Visual walkthrough — pH, pOH, pKa, pKb scales

This page builds the single most important relationship on the p-scale page — the fact that pH + pOH always adds to 14 in water at 25 °C — starting from nothing but a jar of water. We earn every symbol before using it, and every step gets a picture.

By the end you will also see why the twin result ${\text{pK}}_a+{\text{pK}}_b=14$ is the exact same story wearing a different hat.

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Step 1 — What "concentration" means, and why the numbers are horrible

WHAT. Put pure water in a beaker. A tiny fraction of the water molecules break apart. We want to count how many broken pieces float around. The count we use is concentration: how many moles of a particle sit in one litre. We write square brackets around a particle to mean "concentration of that particle", so $[{\text{H}}^{+}]$ = "how much hydrogen ion per litre".

WHY. Acidity is literally "how many hydrogen ions are floating around". More ${\text{H}}^{+}$ → more sour → more acidic. So the whole subject is really about one number: $[{\text{H}}^{+}]$.

PICTURE. Look at the number line below. The concentrations we meet run from $1$ mole/litre (strong acid, left) all the way down to $0.00000000000001=10^{-14}$ mole/litre (strong base, right). That is fourteen zeros. On a normal ruler these numbers are unusable — they all pile up against zero.

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Step 2 — Where the two ions come from (the water see-saw)

WHAT. A water molecule, ${\text{H}}_2\text{O}$, occasionally splits: ${\text{H}}_2\text{O}\rightleftharpoons \underset{\text{acid piece}}{\underbrace{{\text{H}}^{+}}}+\underset{\text{base piece}}{\underbrace{{\text{OH}}^{-}}}$ The double arrow $\rightleftharpoons$ means the split happens both ways at once — some molecules break apart while others rejoin. This idea comes from Arrhenius acids and bases.

WHY. Every time one ${\text{H}}^{+}$ is created, exactly one ${\text{OH}}^{-}$ is created with it — they are born as a pair. So $[{\text{H}}^{+}]$ and $[{\text{OH}}^{-}]$ are not independent; they are linked, like the two ends of a see-saw. Push one down, the other must come up. This linkage is the seed of the "+ = 14" rule.

PICTURE. The see-saw below: ${\text{H}}^{+}$ on the left seat, ${\text{OH}}^{-}$ on the right. Their product stays fixed (Step 3), so raising one lowers the other.

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Step 3 — The unbreakable product $K_w$

WHAT. Measurement shows that in any watery solution at 25 °C, if you multiply the two concentrations together you always get the same tiny number: $K_w=\underset{\text{acid piece}}{\underbrace{[{\text{H}}^{+}]}}\times \underset{\text{base piece}}{\underbrace{[{\text{OH}}^{-}]}}=10^{-14}$ $K_w$ is called the ionic product of water — see Kw.

WHY. This is the mathematical rule behind the see-saw. The product is nailed down at $10^{-14}$. So the two concentrations can trade values, but their product cannot escape. In pure water they must be equal, and the only pair of equal numbers whose product is $10^{-14}$ is $10^{-7}$ and $10^{-7}$.

$[{\text{H}}^{+}]=[{\text{OH}}^{-}]=10^{-7}\text{ M}(\text{pure, neutral water})$

PICTURE. A curve of all allowed pairs. Every point on the orange curve has $[{\text{H}}^{+}]\times [{\text{OH}}^{-}]=10^{-14}$. Slide left and $[{\text{OH}}^{-}]$ shoots up; slide right and $[{\text{H}}^{+}]$ shoots up. The neutral point sits exactly in the middle where they cross.

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Step 4 — The tool: what a logarithm actually does

WHAT. We now introduce ${\log }_{10}$. The logarithm asks one question: ${\log }_{10}(X)=\text{"10 raised to WHAT power gives }X\text{?"}$ For example ${\log }_{10}(1000)=3$ because $10^3=1000$; and ${\log }_{10}(10^{-7})=-7$ because $10^{-7}$ is already ten-to-the-minus-seven.

