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IB Chemistry pKa and pH: How They Link and How to Calculate

IB Chemistry pKa and pH: How They Link and How to Calculate

10 min readOliver Kidd (Co-founder)28 Oct 2026

In IB Chemistry, pH tells you how acidic a particular solution is. pKa tells you how strong a particular acid is. The two meet in one equation, pH = pKa + log([A⁻]/[HA]), and once you can use it with confidence, buffers, indicators and titration curves all become much easier. This guide walks you through the ideas, the calculations and the exam traps, step by step.

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TL;DR

  • pH describes a solution. pKa describes an acid. A lower pKa means a stronger acid.
  • When pH = pKa, the weak acid HA and its conjugate base A⁻ are present in equal amounts.
  • You cannot find a solution’s pH from pKa alone. You also need a concentration or a ratio of acid to base.
  • For a pure weak acid solution, use Ka and an ICE table. For a buffer, use the Henderson-Hasselbalch equation.
  • Indicators change colour roughly within pKa ± 1, which is how you choose one for a titration.

Table of Contents

What pH and pKa each measure

pH is the negative base-10 logarithm of the hydronium ion concentration: pH = −log₁₀[H₃O⁺]. It describes the acidity of a particular aqueous solution, as Chemistry LibreTexts explains. It is not a fixed property of the acid itself. Dilute the same acid and the pH changes.

pKa is different. It is defined as pKa = −log₁₀Ka, and it measures acid strength. A lower pKa means a stronger acid and more ionisation. Unlike pH, pKa belongs to the acid. It applies under stated conditions such as temperature, pressure and solvent.

A simple way to hold the difference: pH is a reading on the solution, pKa is a label on the acid. Many lost marks come from mixing these up, for example by saying a solution “has a pKa”.

What pH and pKa each measure — overview diagram

The link is the Henderson-Hasselbalch equation:

pH = pKa + log₁₀([A⁻]/[HA])

It comes from taking logarithms of the Ka expression, as this general chemistry text shows. Rearranged, it also tells you the ratio of the two forms: log([A⁻]/[HA]) = pH − pKa, as OpenStax sets out.

What it tells you at a glance

Condition What it means
pH = pKa [HA] = [A⁻], the log term is zero
pH < pKa The protonated form HA predominates
pH > pKa The deprotonated form A⁻ predominates

The last two rows are the predominant-form rule. They are quick to use in exam questions that ask which species is mainly present at a given pH.

At pH = pKa a buffer also works best. Maximum buffering capacity occurs there, because the system can absorb small additions of acid or base most effectively.

Can you calculate pH from pKa alone?

No. pKa gives you Ka, since Ka = 10^(−pKa), but a pH needs one more piece of information. Which piece depends on the question:

  • A weak acid in water: you need its initial concentration.
  • A buffer: you need the ratio [A⁻]/[HA], or the amounts of each.
  • A known pH: you can find the ratio of forms, but not the concentrations themselves.

If a question gives only a pKa and asks for a pH, reread it. A concentration or a ratio is hiding somewhere, often in a mass, a volume or a titration point.

Worked example 1: pH of a buffer

A buffer contains 0.20 mol dm⁻³ of a weak acid’s conjugate base A⁻ and 0.10 mol dm⁻³ of the acid HA. The pKa is 4.80.

  1. Write the equation: pH = pKa + log([A⁻]/[HA]).
  2. Substitute: pH = 4.80 + log(0.20/0.10).
  3. Evaluate: log 2 = 0.30, so pH = 5.10.

A sense check helps. There is more base than acid, so the pH should sit above the pKa. It does.

As a general method, use equilibrium concentrations, or starting concentrations when the extent of ionisation is small enough to justify the approximation. In most buffer questions it is.

Worked example 2: pH of a weak acid solution

Here there is no conjugate base added, so Henderson-Hasselbalch is the wrong tool. Take 0.10 mol dm⁻³ HA with pKa 4.80.

