pH, pOH, pKa, pKb scales: Question bank — Chemistry deep dive

This is a concept-trap drill for the p-scales topic. No heavy arithmetic here — every item targets a way of thinking that trips people up: signs, log behaviour, the difference between a property and a state, and what happens at the edges (very dilute, very concentrated, non-25°C). Cover the Question ::: Answer on the right, say your reasoning out loud, then reveal.

If any word below feels unfamiliar, build it back up from 2.6.1-Arrhenius-acids-and-bases, 2.6.5-Ionic-product-of-water-Kw, and the Hinglish note before drilling.

Definition Every symbol used on this page, in plain words

You should not need to leave this page to read the drill. Here is the whole cast:

$[H^{+}]$ — the concentration of hydrogen ions (protons) in moles per litre (M). More $H^{+}$ = more acidic.
$[OH^{-}]$ — the concentration of hydroxide ions in M. More $OH^{-}$ = more basic.
$C$ — the shorthand for the nominal (analytical) concentration of an acid or base in M, i.e. how much you dissolved before any dissociation.
The "p" prefix — writing $p$ in front of any quantity $X$ always means $p X = - lo g_{10} (X)$ : a negative base-10 logarithm, never natural log, never positive. So pH, pOH, pKa, pKb, p $K_{w}$ are all the same operation applied to different quantities.
$pH = - lo g_{10} [H^{+}]$ — a small number that reports how acidic the solution is; the minus sign makes low pH mean high $H^{+}$ .
$pOH = - lo g_{10} [OH^{-}]$ — the same idea for hydroxide.
$K_{w} = [H^{+}] [OH^{-}]$ — the ion product of water; at 25°C it equals $1 0^{- 14}$ (see 2.6.5-Ionic-product-of-water-Kw).
$K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ HA ]}$ — the acid dissociation constant for a weak acid $HA ⇌ H^{+} + A^{-}$ ; big $K_{a}$ = strong acid.
$K_{b} = \frac{[ BH ^{+} ] [ OH ^{-} ]}{[ B ]}$ — the base dissociation constant; big $K_{b}$ = strong base.
$pK_{a} = - lo g_{10} K_{a}$ and $pK_{b} = - lo g_{10} K_{b}$ — the log-compressed versions; low pKa = strong acid.
Activity ( $a$ ) — the effective concentration an ion behaves with once it feels its crowded neighbours; it equals an activity coefficient $γ$ times the true concentration, $a = γ [ion]$ . In dilute solutions $γ \approx 1$ so $a \approx$ concentration, and the simple pH formulas hold.
A conjugate pair is an acid $HA$ and the base $A^{-}$ it becomes after losing its proton.

True or false — justify

A solution with pH = 3 has ten times more hydrogen ions than one with pH = 4

True — each pH unit is one power of ten in

[H^{+}]

, and lower pH means more

H^{+}

, so pH 3 is

10 \times

more concentrated than pH 4.

pH can never be negative

False — pH is

- lo g_{10} [H^{+}]

, so if

[H^{+}] > 1

M (e.g. 10 M strong acid) the log is positive and pH is negative; the "0 to 14" range is a convenience, not a law.

pH can never exceed 14

False — start from

K_{w} = [H^{+}] [OH^{-}] = 1 0^{- 14}

; a strongly basic solution with, say,

[OH^{-}] = 1

M forces

[H^{+}] = K_{w} / [OH^{-}] = 1 0^{- 14}

M, and pushing

[OH^{-}]

past 1 M drops

[H^{+}]

below

1 0^{- 14}

, so

pH = - lo g_{10} [H^{+}]

climbs above 14.

At 25°C, pure water always has pH = 7

True at 25°C only —

K_{w} = 1 0^{- 14}

gives

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

, so pH 7; but

K_{w}

rises with temperature, so hot pure water has pH < 7 while still being neutral.

Hot pure water with pH = 6 is acidic

False — it is still neutral because

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

; neutrality means equal ion concentrations, not pH = 7, and

K_{w}

shifts the neutral point with temperature.

A strong acid always has a low pH

False — strength means fully dissociated, but a

1 0^{- 8}

M HCl solution is so dilute its pH is near 7; concentration, not strength alone, sets the pH.

