Buffer pH Calculator
Calculate the pH of a buffer solution using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]).
Common pKa Values Reference
Buffer pH and the Henderson-Hasselbalch equation
The pH of a buffer is set by two things: the pKa of the weak acid, and the ratio of conjugate base to acid in solution. Henderson-Hasselbalch packages both into one line:
pH = pKa + log([A⁻]/[HA])
This is just the Ka expression — Ka = [H⁺][A⁻]/[HA] — solved for pH after taking the negative log of both sides. The calculator takes pKa, [HA], and [A⁻] as inputs and returns pH along with the substituted log term.
The structure of the equation tells you everything about how a buffer behaves. When [A⁻] = [HA], the log is zero and pH equals pKa exactly — this is the center of the buffer’s range and where capacity is highest. As the ratio shifts away from 1, pH drifts above or below pKa, but only logarithmically: a tenfold excess of conjugate base raises pH by exactly one unit, and a hundredfold excess raises it by two. That slow drift is what makes a buffer resist pH change.
The formula assumes the equilibrium concentrations of HA and A⁻ are close to the analytical concentrations you mixed. That holds when both species are present at comparable concentrations and the acid is genuinely weak. Once the ratio gets past 10:1 in either direction, you’re at the edge of the useful range; past 100:1 the approximation has broken down and you need the full equilibrium expression.
Worked examples
Acetate buffer: 0.10 M acetic acid (pKa 4.76) and 0.15 M acetate. pH = 4.76 + log(0.15/0.10) = 4.76 + 0.176 = 4.94
Phosphate at physiological pH: 0.050 M H₂PO₄⁻ (pKa 7.20) and 0.050 M HPO₄²⁻. pH = 7.20 + log(1) = 7.20 — the equimolar case where pH equals pKa.
Ammonia/ammonium: 0.20 M NH₄⁺ (pKa 9.25) and 0.30 M NH₃. pH = 9.25 + log(0.30/0.20) = 9.25 + 0.176 = 9.43