How this instrument works
A buffer solution is a mix of a weak acid and its conjugate base, sitting together in a way that resists pH change when small amounts of acid or base are added. The Henderson-Hasselbalch equation is the standard shortcut for finding a buffer's pH without solving the full equilibrium expression from scratch: pH = −log₁₀(Ka) + log₁₀([A⁻]/[HA]), where [A⁻] is the conjugate base concentration and [HA] is the weak acid concentration.
The first term, −log₁₀(Ka), is just pKa written a different way — it's the pH at which the acid and base would sit in exactly equal concentration. The second term is a correction that shifts the pH up or down from that reference point depending on which side of the acid/base balance the buffer actually leans toward. Equal concentrations make the log term zero, so pH lands exactly on pKa; extra base pushes pH higher, extra acid pulls it lower.
This instrument takes Ka directly rather than pKa, which matters because lab measurements and literature values are often reported as Ka (sometimes in scientific notation like 1.78 × 10⁻⁵) rather than as the already-converted pKa. Feeding Ka in raw and letting the instrument take the log for you removes one manual conversion step — and one common place to introduce a sign error.
- Enter the acid dissociation constant, Ka, for the weak acid in the buffer.
- Enter the conjugate base concentration, [A⁻], in mol/L.
- Enter the weak acid concentration, [HA], in mol/L.
- Read the buffer pH, computed from the Henderson-Hasselbalch equation.
Worked example — a carbonate buffer at equal acid and base
A carbonate buffer has a Ka of 3.981 × 10⁻⁷ (that's pKa = 6.4), mixed with equal concentrations of acid and conjugate base — 6 mol/L of each. Because [A⁻] and [HA] are identical, the log ratio term is log₁₀(6/6) = log₁₀(1) = 0, so the entire second term of Henderson-Hasselbalch drops out and pH = −log₁₀(3.981×10⁻⁷) = 6.4 exactly, landing precisely on the buffer's pKa.
Now unbalance it: keep the same acid, but use 0.2 mol/L of conjugate base against only 0.05 mol/L of acid — a 4:1 excess of base over acid. The log ratio is log₁₀(4) ≈ 0.602, and since the acid here is acetic acid (Ka = 1.78×10⁻⁵, pKa 4.75), the resulting pH climbs to about 5.35 — well above pKa, exactly as expected when conjugate base dominates the mixture.
Questions
What is the Henderson-Hasselbalch equation used for?
It's the standard formula for calculating the pH of a buffer solution directly from the weak acid's Ka (or pKa) and the ratio of conjugate base to weak acid concentrations, without needing to solve the full acid-dissociation equilibrium expression. It's also used in reverse — given a target pH and a chosen buffer, to work out what ratio of acid to base is needed to prepare it.
Why does pH equal pKa when acid and base concentrations are equal?
Because the Henderson-Hasselbalch equation's second term is log₁₀([A⁻]/[HA]), and when those two concentrations are equal, that ratio is exactly 1. The logarithm of 1 is 0, so the whole term vanishes and pH is left equal to just −log₁₀(Ka), which is the definition of pKa. This is also why pKa is described as the pH at the midpoint of a buffer's titration curve — it's precisely where acid and base are balanced 1:1.
What's the difference between this calculator and a pKa-based buffer pH tool?
They compute the identical pH from the identical underlying relationship — the only difference is which form of the acid strength you type in. This instrument takes Ka directly (useful when your source data is reported that way, often in scientific notation), while a pKa-based version skips straight to the already-converted −log₁₀(Ka) value. Convert between them with pKa = −log₁₀(Ka) if you have one and need the other.
Does the Henderson-Hasselbalch equation work for any acid concentration?
It's an approximation that holds well when the buffer concentrations are reasonably high relative to Ka — the typical situation for a working buffer solution. At very dilute concentrations, or when the acid or base is unusually strong or weak, the simplifying assumptions behind the equation (that the equilibrium concentrations are close to the starting concentrations) start to break down, and a full equilibrium calculation gives a more accurate answer.
What happens to pH if I have more acid than conjugate base?
The pH drops below the pKa. With [HA] greater than [A⁻], the ratio [A⁻]/[HA] is less than 1, so its logarithm is negative, and that negative value gets added to pKa, pulling the pH down. A buffer with a 1:4 base-to-acid ratio, for instance, sits noticeably more acidic than its own pKa — the exact behavior the acetic acid example above demonstrates in reverse.