Lowering pKa means the acid is stronger

True —

pK_{a} = - lo g_{10} K_{a}

, so a smaller pKa corresponds to a larger

K_{a}

, i.e. more dissociation and a stronger acid.

If pKa = 4.5 then the solution's pH is 4.5

False — pKa is a fixed property of the acid; pH is a state that depends on concentration. They coincide only at the buffer midpoint where

[HA] = [A^{-}]

pH + pOH = 14 holds at every temperature

False — it equals

p K_{w}

, which is 14 only at 25°C; at other temperatures

K_{w}

changes and the sum shifts.

For a conjugate acid–base pair,

pK_{a} + pK_{b} = 14

at 25°C

True — multiplying

K_{a} K_{b}

collapses to

K_{w} = 1 0^{- 14}

, so taking logs gives

pK_{a} + pK_{b} = p K_{w} = 14

Adding water to a strong acid can push its pH above 7

False — dilution moves pH toward 7 but never past it; as you dilute, water's own

H^{+}

dominates and pH approaches (but does not cross) 7 from below.

Doubling the concentration of a strong acid drops pH by exactly 1

False — a

10 \times

increase drops pH by 1; doubling drops it by

lo g_{10} 2 \approx 0.3

A diprotic acid has a single pKa

False — it releases two protons in two steps, so it has two constants,

pK_{a 1}

and

pK_{a 2}

, with

pK_{a 1} < pK_{a 2}

because pulling a second proton off an already-negative ion is harder.

Spot the error

"pH =

lo g_{10} [H^{+}]

, so more acid gives a bigger pH."

The negative sign is missing. Numeric check: for

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

M (concentrated) the wrong formula gives pH

= - 1

, and for

1 0^{- 6}

M (dilute) it gives pH

= - 6

— so the concentrated one gets the larger (less negative) number, backwards for acidity. The correct

pH = - lo g_{10} [H^{+}]

gives 1 and 6, ranking them the right way.

"pOH =

- ln [OH^{-}]

Wrong logarithm. The correct definition is

pOH = - lo g_{10} [OH^{-}]

; p-scales always use base-10, and using

ln

inflates every value by a factor of

ln 10 \approx 2.303

lo g (2.0 \times 1 0^{- 3}) = lo g 2.0 \times lo g 1 0^{- 3}

The product rule adds logs, it doesn't multiply them:

lo g (ab) = lo g a + lo g b

, giving

0.30 + (- 3) = - 2.70

"A stronger acid has a stronger conjugate base."

Reversed. Since

pK_{a} + pK_{b} = 14

, a low pKa (strong acid) forces a high pKb — a weaker conjugate base.

"At the buffer midpoint the solution is 100% dissociated."

pH = pK_{a}

the acid is 50% dissociated:

[HA] = [A^{-}]

, an equal mix, which is why buffering is maximal there — the flat middle of the titration curve in the figure above.

"pKa changes when you dilute the acid."

pKa is a temperature-fixed constant of the substance; dilution changes pH and the degree of dissociation, but never

K_{a}

or pKa.

"If

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

the solution must have pH 7."

Equal ion concentrations means neutral, but the actual pH value equals

\frac{1}{2} p K_{w}

, which is 7 only at 25°C.

"At very high concentration, pH is just

- lo g_{10}

of the molarity."

Only true when ions are dilute enough to ignore each other; in concentrated solutions ions crowd and screen one another, so pH tracks the activity

a = γ [H^{+}]

(effective concentration), not the raw molarity — the simple formula breaks down.

Why questions

Why is the p-scale logarithmic instead of linear?

Because

[H^{+}]

spans about 14 orders of magnitude (

1 0^{0}

1 0^{- 14}

M); logs compress that huge range into small numbers and turn multiplicative changes into simple additions — exactly the squeeze pictured in the first figure.

Why does the p-operator carry a minus sign?

So that higher pX corresponds to lower X. Numeric demonstration: concentrated acid

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

gives pH 1, dilute

[H^{+}] = 1 0^{- 6}

gives pH 6 — the concentrated one gets the smaller number. Drop the minus and you get

- 1

- 6

, which would rank the dilute solution as more acidic. The minus is what keeps the ordering intuitive.

