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

Before we start, three plain words we will use over and over:

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Step 1 — Water is a restless crowd

WHAT. Picture a bucket of pure water. Nothing is added. Yet the molecules are not sitting still — they jostle, and now and then one snatches a positive proton from a neighbour.

WHY. Water is polar: the hydrogen end is slightly positive (${\delta }^{+}$), the oxygen end slightly negative (${\delta }^{-}$). Opposites attract, so a ${\delta }^{+}$ hydrogen is tugged toward a neighbour's ${\delta }^{-}$ oxygen. Every so often the tug is strong enough to rip the proton clean off. This is called autoionization (the water ionizing itself).

PICTURE. In the figure, two neutral water molecules collide. One donates a proton and becomes ${\text{OH}}^{-}$ (short one hydrogen, now negative). The other accepts it and becomes ${\text{H}}_3{\text{O}}^{+}$ (an extra hydrogen, now positive). We often shorten "${\text{H}}_3{\text{O}}^{+}$" to just "${\text{H}}^{+}$" for tidiness — same idea, a spare positive proton floating in the water.

$\underset{\text{donor}}{\underbrace{{\text{H}}_2\text{O}}}+\underset{\text{acceptor}}{\underbrace{{\text{H}}_2\text{O}}}\rightleftharpoons \underset{\text{gained a proton}}{\underbrace{{\text{H}}_3{\text{O}}^{+}}}+\underset{\text{lost a proton}}{\underbrace{{\text{OH}}^{-}}}$

The double arrow $\rightleftharpoons$ is the key: the reaction runs both ways at once. Pairs split apart and recombine constantly. See Le Chatelier's principle for what "both ways at once" really means.

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Step 2 — "Both ways at once" freezes into a ratio

WHAT. When the forward splitting and the backward recombining happen at the same rate, the crowd stops changing overall — this steady state is called equilibrium. At equilibrium there is a fixed relationship between the crowds of each species.

WHY. For any reversible reaction, chemists found experimentally that at equilibrium one particular combination of concentrations is a constant — call it $K_c$ (the equilibrium constant). We build it as products on top, reactants on the bottom, each raised to the power of how many copies appear in the equation.

PICTURE. The figure shows a balance beam. On the right pan (top of the fraction) sit the products ${\text{H}}^{+}$ and ${\text{OH}}^{-}$. On the left pan (bottom) sit two water molecules — two, because water appears twice on the reactant side, so it is squared.

$K_c=\frac{\stackrel{\text{one }{\text{H}}^{+}}{\overbrace{[{\text{H}}^{+}]}}\stackrel{\text{one }{\text{OH}}^{-}}{\overbrace{[{\text{OH}}^{-}]}}}{\underset{\text{two waters}\Rightarrow \text{power }2}{\underbrace{[{\text{H}}_2\text{O}{]}^2}}}$

Why the exponent-equals-coefficient rule? Because two water molecules must both be present for one split to occur, so the chance scales with concentration twice over — hence the square. This same machinery reappears in Weak acids Ka and Solubility product Ksp.

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Step 3 — The water on the bottom barely moves

WHAT. We measure how crowded liquid water is with itself. One litre of water weighs $1000\text{g}$; one mole weighs $18\text{g}$. So:

$[{\text{H}}_2\text{O}]=\frac{1000\text{g/L}}{18\text{g/mol}}=55.5\text{M}$

WHY. This number is gigantic compared to the ions. Only about $10^{-7}\text{M}$ of water ever splits. Watch the subtraction in full digits:

$55.5000000\text{M}-0.0000001\text{M}=55.4999999\text{M}$

The fractional change is $\frac{10^{-7}}{55.5}\approx 1.8\times 10^{-9}$ — under two parts per billion. A quantity that shifts this little is, for all practical purposes, a constant (and its activity, from Step 2, stays $\approx 1$). Constants can be moved around freely.

PICTURE. The figure plots both amounts on a true logarithmic scale — honest proportion, no broken axis. The gap between $55.5$ and $10^{-7}$ spans nearly nine decades; a faint marker shows how the tiny $10^{-7}$ shift barely dents the $55.5$ bar.

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Step 4 — Fold the constant water into a brand-new constant

WHAT. A constant ($K_c$) times another constant ($[{\text{H}}_2\text{O}{]}^2$) is still just a constant. So multiply both sides by $[{\text{H}}_2\text{O}{]}^2$ to clear the denominator, and rename the whole left side.

WHY. We would rather carry around one clean number than a product of two. Merging them gives us a purpose-built constant for water: $K_w$, the ionic product of water.

PICTURE. The figure shows the fraction from Step 2, then the $55.5^2$ block sliding up from the bottom and fusing with $K_c$ into a single labelled box $K_w$.

$\underset{\text{constant}\times \text{constant}=\text{new constant }{K}_w}{\underbrace{K_c\times [{\text{H}}_2\text{O}{]}^2}}=[{\text{H}}^{+}][{\text{OH}}^{-}]$

$\boxed{K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]}$

Notice what survived: only the ion crowds. The water squared is hidden inside $K_w$ forever. That is why $K_w$ has no water term.

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Step 5 — Read the actual number off an experiment

WHAT. We can't derive the size of $K_w$ from pure logic — we must measure it. Conductivity meters (how well water carries electric current tells you how many ions are present) show that at $25{}^{\circ }\text{C}$ pure water has:

$[{\text{H}}^{+}]=1.0\times 10^{-7}\text{M},[{\text{OH}}^{-}]=1.0\times 10^{-7}\text{M}$

WHY equal? Look back at Step 1: every split makes exactly one ${\text{H}}^{+}$ and exactly one ${\text{OH}}^{-}$, born as a pair. In pure water with nothing else added, their crowds must match.