WHY THIS TOOL AND NOT ANOTHER. We need something that turns a factor of ten into a step of one, so the horrible 14-zero range in Step 1 becomes the friendly range 0–14. The logarithm does exactly that: every time $X$ shrinks by $10\times$, ${\log }_{10}(X)$ drops by exactly 1. And crucially it obeys the product rule ${\log }_{10}(a\times b)={\log }_{10}(a)+{\log }_{10}(b),$ which turns multiplication into addition — precisely what we need to break the $K_w$ product into a sum.

We use base 10 (not natural log $\ln$) purely because our concentrations are written as powers of ten. Using $\ln$ would scale every answer by $2.303$ and break the clean "0 to 14" scale.

PICTURE. Left: the exponential scale (values pile against zero). Right: after ${\log }_{10}$, the same values spread out evenly. That is the "compression" doing its job.

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Step 5 — The "p" flip: defining pH and pOH

WHAT. Applying ${\log }_{10}$ to $10^{-7}$ gives $-7$ — a negative number, awkward for everyday talk. So we bolt a minus sign onto the front and call the operation p: $\text{pX}=-{\log }_{10}(X)$ Then $\text{pH}=-{\log }_{10}[{\text{H}}^{+}],\text{pOH}=-{\log }_{10}[{\text{OH}}^{-}].$

WHY THE MINUS. The minus flips the sign so neutral water reads $\text{pH}=-{\log }_{10}(10^{-7})=+7$ instead of $-7$. It also flips the direction: as $[{\text{H}}^{+}]$ gets bigger (more acidic), $\text{pH}$ gets smaller. Low pH = acidic, high pH = basic — the familiar convention, delivered by one minus sign.

PICTURE. The concentration axis (top, running $10^0\to 10^{-14}$) is redrawn as the pH axis (bottom, running $0\to 14$). The red minus-flip arrow shows how $10^{-7}$ lands on the value 7 right in the middle.

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Step 6 — Snapping it together: pH + pOH = 14

WHAT. Take the unbreakable product from Step 3 and hit it with ${\log }_{10}$:

$\underset{\text{the fixed product}}{\underbrace{[{\text{H}}^{+}][{\text{OH}}^{-}]}}=10^{-14}$

Apply ${\log }_{10}$ to both sides: ${\log }_{10}([{\text{H}}^{+}][{\text{OH}}^{-}])={\log }_{10}(10^{-14})=-14$

Use the product rule from Step 4 to split the left side into a sum: $\underset{=-\text{pH}}{\underbrace{{\log }_{10}[{\text{H}}^{+}]}}+\underset{=-\text{pOH}}{\underbrace{{\log }_{10}[{\text{OH}}^{-}]}}=-14$

Multiply every term by $-1$ (this is the "p"-flip from Step 5): $\boxed{\text{pH}+\text{pOH}=14}$

WHY. Each move earns its keep: $\log$ turned the product $K_w$ into a sum; the product rule separated the two ions; the minus sign converted both logs into the p-quantities. The fixed product $10^{-14}$ became the fixed sum $14$.

PICTURE. The see-saw returns — but now measured in p-units. The plank is a straight line of length 14. If pH sits at 4, pOH must sit at 10; they always balance to 14.

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Step 7 — The same trick gives pKa + pKb = 14

WHAT. A weak acid $\text{HA}$ and its conjugate base ${\text{A}}^{-}$ have their own constants: $K_a=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]},K_b=\frac{[\text{HA}][{\text{OH}}^{-}]}{[{\text{A}}^{-}]}.$ Multiply them: $K_a\times K_b=\frac{[{\text{H}}^{+}]\cancel{[{\text{A}}^{-}]}}{\cancel{[\text{HA}]}}\cdot \frac{\cancel{[\text{HA}]}[{\text{OH}}^{-}]}{\cancel{[{\text{A}}^{-}]}}=[{\text{H}}^{+}][{\text{OH}}^{-}]=K_w=10^{-14}.$

WHY. Every fragile bracket ($[\text{HA}]$ and $[{\text{A}}^{-}]$) cancels in a diagonal criss-cross, leaving behind exactly the water product $K_w$ from Step 3. So $K_a\cdot K_b$ is the same fixed product — which means the identical $\log$-then-flip machinery gives: ${\text{pK}}_a+{\text{pK}}_b=14.$

PICTURE. The cancellation drawn as coloured tiles pairing up and vanishing, leaving only $[{\text{H}}^{+}][{\text{OH}}^{-}]$.