Why do

K_{a}

and

K_{b}

of a conjugate pair multiply to

K_{w}

Adding their two equilibrium equations cancels

HA

and

A^{-}

, leaving the water autoionization

H^{+} + OH^{-} ⇌

nothing, so the constants multiply to

[H^{+}] [OH^{-}] = K_{w}

(see 2.6.5-Ionic-product-of-water-Kw).

Why is

pH = pK_{a}

the point of maximum buffering?

There

[HA] = [A^{-}]

, so the mixture can absorb added acid or base almost symmetrically before the ratio shifts much — this is the flattest part of the titration curve shown above (see 2.6.8-Henderson-Hasselbalch-equation).

Why can't strong-acid pH simply be

- lo g_{10} C

(where

C

is the concentration you dissolved) at very low concentration?

Because water's own

1 0^{- 7}

M of

H^{+}

is no longer negligible; you must add the acid's and water's contributions, which is why

1 0^{- 8}

M HCl gives pH slightly under 7, not 8 (see 2.6.11-pH-calculationsfor-strong-acids-bases).

Why is a 2.6-unit pKa difference worth about a

400 \times

strength ratio?

Each pKa unit is one power of ten in

K_{a}

, so

1 0^{2.6} \approx 398

; the exponent, not the difference itself, sets the multiplicative gap.

Why do we define pKa at all instead of just quoting

K_{a}

K_{a}

values sprawl from

1 0^{2}

down to

1 0^{- 50}

; taking

- lo g_{10}

turns them into compact, comparable numbers where a smaller pKa cleanly signals a stronger acid.

Edge cases

What is the pH of

1 0^{- 8}

M HCl?

Close to 7 (about 6.96), not 8 — the acid is too dilute to overpower water's own ionization, so the solution stays just barely acidic (see 2.6.11-pH-calculationsfor-strong-acids-bases).

Can pH ever equal exactly the concentration's exponent for a weak acid?

No in general — a weak acid only partially dissociates, so

[H^{+}]

is much less than the nominal concentration

C

; you need the

K_{a}

equilibrium, not

- lo g_{10} C

(see 2.6.12-pH-calculations-for-weak-acids-bases).

How many pH-defining constants does a polyprotic acid like

H_{3} PO_{4}

have?

Three —

pK_{a 1}, pK_{a 2}, pK_{a 3}

, one per proton lost, each larger than the last; a titration curve of such an acid shows a separate flat buffer region around each pKa.

Is pOH meaningful for a solution containing no added base?

Yes — even pure acid has some

OH^{-}

from water, so

[OH^{-}] = K_{w} / [H^{+}]

is always positive and pOH is always defined.

What happens to the neutral pH as temperature rises?

K_{w}

increases, so the neutral pH drops below 7 (e.g. ~6.14 at 100°C); the water is still neutral because

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

, only the numeric label moves.

Does

pK_{a} + pK_{b} = 14

apply to a strong acid and its conjugate?

Formally yes with

p K_{w} = 14

, but a strong acid's

K_{a}

is so large its "pKa" is negative and its conjugate is a spectator with almost no base strength — the relation still balances the books.

Is a solution of pH exactly 7 always neutral?

Only at 25°C — pH 7 means neutral only when

\frac{1}{2} p K_{w} = 7

; at other temperatures a pH-7 solution can be slightly acidic or basic depending on

K_{w}

Why does the simple

pH = - lo g_{10} [H^{+}]

formula fail for very concentrated acids?

Because at high concentration ions interact and screen each other, so the effective concentration (the activity

a = γ [H^{+}]

) differs from the counted molarity; rigorously pH is

- lo g_{10} a

, and only in dilute solutions does

γ \approx 1

so activity ≈ concentration.

What limits how negative or how large pH can get in practice?

The same activity effects — beyond a few molar, ion interactions make the simple

- lo g_{10} [H^{+}]

formula unreliable, so extreme pH values must be defined through the activity

a = γ [H^{+}]

rather than raw concentration.

Recall One-line self-test

Cover everything and answer: does pH measure strength or state, and what single relation converts $K_{a}$ to $K_{b}$ ? Answer ::: pH measures state (depends on concentration/equilibrium), while strength is the fixed pKa; the converter is $K_{a} K_{b} = K_{w}$ , i.e. $pK_{a} + pK_{b} = p K_{w} = 14$ at 25°C.