the equilibrium shifts right to make more ions, so the fraction dissociated rises. This is Ostwald's dilution law — see Ostwald's dilution law.

True or false: diluting a weak acid increases its

[H^{+}]

False. Even though

α

rises, the total amount of acid falls faster, so

[H^{+}] = c α

actually decreases on dilution (pH rises toward 7).

True or false: a larger

K_{a}

always means a larger

α

False.

α = K_{a} / c

, so

α

depends on both

K_{a}

and

c

. A high-

K_{a}

acid at high concentration can have smaller

α

than a low-

K_{a}

acid that is very dilute.

True or false: adding water to a solution changes the value of

K_{a}

False.

K_{a}

is a temperature-dependent constant of the acid itself; dilution never changes it. Only

α

and the concentrations move.

True or false: raising temperature can change

α

even at fixed concentration.

True.

K_{a}

itself is temperature-dependent, and since

α = K_{a} / c

, changing

K_{a}

shifts

α

. Dissociation of most weak acids is slightly endothermic, so heating tends to raise

α

True or false: at infinite dilution, both a strong and a weak acid approach

α = 1

True. As

c \to 0

, the term

K_{a} / c

blows up and the equilibrium is driven fully to ions, so even a weak acid becomes essentially fully dissociated. Strength differences vanish in the ultra-dilute limit.

Spot the error

"0.1 M acetic acid gives

[H^{+}] = 0.1

M because each molecule has one

H

Error: stoichiometry is being used where equilibrium rules. Only the fraction

α

dissociates, so

[H^{+}] = c α \approx 0.1 \times 0.013 \approx 1.3 \times 1 0^{- 3}

M, not 0.1 M. See pH and pOH calculations.

α = K_{a} / c

, since

K_{a} \approx c α^{2}

Error: they solved incorrectly. From

K_{a} \approx c α^{2}

you divide then take the square root:

α = K_{a} / c

. Forgetting the square root gives a wildly wrong (and unit-inconsistent) answer.

α

has units of mol/L because it comes from concentrations."

Error:

α

is a ratio of amounts, so the units cancel — it is dimensionless. If your

α

carries units, a mistake has crept in.

"We used

1 - α \approx 1

, so this formula works for any acid."

Error: the approximation only holds when

α

is small (roughly

α < 5%

, i.e.

c / K_{a} > 100

). For a fairly strong or very dilute weak acid it fails, and you must use the full quadratic

c α^{2} + K_{a} α - K_{a} = 0

"A weak base has

α = [H^{+}] / c

Error: for a base you track hydroxide, so

α = [OH^{-}] / c

. Using

[H^{+}]

measures the tiny water-leftover ion, not the base's dissociation.

"Since

H_{2} SO_{4}

is strong, it releases exactly

2 c

moles of

H^{+}

fully."

Error: only the first proton dissociates completely; the second is a weaker step governed by its own

K_{a}

. Polyprotic acids do not release all protons equally — see Polyprotic acids.

"Adding

NaCl

(a salt of the same anion... say acetate) doesn't affect acetic acid's

α

Error: adding acetate (

A^{-}

) pushes the equilibrium left and lowers

α

— the Common ion effect. Any shared ion suppresses dissociation.

Why questions

Why does

α

rise on dilution but

[H^{+}]

fall?

Dilution reduces every concentration, shifting equilibrium right so a larger fraction splits (α up). But the total pool of acid shrank, so the absolute count of ions per litre still drops (H⁺ down). Fraction and amount move in opposite directions.

Why can't we just compare

K_{a}

values to rank the pH of two real solutions?

Because pH depends on

[H^{+}] = c α

, and

α

folds in concentration. A weak-but-concentrated acid can have lower pH than a stronger-but-dilute one. You need both

K_{a}

and

c

Why is the conjugate base of a strong acid a very weak base?

HA

gives up its proton eagerly, the leftover

A^{-}

has almost no tendency to grab it back — a stable ion makes a feeble base. Strength of acid and weakness of its conjugate are two sides of the same equilibrium (relevant to Hydrolysis of salts).

Why does the "rule of 100" (

c / K_{a} > 100

) let us drop the

1 - α

term?

c / K_{a} > 100

then

α < 0.1

, so

1 - α > 0.9

and treating it as 1 introduces under ~5% error. Below that ratio the neglected

α

is too big to ignore and the quadratic is required.

Why does concentrated vinegar not burn skin the way concentrated HCl does?

Vinegar is a weak acid: only ~1% of its molecules are dissociated, so free

[H^{+}]

is low. HCl is fully dissociated, delivering a hundred-fold-plus higher

[H^{+}]

at the same molarity.

Why is

K_{a}

preferred over

α

for comparing intrinsic acid strength?

K_{a}

is fixed for the acid at a given temperature, so it is a clean fingerprint of strength.

α

slides with concentration, so two beakers of the same acid can show different

α

— making it useless as a strength label. See Acid-base equilibria and Ka, Kb.

Edge cases

Edge case: what is

α

for an extremely dilute solution of a weak acid (say

1 0^{- 7}

M)?

The simple

α = K_{a} / c

breaks and can even give

α > 1

(nonsense). At such dilution the

H^{+}

from water autoionisation (

1 0^{- 7}

M) is comparable to that from the acid and must be included; naive formulas fail.

Edge case: can the exact formula ever give

α > 1

No — the exact root

α = \frac{- K _{a} + K _{a}^{2} + 4 c K _{a}}{2 c}

is bounded above by 1 for all real positive

c, K_{a}

. If your working ever spits out

α > 1

, you used the approximate formula outside its valid range.

Edge case: what happens to

α

c \to \infty

(very concentrated weak acid)?

α \to 0

. From

α \approx K_{a} / c

, huge

c

crushes the fraction toward zero — the crowded ions recombine, so proportionally almost nothing stays dissociated (though absolute

[H^{+}]

is still large).

Edge case: for a strong acid at very low concentration (

\sim 1 0^{- 8}

M), is pH

= 8

No — that would imply the acid made the solution basic, which is impossible. Water's own

1 0^{- 7}

M of

H^{+}

dominates, so pH stays just below 7. You must add the acid's contribution to water's, not ignore water.

Edge case: is

α

meaningful for a strong acid at all?

Barely — it is essentially pinned at 1, so it carries no useful information.

α

is a diagnostic made for weak acids/bases, where the interesting partial splitting happens.

Edge case: two acids share the same

α

in their beakers. Same strength?

Not necessarily. Equal

α

can arise from a strong acid at high

c

and a weak acid at low

c

, since

α = K_{a} / c

trades off the two. Equal

α

says nothing about equal

K_{a}

Recall Quick self-test

One-line each — say the reason aloud before revealing. On dilution: α does what, [H⁺] does what? ::: α rises, [H⁺] falls. Does K_a change on dilution? ::: No — it is a constant of the acid at fixed temperature. The relation you keep abusing? ::: $α = K_{a} / c$ , valid only when $α < 5%$ ( $c / K_{a} > 100$ ). When must you include water's $H^{+}$ ? ::: In very dilute solutions where acid $[H^{+}]$ nears $1 0^{- 7}$ M.