  1. Convert: Ka = 10^(−4.80) = 1.6 × 10⁻⁵.
  2. Set up an ICE table. The initial pH of a weak acid solution is found this way, using Ka.
  3. Assume x is small: Ka ≈ x²/0.10, so x = [H⁺] = √(1.6 × 10⁻⁵ × 0.10) = 1.3 × 10⁻³ mol dm⁻³.
  4. pH = −log(1.3 × 10⁻³) = 2.90.

The shortcut is pH = ½(pKa − log c), here ½(4.80 + 1) = 2.90. Check that x is far smaller than the starting concentration before you rely on the approximation. Here it is about 1% of 0.10, so it holds.

pH and pKa in titrations

A common slip is to say pH = pKa at the equivalence point. It does not. At the equivalence point of a weak acid titrated with a strong base, the pH is above 7 because the conjugate base hydrolyses water.

The point where pH = pKa is the half-equivalence point. There, half the acid has been converted to its conjugate base, so [HA] = [A⁻]. Reading that pH off a curve gives you the pKa directly.

Titration curve marking half and full equivalence

Acid strength also shapes the curve, including the starting pH and the region near equivalence. The amount of base needed to reach equivalence, however, does not depend on acid strength. For a full tour of curve shapes and marking points, see our guide to IB Chemistry titration curves.

Choosing an indicator with pKa

An indicator is itself a weak acid, HIn. Its colour depends on the ratio [In⁻]/[HIn], and most indicators change colour over about pKa ± 1.

So the rule is simple. Pick the indicator whose pKa range sits on the steep part of your curve. For a weak acid and strong base, that means a range above 7. This skill appears in school IB courses too. For example, a US district’s IB Chemistry unit planner lists identifying an appropriate indicator from the salt and the indicator’s pH range as a learning objective, alongside interpreting Ka and pKa values.

Common exam mistakes

  • Using Henderson-Hasselbalch for a weak acid with no conjugate base present.
  • Swapping the ratio, writing [HA]/[A⁻] instead of [A⁻]/[HA].
  • Forgetting that pKa = −log Ka, so a smaller number means a stronger acid.
  • Quoting pH = pKa at equivalence instead of half-equivalence.
  • Skipping the check that the small-change approximation is valid.

Our acids and bases revision plan maps these ideas to exam questions over three weeks. If equilibrium expressions still feel shaky, start with Kc rules and two examples, since Ka is just a special case.

Our perspective

We see the same pattern again and again. Students who struggle with pKa and pH rarely lack maths. They lack a clear picture of which quantity belongs to what: pKa to the acid, pH to the solution, and the ratio as the bridge between them.

The IB’s chemistry course promotes concept-based teaching and learning to build critical thinking, and this topic is a good example. Memorising the equation is not enough. You need to be able to say what a result means: which form predominates, why a buffer resists change, why an indicator turns when it does.

That is how we build Tiber Tutor. Our notes, animated videos, exam-style questions and mark schemes are written by IB examiners and linked together, so a mistake on a buffer question points you straight back to the notes that fix it. Our performance tracking shows which sub-topics need work, so you spend your time where it counts. We believe it is the best-value IB STEM platform available, because the examiner-written content, the mock exams and the analytics all sit in one membership.

— Oliver

Practise it with exam-style questions

Reading a worked example is the first step. Doing the question under exam conditions is the second. IB Chemistry resources on Tiber Tutor include topic tests, examiner mark schemes, student-led answer review and mock exams, all with progress tracking so you can see your confidence grow. You can try everything free for 7 days. After that, the Per-Subject Plan is $9 per month and the All-Access Plan is $19 per month. See the pricing page for details.

FAQ

How do you calculate pH from pKa?

You need more than the pKa. For a buffer, use pH = pKa + log([A⁻]/[HA]). For a pure weak acid, convert pKa to Ka, then use an ICE table with the starting concentration to find [H⁺].

It is the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). It comes from taking logarithms of the Ka expression and rearranging.

What happens when pH equals pKa?

The concentrations of the weak acid HA and its conjugate base A⁻ are equal, so the log term is zero. In a titration of a weak acid with a strong base, this occurs at the half-equivalence point.

Does pH equal pKa at the equivalence point?

No. For a weak acid titrated with a strong base, the equivalence point pH is above 7 because the conjugate base hydrolyses. The pH equals the pKa at half-equivalence.

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