The last page of the topic, and it needs almost no new chemistry. Rearrange the Ka expression you have been using since page 11, take a log, and out drops the equation that tells you the pH of any buffer — and what happens to it when you attack it with acid.
For a basic buffer: pOH = pKb + log10([salt] ÷ [base]), then pH = 14.00 − pOH.
When [salt] = [acid], log(1) = 0, so pH = pKa.
The pH depends on the ratio, so you can use moles instead of concentrations when both are in the same solution.
Adding acid increases the acid and decreases the salt by the same number of moles. Adding alkali does the reverse.
A ratio change of ten times only moves the pH by one unit.
Choose the weak acid whose pKa is closest to the pH you want, then fine-tune with the ratio.
Where the equation comes from
Start with Ka for the weak acid in the buffer. The A− in solution comes almost entirely from the salt, so we write [salt] for it, and the HA is almost entirely undissociated, so we write [acid] for that.
Two lines of rearranging
Ka = [H+][salt] ÷ [acid]
[H+] = Ka × [acid] ÷ [salt]
pH = pKa + log10([salt] ÷ [acid])
You can work with either form. The middle one is often quicker if the question gives you Ka; the bottom one is quicker if it gives you pKa. They give identical answers.
This is why you cannot make a pH 9 buffer out of ethanoic acid. Its pKa is 4.76, and no sensible ratio will drag it four units away from that.
Watch the order inside the log. It is salt over acid, so more salt pushes the pH up. If you write it upside down every answer will be reflected about the pKa, which is an easy mistake to spot: check that more salt gave you a higher pH.
Adding acid or alkali to a buffer
These questions look harder than they are. The added acid converts some salt into acid; the added alkali does the reverse. You just keep a tally of the moles and then put the new numbers back into the equation.
That last line is the whole point of the topic. The same 0.010 mol of HCl in a litre of pure water would take the pH from 7 down to 2. Here it moved by less than a tenth of a unit.
🧩 Buffer pH after adding acid or alkali
Work out the moles of acid and salt you started with.
Work out the moles of H+ or OH− added.
Adjust the tally. Added H+: acid goes up, salt goes down. Added OH−: acid goes down, salt goes up.
Put the new moles straight into pH = pKa + log([salt] ÷ [acid]). The volume cancels, so you never need to convert to concentrations.
Worked examples
WORKED EXAMPLE
Calculate the pH of a buffer containing 0.305 mol dm−3 ethanoic acid and 0.520 mol dm−3 sodium ethanoate. Ka = 1.74 × 10−5.
Step 1: use the rearranged Ka expression[H+] = Ka × [acid] ÷ [salt]Step 2: substitute[H+] = 1.74 × 10−5 × 0.305 ÷ 0.520 = 1.021 × 10−5Step 3: take the logpH = −log10(1.021 × 10−5) = 4.991Step 4: check it against pKa
pKa is 4.76 and there is more salt than acid, so the pH should be slightly above it. It is.
pH = 4.99the same sum through Henderson–Hasselbalch: 4.76 + log(0.520/0.305) = 4.99
WORKED EXAMPLE
A buffer contains 0.100 mol of ethanoic acid and 0.100 mol of sodium ethanoate in 1.00 dm3. Calculate the new pH after adding 0.010 mol of HCl. pKa = 4.76.
Step 1: note the starting pH
Equal moles means log(1) = 0, so it starts at pH 4.76.
Step 2: adjust the tally for the added acid
The H+ reacts with ethanoate and turns it into ethanoic acid.
acid: 0.100 + 0.010 = 0.110 molsalt: 0.100 − 0.010 = 0.090 molStep 3: put the new amounts back inpH = 4.76 + log10(0.090 ÷ 0.110) = 4.76 + log10(0.818)= 4.76 − 0.087 = 4.673pH = 4.67the pH fell by only 0.09; the same acid in pure water would give pH 2.00
WORKED EXAMPLE
Calculate the pH of a buffer made from 0.200 mol dm−3 ammonia and 0.150 mol dm−3 ammonium chloride. pKb(NH3) = 4.75.
Step 1: spot that this is a basic buffer
Weak base plus its salt, so work in pOH first.
Step 2: use the base version of the equationpOH = pKb + log10([salt] ÷ [base])= 4.75 + log10(0.150 ÷ 0.200) = 4.75 + log10(0.750)= 4.75 − 0.125 = 4.625Step 3: convert to pHpH = 14.00 − 4.625 = 9.375pH = 9.38a basic buffer should land above 7, and 9.38 does — always check that
WORKED EXAMPLE
You need a buffer at pH 4.8. Explain how you would choose the acid, using the table of pKa values from earlier in the topic.
Step 1: remember what the ratio can and cannot doA ten-fold ratio only shifts the pH by 1, so the pKa must be close to 4.8 already.Step 2: scan the pKa values
Methanoic 3.75, benzoic 4.20, ethanoic 4.76, carbonic 6.35. Ethanoic is nearest.
Step 3: fix the ratio4.8 = 4.76 + log([salt]/[acid]), so log([salt]/[acid]) = 0.04[salt]/[acid] = 100.04 = 1.1Use ethanoic acid with sodium ethanoate, roughly 1.1 mol of salt per mol of acidnear-equal amounts also give the best capacity, so this is a good buffer to build
💡 Exam tip
The ratio is salt over acid inside the log. Getting it upside down flips your answer about the pKa.
Use moles, not concentrations, when everything is in one solution. The volume cancels and you save a step.
Always sanity check against pKa: more salt than acid means the pH is above pKa.
For a basic buffer, finish with pH = 14.00 − pOH. Handing in the pOH is the classic error.
Convert pK to K or K to pK at the very start, and stick to one form.
When asked to design a buffer, pick the acid with the nearest pKa, then adjust the ratio.
⚠ Common mix-up
Inverting the ratio. [acid]/[salt] belongs in the [H+] version, [salt]/[acid] in the pH version.
Adding the H+ to the salt instead of the acid. Added acid makes more acid and less salt.
Converting to concentrations unnecessarily, then forgetting that the volume changed for both.
Using [H+] = √(Ka × c) on a buffer. That formula is for a weak acid on its own, where [H+] = [A−].
Expecting a big pH change. If your answer moved by more than a few tenths, check your tally.
Trying to build a buffer far from the pKa. Two units away needs a 100 : 1 ratio and has almost no capacity.
That is the whole of Proton Transfer. Look back at where it started: an acid is a proton donor. Everything since — pH, Kw, titration curves, indicators, buffers — is that one sentence followed carefully to its conclusions. Up next: Electron Transfer Reactions, where a different particle starts moving.
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