Every alkali question you have done so far has taken the long way round: hydroxide concentration, then the ionic product, then hydrogen ion concentration, then a log. There is a shortcut. Give the hydroxide side its own scale and the whole journey collapses into a single subtraction.
📘 What you need to know
pOH = −log10[OH−], exactly the same idea as pH but for hydroxide ions.
Rearranged: [OH−] = 10−pOH.
The two scales are locked together: pH + pOH = pKw.
At 298 K, pKw = 14.00, so pH + pOH = 14.00.
pKw = −log10Kw, so if the temperature changes, so does the 14.
A low pOH means a strongly alkaline solution — the scale runs the opposite way to pH.
You can start from any one of [H+], [OH−], pH or pOH and reach the other three.
The same trick, applied to hydroxide
The letter “p” in front of a quantity just means “take the negative log to base 10 of it”. So pH is −log[H+], pOH is −log[OH−], and pKw is −logKw. Once you see “p” as an instruction rather than a letter, all these definitions become one definition.
The pOH pair
pOH = −log10[OH−] and [OH−] = 10−pOH
Where the 14 comes from
You already know that [H+][OH−] = Kw. Take the negative log of both sides. A log of a product is the sum of the logs, so:
Deriving the link
−log([H+] × [OH−]) = −logKw
(−log[H+]) + (−log[OH−]) = −logKw pH + pOH = pKw = 14.00 at 298 K
So the 14 is not a magic number either. It is just −log(1.00 × 10−14).
Because it comes from Kw, the 14 moves with temperature. At 100 °C, Kw = 5.62 × 10−13, so pKw = 12.25 and pH + pOH = 12.25. If a question gives you a temperature other than 298 K, do not reach for 14 automatically.
The top edge is nearly always the quickest route. If you know [OH−], go up to pOH and across — two steps instead of three.
The two scales are mirror images
Because they always add to 14, the two scales run in opposite directions. Strongly alkaline solutions have a high pH and a low pOH. Strongly acidic solutions have a low pH and a high pOH.
Neutral water at 298 K is the only point where pH and pOH are both 7. Heat the water and both fall together, but they still add up to pKw.
🧩 The fast route for any alkali question
Write down [OH−] from the concentration of the strong base.
pOH = −log10[OH−].
pH = 14.00 − pOH.
Sanity check: an alkali should land above 7. If it did not, you probably logged the wrong concentration.
Compare the two routes. The long way for 0.0500 mol dm−3 NaOH: divide Kw by 0.0500, then log it. The short way: log 0.0500 to get pOH 1.30, then 14.00 − 1.30 = 12.70. Same answer, one fewer chance to mistype an exponent.
Worked examples
WORKED EXAMPLE
Calculate the pH of 0.150 mol dm−3 KOH(aq) at 298 K using pOH.
Step 1: KOH is a strong base, one OH− per formula unit[OH−] = 0.150 mol dm−3Step 2: take the negative logpOH = −log10(0.150) = 0.8239Step 3: subtract from 14.00pH = 14.00 − 0.82 = 13.18pH = 13.18a strong alkali near 0.1 mol dm⁻³ should always come out around 13
WORKED EXAMPLE
A solution has pH 3.40 at 298 K. Find its pOH and its hydroxide ion concentration.
Step 1: cross the top of the squarepOH = 14.00 − 3.40 = 10.60Step 2: undo the log[OH−] = 10−10.60 = 2.51 × 10−11 mol dm−3Step 3: check it makes sense
pH 3.40 is acidic, so hydroxide should be tiny. It is.
pOH = 10.60 and [OH−] = 2.51 × 10−11 mol dm−3two steps instead of the three you would need going round the bottom
WORKED EXAMPLE
At 60 °C, Kw = 9.61 × 10−14. A solution at this temperature has pOH 4.20. Calculate its pH.
Step 1: do not assume 14 — work out pKwpKw = −log10(9.61 × 10−14) = 13.02Step 2: use the general relationshippH + pOH = pKwStep 3: substitutepH = 13.02 − 4.20 = 8.82pH = 8.82using 14 here would have given 9.80 — a whole unit out
💡 Exam tip
For any strong base question, go straight to pOH. It is the shortest and safest path.
Write the relationship as pH + pOH = pKw, then substitute 14.00 only if the temperature is 298 K.
Remember “p” means −log10 of whatever follows. That one idea covers pH, pOH, pKw, pKa and pKb.
For Ba(OH)2 or Ca(OH)2, remember there are two hydroxides per formula unit, so double the concentration first.
Quote pOH to 2 decimal places, the same as pH.
Check the direction at the end: pOH below 7 means alkaline, pOH above 7 means acidic. It is the opposite way round to pH.
⚠ Common mix-up
Calling −log[OH−] the pH. That is the pOH. You still need to subtract it from 14.
Reading the pOH scale like the pH scale. A pOH of 1 is very alkaline, not very acidic.
Using 14 at every temperature. The sum equals pKw, which changes when the water warms up.
Forgetting the 2 in Ba(OH)2. 0.05 mol dm−3 Ba(OH)2 gives 0.10 mol dm−3 OH−.
Subtracting concentrations instead of p values. The rule works on the logged numbers, never on the raw concentrations.
Applying pOH to a weak base concentration directly. A weak base is only partly ionised, so [OH−] is not the label concentration.
Up next: Acid and Base Dissociation Constants. That last mix-up is the doorway — to handle weak acids and bases properly we need a number that measures exactly how far each one dissociates.
Want this explained one-to-one?
Book a free session with an experienced IB Chemistry tutor and get your trickiest topics made simple.
