Henderson–Hasselbalch
pH of a buffer solution.
Formula
Variables
Example
A buffer with equal [A⁻] and [HA] has pH = pKₐ (the log term is zero).
How it works
The Henderson–Hasselbalch equation links the pH of a weak-acid buffer to the balance between its conjugate base and acid forms. It comes from rearranging the weak-acid equilibrium expression, so it works best when both members of the buffer pair are present in appreciable amounts and the acid does not dissociate completely. The logarithm shows that pH depends on the ratio [A⁻]/[HA], not their absolute amounts, which is why buffers resist pH change when that ratio stays similar.
When to use it
Use this when analyzing or preparing buffer solutions made from a weak acid and its conjugate base, especially in lab chemistry, biochemistry, and titration problems. It is also useful for finding how much acid or base has been added to a buffer and estimating the resulting pH.
Worked example, step by step
A buffer contains 0.20 M lactic acid (HA) and 0.050 M lactate (A⁻). The acid has pK_a = 3.86. Step 1: Write the equation: pH = pK_a + log([A⁻]/[HA]) Step 2: Substitute values: pH = 3.86 + log(0.050/0.20) Step 3: Simplify the ratio: 0.050/0.20 = 0.25 Step 4: Take the logarithm: log(0.25) = -0.602 Step 5: Add to pK_a: pH = 3.86 + (-0.602) = 3.258 Step 6: Round appropriately: pH ≈ 3.26
Common mistakes to avoid
- Reversing the ratio and using [HA]/[A⁻] instead of [A⁻]/[HA]
- Using the formula for a strong acid or strong base solution, where it does not apply
- Confusing pK_a with pH or forgetting that the logarithm is base 10
Did You Know?
The Henderson–Hasselbalch equation is how your blood keeps its pH nearly constant to keep you alive.
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