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Step 8 — Edge and degenerate cases (never get surprised)

WHAT & WHY. We must check the corners of the scale so no scenario ambushes the reader.

• Neutral water: $[{\text{H}}^{+}]=[{\text{OH}}^{-}]=10^{-7}$, so $\text{pH}=\text{pOH}=7$, and $7+7=14$. ✔
• Strong acid, 1 M: $[{\text{H}}^{+}]=1=10^0$, so $\text{pH}=0$; then $\text{pOH}=14$. The see-saw is fully tilted. Still sums to 14. ✔
• Strong base, 1 M OH⁻: $\text{pOH}=0$, $\text{pH}=14$ — the mirror image.
• Below-zero / above-14 pH: With $[{\text{H}}^{+}]>1$ M (very concentrated acid) $\text{pH}$ goes negative; the 0–14 range is a habit, not a law. The sum rule still holds.
• Temperature ≠ 25 °C: $K_w$ changes with temperature (water splits more when hot). At 60 °C, $K_w\approx 10^{-13}$, so neutral pH $\approx 6.5$ and the sum becomes $\text{pH}+\text{pOH}=13$. The derivation is identical; only the constant moves.
• Truly pure infinitely-dilute limit: you can never make $[{\text{H}}^{+}]=0$ — water itself always supplies $\approx 10^{-7}$ M. So $\text{pH}$ can approach but never reach $+\infty$.

PICTURE. The full see-saw drawn at four tilts (pH = 0, 7, 14, and a hot-water case summing to 13) so every extreme is visible at once.

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The one-picture summary

Everything above, compressed: a jar of water → the fixed product $K_w=10^{-14}$ → apply ${\log }_{10}$ (product becomes a sum) → apply the minus flip (concentrations become p-values) → out drops $\text{pH}+\text{pOH}=14$, and the identical path with cancelling brackets gives ${\text{pK}}_a+{\text{pK}}_b=14$.

Recall Feynman retelling — say it back in plain words

Water quietly splits into two pieces: a sour piece (${\text{H}}^{+}$) and a soapy piece (${\text{OH}}^{-}$). They're born in pairs, and nature fixes their product at a tiny $10^{-14}$ — a see-saw whose two ends must multiply to a constant. Those raw numbers have fourteen zeros, so we shrink them with a logarithm, which has a magic power: it turns "multiply" into "add". Multiplying the two concentrations becomes adding their logs. Slap a minus sign on each log so the numbers come out friendly (neutral water reads 7, not −7), and the fixed product of $10^{-14}$ turns into a fixed sum of 14. That's the whole trick: $\text{pH}+\text{pOH}=14$. Run the exact same play on a weak acid and its partner base — their two constants multiply, the fragile middle terms cancel, and you're left with the same $10^{-14}$. Log it, flip it, and you get ${\text{pK}}_a+{\text{pK}}_b=14$. Same see-saw, same 14, different costume.

Recall Quick self-test

Why do we use log base 10 and not $\ln$? ::: Because concentrations are written as powers of ten; base-10 makes each ×10 a clean step of 1, and keeps the scale at 0–14. A solution has pOH = 9. What is its pH at 25 °C? ::: $14-9=5$. Why does a small ${\text{pK}}_a$ force a large ${\text{pK}}_b$? ::: Their sum is fixed at 14, so one small means the other is large. At 60 °C the neutral pH is about 6.5, not 7. Why? ::: $K_w$ grows with temperature (more water splits), so neutral pH drops and the sum falls below 14.