Buffers are designed to resist pH change, so predicting their pH quickly is essential. The Henderson–Hasselbalch equation links buffer pH to pKa and the conjugate pair ratio.

Core idea: buffer pH is controlled by a ratio

A buffer contains appreciable amounts of a weak acid and its conjugate base (or a weak base and its conjugate acid). The Henderson–Hasselbalch equation expresses that the pH is set by how much base-form you have compared to acid-form, not by either concentration alone.

Key terms you must recognise

A buffer’s two components must be present together; otherwise, the “ratio control” on pH is lost.

Lower pKa means a stronger acid; higher pKa means a weaker acid.

Henderson–Hasselbalch equation (what you use on buffers)

Use Henderson–Hasselbalch when the solution contains both $HA$ and $A^-$ in meaningful amounts (a true buffer situation).

This equation is base-10 logarithmic and depends on the ratio $[A^-]/[HA]$.

These titration curves plot pH versus volume of titrant added, contrasting a strong acid with a weak acid titrated by a strong base. The weak-acid curve shows an extended buffer region where both $HA$ and $A^-$ are present, so Henderson–Hasselbalch applies and the midpoint (half-equivalence) corresponds to $[A^-]=[HA]$ (therefore $pH=pK_a$). Source

How to apply the equation correctly

1) Identify the conjugate pair

• Confirm the buffer pair is of the form weak acid $HA$ and conjugate base $A^-$.

• Match the given pKa to the correct acid $HA$ (not to $A^-$).

2) Determine the needed ratio term

• Use the ratio $\dfrac{[A^-]}{[HA]}$ exactly as written.

• If the buffer is made by mixing solutions, it is often more reliable to use moles:

  • When both species are in the same final solution volume, $\dfrac{[A^-]}{[HA]}=\dfrac{n_{A^-}}{n_{HA}}$ because the volume cancels.

• Maintain consistent units for concentrations (typically molarity).

3) Interpret the equation (quick qualitative checks)

• If $[A^-]=[HA]$, then $\log(1)=0$, so pH = pKa.

• If $[A^-] > [HA]$, then $\log!\left(\dfrac{[A^-]}{[HA]}\right) > 0$, so pH > pKa (more basic).

• If $[A^-] < [HA]$, then the log term is negative, so pH < pKa (more acidic).

• Increasing both $[A^-]$ and $[HA]$ by the same factor keeps the ratio (and thus the pH) unchanged.

Common pitfalls (and how to avoid them)

• Swapping the ratio: Writing $\log([HA]/[A^-])$ flips the sign and gives the wrong pH shift direction.

• Using the wrong pKa: Ensure the provided $pK_a$ corresponds to $HA \rightleftharpoons H^+ + A^-$.

• Forgetting it is a buffer requirement: If one component is essentially absent (all $HA$ or all $A^-$), Henderson–Hasselbalch no longer describes the solution well.

• Confusing concentration with amount: When solutions are mixed, compare final concentrations or use moles so dilution is handled correctly.

• Log rules errors: Remember $\log(x)$ changes slowly; large pH changes require large ratio changes.