Work out the pH of a weak acid properly and you end up solving a quadratic. Nobody wants that in an exam. Three sensible assumptions knock the algebra down to a single square root, and the answers stay accurate to two decimal places.
📘 What you need to know
Assumption 1: so little of a weak acid dissociates that [HA] at equilibrium ≈ the concentration you started with.
Assumption 2: each molecule that splits gives one of each ion, so [H+] = [A−].
Assumption 3: the H+ from water is negligible next to the H+ from the acid.
Those three turn Ka into Ka = [H+]2 ÷ [HA], so [H+] = √(Ka × c).
For a weak base the mirror image applies: [OH−] = √(Kb × c), then find pOH and subtract from 14.00.
Going the other way, Ka = [H+]2 ÷ c lets you get Ka from a measured pH.
Assume 298 K unless the question says otherwise.
The assumptions hold while less than about 5% of the acid has dissociated.
The assumption that does all the work
Take 0.100 mol dm−3 ethanoic acid. Solve it exactly and you find [H+] = 1.31 × 10−3 mol dm−3. That is only 1.3% of the acid you put in, which means 98.7% of the molecules never split at all.
Subtracting 1.3% from the starting concentration would change the final pH by about 0.01 — less than the rounding in your answer. That is why we simply do not bother.
Building the shortcut
Start from the proper expression and apply the assumptions one at a time.
From the full expression to the shortcut
Ka = [H+][A−] ÷ [HA]
since [H+] = [A−] → Ka = [H+]2 ÷ [HA]
since [HA] ≈ c → [H+] = √(Ka × c)
Notice the square. It is there because the same unknown appears twice on the top of the expression — once as H+ and once as A−. Students who forget the square end up taking no root at all and are typically two pH units out.
If the question hands you pKa or pKb, your very first line should be K = 10−pK. Trying to carry a pK value into the square root does not work.
🧩 Getting Ka out of a measured pH
Turn the pH into [H+] = 10−pH.
Use [A−] = [H+], because each molecule that split gave one of each.
Use [HA] ≈ c, the concentration in the question.
Put them together: Ka = [H+]2 ÷ c, and take −log if pKa is wanted.
Worked examples
WORKED EXAMPLE
Calculate the pH of 0.200 mol dm−3 propanoic acid at 298 K. Ka = 1.34 × 10−5.
Step 1: state which route you are taking
Weak acid with c and Ka given, so use [H+] = √(Ka × c).
Step 2: substitute[H+] = √(1.34 × 10−5 × 0.200) = √(2.68 × 10−6)= 1.637 × 10−3 mol dm−3Step 3: take the logpH = −log10(1.637 × 10−3) = 2.786…Step 4: check the assumption held1.637 × 10−3 ÷ 0.200 = 0.8%, comfortably under 5%.pH = 2.79a strong acid at this concentration would be pH 0.70 — the gap is the weakness
WORKED EXAMPLE
A 0.0500 mol dm−3 solution of a weak acid has pH 3.20. Calculate Ka and pKa.
Step 1: pH back to [H+][H+] = 10−3.20 = 6.31 × 10−4 mol dm−3Step 2: apply the two assumptions
[A−] equals that same value, and [HA] stays at 0.0500.
Step 3: substitute into the expressionKa = (6.31 × 10−4)2 ÷ 0.0500= 3.98 × 10−7 ÷ 0.0500 = 7.96 × 10−6Step 4: take the negative log for pKapKa = −log10(7.96 × 10−6) = 5.10Ka = 7.96 × 10−6 and pKa = 5.10square the concentration before dividing, not after — order matters here
WORKED EXAMPLE
Calculate the pH of 0.0350 mol dm−3 methylamine, CH3NH2, which has pKb = 3.35 at 298 K.
Step 1: convert pKb into Kb before anything elseKb = 10−3.35 = 4.47 × 10−4Step 2: use the base version of the shortcut[OH−] = √(4.47 × 10−4 × 0.0350) = √(1.563 × 10−5)= 3.954 × 10−3 mol dm−3Step 3: find pOHpOH = −log10(3.954 × 10−3) = 2.403Step 4: the extra step for a basepH = 14.00 − 2.403 = 11.597pH = 11.60stopping at 2.40 and calling it the pH is the single most common error here
💡 Exam tip
State your assumptions if the question says “state any assumptions”. There are usually easy marks for naming them.
Convert pK to K first. Every time.
For a base, write pOH clearly and then subtract. Labelling the line stops you handing in the pOH as your final answer.
Take the square root, not half. √(2.68 × 10−6) is 1.64 × 10−3, not 1.34 × 10−6.
Sanity check the range: a weak acid should land around pH 2 to 5, and a weak base around pH 9 to 12.
Keep every digit on the calculator until the last line. Rounding partway through will shift the second decimal place.
⚠ Common mix-up
Using [H+] = c for a weak acid. That is the strong acid rule and it does not apply here.
Forgetting the square root. This is by far the most frequent slip in the whole HL section.
Giving pOH as the pH. For a base you must do the final subtraction.
Feeding pKa straight into the square root. Convert it to Ka first.
Applying these shortcuts to a strong acid. A strong acid has no Ka worth using — just take the log of the concentration.
Using them on a buffer. A buffer already contains plenty of A−, so [H+] = [A−] is false there.
Up next: Salt Hydrolysis. We now have the tools to answer a question that has been hanging around since the titration pages: why does a solution of a perfectly ordinary salt come out acidic or alkaline?
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