Ionic product of water Kw = 10⁻¹⁴ at 25 °C: Question bank

Before we start, the notation so no symbol is used blind:

$[H^{+}]$ means "concentration of hydrogen ions, in moles per litre (M)".
$K_{c}$ is the ordinary equilibrium constant of a reaction: $K_{c} = \frac{[ products ]}{[ reactants ]}$ with each concentration raised to its coefficient. For water it is $K_{c} = \frac{[ H ^{+} ] [ OH ^{-} ]}{[ H _{2} O ] ^{2}}$ .
$K_{w} = [H^{+}] [OH^{-}]$ is the fixed product you get after folding the constant $[H_{2} O]^{2}$ into $K_{c}$ ; at 25 °C it equals $1 0^{- 14}$ .
pH $= - lo g [H^{+}]$ and pOH $= - lo g [OH^{-}]$ (the $lo g$ is base 10).
pKw $= - lo g K_{w}$ — the same "take minus-log" trick applied to $K_{w}$ . At 25 °C, $p K_{w} = - lo g (1 0^{- 14}) = 14$ . Since $- lo g (ab) = - lo g a - lo g b$ , this is exactly why $pH + pOH = p K_{w}$ .
ΔH° is the standard enthalpy change of a reaction — roughly the heat absorbed (if positive) or released (if negative) per mole. For water's autoionization $Δ H^{\circ} \approx + 56$ kJ/mol, the "+" meaning heat is absorbed (endothermic).

Two pictures anchor the whole page — the $K_{w}$ -vs-temperature curve and the acid-added seesaw. Refer back to them as you work the questions.

Figure — Ionic product of water Kw = 10⁻¹⁴ at 25 °C

True or false — justify

Neutral water always has pH = 7.

False — neutral means

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

, which only gives pH 7 at 25 °C; at 50 °C neutral water has pH ≈ 6.63 because

K_{w}

has grown (see the s01 curve).

Kw = 10⁻¹⁴ only applies to pure water.

False —

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

holds in any aqueous solution at 25 °C, acidic or basic; only pure water additionally has

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

A pH of 6.63 always means the solution is slightly acidic.

False — at 50 °C that is exactly neutral water; "acidic" is defined by

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

, not by pH < 7.

If you add acid, the amount of OH⁻ in solution literally drops to zero.

False — OH⁻ becomes very small but never zero; it settles at

K_{w} / [H^{+}]

, so a nonzero product is always maintained (the seesaw in s02).

Raising temperature makes water more acidic.

False — heating raises both

[H^{+}]

and

[OH^{-}]

equally in pure water, so it stays neutral; it does lower the neutral pH but not the acidity.

Kw is an equilibrium constant, so it has units of M².

False (by convention) — like other equilibrium constants it is a ratio of activities to a standard state, so it is treated as dimensionless; students often wrongly attach M².

Adding NaCl (a neutral salt) to water changes Kw.

False (at the activity level) —

K_{w}

(activity product) depends only on temperature; a spectator salt doesn't shift it, though at high salt levels the concentration product can look different because activity coefficients drift from 1.

In a 0.01 M HCl solution, all of the [H⁺] comes from HCl and none from water.

False — HCl supplies

0.01

M, but water still contributes ~

1 0^{- 12}

M of H⁺; it's negligible for the total, yet it is exactly this water contribution that keeps OH⁻ from being zero.

pH + pOH = 14 works at every temperature.

False — it equals

p K_{w}

, which is 14 only at 25 °C; at 50 °C,

p K_{w} \approx 13.26

, so pH + pOH ≈ 13.26.

Spot the error

"[H₂O] cancels out of Kc, so we just ignore it."

Wrong wording —

[H_{2} O]^{2}

doesn't cancel; it is absorbed into the constant:

K_{w} = K_{c} \times (55.5)^{2}

, where

K_{c} = \frac{[ H ^{+} ] [ OH ^{-} ]}{[ H _{2} O ] ^{2}}

is the ordinary equilibrium constant. It's folded in, not deleted.

"Since only 1 in 10⁹ molecules ionizes, [H⁺] = 10⁻⁹ M."

Wrong number — you must multiply the fraction by water's molarity:

55.5 M \times 1 0^{- 9} \approx 5.55 \times 1 0^{- 8}

M. The "1 in 10⁹" is itself only a rough figure; the measured

[H^{+}]

1 0^{- 7}

M, corresponding to an ionized fraction nearer

1.8 \times 1 0^{- 9}

"[OH⁻] = Kw / [H⁺] = 10⁻¹⁴ / 10⁻³ = 10⁻¹ M."

Arithmetic slip in the exponent —

- 14 - (- 3) = - 11

, so the answer is

1 0^{- 11}

M, not

1 0^{- 1}

M. Subtract exponents when dividing.

"Water is a strong electrolyte because it makes H⁺ and OH⁻."

