AP® Chemistry · Unit 8 · Topic 8.9
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation says a buffer's pH equals the pKa of its weak acid plus the log of the ratio of base to acid concentrations. It turns the messy equilibrium math of a buffer into a quick lookup.
Start from the equilibrium for a weak acid: HA <→ H+ + A-, with Ka = [H+][A-]/[HA]. Rearranging for [H+] gives [H+] = Ka x [HA]/[A-]. Take the negative log of both sides. The negative log of [H+] is pH, the negative log of Ka is pKa, and the negative log of a ratio flips it, turning [HA]/[A-] into [A-]/[HA] once you distribute the sign. That leaves pH = pKa + log([A-]/[HA]), or in general form, pH = pKa + log([base]/[acid]).
That ratio is why concentration alone does not set a buffer's pH. Both the acid and its conjugate base live in the same volume of solution, so if you double the volume, both concentrations get cut in half together. The ratio between them stays exactly the same, and so does the pH. What actually shifts the pH is changing how much acid or base you have relative to each other, not how spread out they are in water.
This equation is the algebraic shortcut behind everything else you do with buffers, including picking a buffer recipe for a target pH or reading the flat middle region of a titration curve. At the half-equivalence point of a weak acid titration, [A-] equals [HA], the log term is zero, and pH equals pKa exactly. That single fact ties this equation directly to the titration curves you study next.
Key terms
- pKa
- The negative log base ten of the acid dissociation constant Ka, a fixed number for a given weak acid.
- Buffer
- A solution containing a weak acid and its conjugate base together, which resists large pH changes.
- Half-equivalence point
- The point in a titration where exactly half the weak acid has been converted to its conjugate base, making pH equal pKa.
- Conjugate base
- The species left behind after a weak acid donates a proton, such as acetate from acetic acid.
Watch it explained
- How it worksWalk me through where the Henderson-Hasselbalch equation comes from starting at Ka.
- Worked exampleFind the pH of a buffer holding 0.20 M acetic acid and 0.30 M sodium acetate.
- Worked exampleFind the mass of sodium acetate needed to make 500 mL of a pH 5.00 buffer in 0.10 M acetic acid.
- Common mistakeWhy isn't diluting a buffer with water a way to change its pH?
Check yourself
End of topic.