PICTURE. A number line of powers of ten. Both ions sit on the same tick at $10^{-7}$. Multiply the two ticks (add the exponents: $-7+-7=-14$) and land on $10^{-14}$.

$K_w=(1.0\times 10^{-7})(1.0\times 10^{-7})=1.0\times 10^{-14}(\text{at }25{}^{\circ }\text{C})$

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Step 6 — The seesaw: why the product never breaks

WHAT. $K_w=10^{-14}$ holds in every aqueous solution, not just pure water. Add acid and $[{\text{H}}^{+}]$ shoots up; instantly $[{\text{OH}}^{-}]$ must shrink so the product stays $10^{-14}$.

WHY. $K_w$ is a constant. A constant product of two numbers means: raise one, the other falls in exact inverse proportion. Extra ${\text{H}}^{+}$ collides with ${\text{OH}}^{-}$ and remakes water, draining the ${\text{OH}}^{-}$ crowd until balance returns.

PICTURE. A literal seesaw plotted on log axes. As $[{\text{H}}^{+}]$ walks right (more acidic), $[{\text{OH}}^{-}]$ walks left (less), and the sum of the exponents — the height of the plank's midpoint — never budges from $-14$.

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Step 7 — The degenerate & edge cases (nothing left unshown)

WHAT. Three boundary situations people trip on:

1. Perfectly neutral water: $[{\text{H}}^{+}]=[{\text{OH}}^{-}]=x$, so $K_w=x^2$. Take the square root: $x=\sqrt{K_w}$.
2. Heat it up: splitting bonds absorbs heat (endothermic, $\Delta H^{\circ }\approx +56\text{kJ/mol}$). By Le Chatelier's principle, adding heat pushes the reaction forward → more ions → $K_w$ grows. At $50{}^{\circ }\text{C}$, $K_w=5.47\times 10^{-14}$.
3. "Neutral" ≠ "pH 7": neutral only means $[{\text{H}}^{+}]=[{\text{OH}}^{-}]$. At $50{}^{\circ }\text{C}$ that equal value is larger, so the neutral pH shifts below 7.

WHY. These show the formula never breaks — it simply rescales with temperature. The $10^{-14}$ is a $25{}^{\circ }\text{C}$ snapshot, not a law of nature.

PICTURE. A curve of $K_w$ climbing with temperature; a dot marks the neutral point at each $T$, drifting to lower pH as we heat.

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The one-picture summary

The whole journey on one canvas: collision → double arrow → fraction → drop the constant water → measure → seesaw. Every symbol traced from a physical picture to the boxed result $K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]=10^{-14}$.

Recall Feynman retelling — say it back in plain words

Water isn't lazy — its molecules keep bumping, and once in a rare while one steals a proton from another. That makes a positive piece (${\text{H}}^{+}$) and a negative piece (${\text{OH}}^{-}$), always born as a matched pair, and they keep recombining too. When making-and-breaking happen at equal speed, the crowds settle: products-over-reactants is a fixed number, $K_c$. Every bracket is secretly an activity — a concentration divided by a $1\text{M}$ reference — so the number is unitless, and a pure liquid like water has activity $\approx 1$. That's why the water on the bottom, overwhelmingly plentiful ($55.5$ moles per litre versus a whisper of ions), never really changes and can be folded into a new constant, $K_w$, leaving just the two ion crowds multiplied together. Nobody can predict the size from thought alone, so we measure it: at room temperature each crowd is $10^{-7}$, and $10^{-7}\times 10^{-7}=10^{-14}$. That product is a locked seesaw — pour in acid and ${\text{H}}^{+}$ rises, so ${\text{OH}}^{-}$ must sink by exactly the same amount to keep the product at $10^{-14}$. Heat the water and you feed the bond-breaking (it's endothermic), so more pairs split and $K_w$ climbs — meaning "neutral" doesn't always land on pH 7. One number, born from one collision, ruling the whole scale.

Recall Quick self-test

Where does the $[{\text{H}}_2\text{O}{]}^2$ term go in $K_w$? ::: It is absorbed into $K_w$ because water is a pure liquid with activity $\approx 1$ (concentration $55.5$ M essentially constant); $K_w=K_c\times 55.5^2$. Why is $K_w$ written without units even though $[{\text{H}}^{+}][{\text{OH}}^{-}]$ looks like ${\text{M}}^2$? ::: Because each bracket is really an activity — a concentration divided by a $1\text{M}$ reference — so the units cancel and $K_w$ is a pure number. Why are $[{\text{H}}^{+}]$ and $[{\text{OH}}^{-}]$ equal in pure water? ::: Each autoionization event makes exactly one of each, born as a pair, with nothing else added. If $[{\text{H}}^{+}]=10^{-2}$ M at 25 °C, what is $[{\text{OH}}^{-}]$? ::: $10^{-14}/10^{-2}=10^{-12}$ M.

Connections

• Parent topic — the full note
• Le Chatelier's principle — why heat raises $K_w$
• pH and pOH scale — the scale this constant defines
• Strong acids and bases — the seesaw in action
• Weak acids Ka — same $K_c$ machinery; $K_w=K_a\times K_b$
• Solubility product Ksp — identical "constant product" logic
• Common ion effect — suppressing ionization