Wrong — water ionizes to an extraordinarily tiny extent (~

1 0^{- 7}

M), so it is a very weak electrolyte; that smallness is the whole point of

K_{w}

"At equilibrium the reaction has stopped, so no more H₂O breaks apart."

Wrong — equilibrium is dynamic: ionization and recombination continue at equal rates, so concentrations are steady but the molecular activity never stops.

"To find [OH⁻] in acid, add [H⁺] and OH⁻."

Wrong operation — you divide

K_{w}

[H^{+}]

; the product is fixed, so raising one factor forces the other down, a division not an addition.

"Kw increases with temperature because ions move faster."

Wrong reason — the real cause is that autoionization is endothermic (

Δ H^{\circ} \approx + 56

kJ/mol, heat absorbed), so by Le Chatelier's principle heat pushes equilibrium toward more ions.

Why questions

Why can we treat [H₂O] as a constant in the derivation?

Because pure water is ~55.5 M and only ~

1 0^{- 7}

M ionizes; the drop from 55.5 to 55.4999999 M is utterly negligible, so

[H_{2} O]

behaves as a fixed number.

Why does adding a strong acid decrease [OH⁻]?

The extra H⁺ combines with existing OH⁻ to form water (the Common ion effect on the water equilibrium), lowering

[OH^{-}]

until the product

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

returns to

1 0^{- 14}

Why is the log (base 10) the right tool for defining pH and pKw?

Concentrations span many powers of ten (

1 0^{0}

1 0^{- 14}

); the base-10 log compresses that huge range into a small readable scale, and its rule

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

turns the product

K_{w}

into the sum

pH + pOH = p K_{w}

Why must [H⁺] equal [OH⁻] in pure water but not in an acid?

In pure water each ionization event makes one H⁺ and one OH⁻ together, so they're born in equal numbers; an added acid introduces H⁺ from another source, breaking that one-to-one balance.

Why does Kw let us find Kb from Ka?

For a conjugate acid–base pair, multiplying their equilibria reproduces the water autoionization, giving

K_{w} = K_{a} \times K_{b}

; see Weak acids Ka. Knowing

K_{w}

and one constant fixes the other.

Why is neutrality defined by [H⁺] = [OH⁻] rather than by pH = 7?

Because

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

is the temperature-independent physical statement of balance; pH = 7 is merely its numerical value at 25 °C and drifts as

K_{w}

changes.

Why does conductivity measurement tell us [H⁺] and [OH⁻] in pure water?

Ions carry electric current, so the (very small) measured conductivity of pure water reveals how many ions are present, yielding

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

Edge cases

Can [H⁺] ever be exactly zero in an aqueous solution?

No — even in strong base,

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

stays a small positive number; a nonzero product forbids either ion from vanishing.

What happens to pH of pure water as temperature rises toward 100 °C?

K_{w}

grows (to

5.13 \times 1 0^{- 13}

), so neutral

[H^{+}]

rises and neutral pH falls to about 6.14 — still neutral, just a lower number (right end of the s01 curve).

Is a solution with pH = pOH necessarily neutral?

Yes — pH = pOH means

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

, the exact definition of neutral, at any temperature (only the shared value differs).

For a very dilute acid (say [H⁺] added = 10⁻⁸ M), is the solution acidic?

Only slightly, and you must not ignore water's own

1 0^{- 7}

M contribution; the total

[H^{+}]

is barely above

1 0^{- 7}

, so pH is just under 7, not 8.

If you cool water below 25 °C, which way does neutral pH move?

Upward —

K_{w}

shrinks (e.g.

1.14 \times 1 0^{- 15}

at 0 °C), so neutral

[H^{+}]

is smaller and neutral pH rises above 7 (about 7.47 at 0 °C).

At the limit of infinite dilution of any solute, what does the solution approach?

Pure water's own equilibrium —

[H^{+}]

and

[OH^{-}]

both tend toward

1 0^{- 7}

M (at 25 °C), since the solute's contribution washes out and activity coefficients approach 1.

At very high salt concentration, why can the measured [H⁺][OH⁻] differ from 10⁻¹⁴?

Because

K_{w}

is really an activity product; when ionic strength is large the activity coefficients

γ

fall below 1, so effective (activity) values differ from raw concentrations even though the true

K_{w}

is unchanged.

Recall One-line summary of every trap

The activity product $[H^{+}] [OH^{-}] = K_{w}$ is fixed only by temperature; neutrality is $[H^{+}] = [OH^{-}]$ , not pH = 7; neither ion is ever zero; and in dilute solutions activity ≈ concentration. Everything else follows.

Connections

Parent: Kw (Hinglish)
Le Chatelier's principle — temperature dependence of Kw
pH and pOH scale — the pH + pOH = pKw relation
Strong acids and bases — complete ionization edge cases
Common ion effect — why added H⁺ suppresses OH⁻
Weak acids Ka — the $K_{w} = K_{a} \times K_{b}$ link