The formula
- Buffer pH
pH = pKa + log([A⁻] / [HA])- pKa from the acid ionization constant
pKa = -log K_a- Equal concentrations
pH = pKa when [A⁻] = [HA]- Ratio needed for a target pH
[A⁻] / [HA] = 10^(pH - pKa)
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
pH | pH of the buffer solution | dimensionless |
pKa | Negative base-10 logarithm of the acid ionization constant | dimensionless |
K_a | Acid ionization constant of the weak acid | dimensionless |
[A⁻] | Concentration of the conjugate base | M |
[HA] | Concentration of the weak acid | M |
When it applies
- The solution is a genuine buffer, holding both a weak acid and its conjugate base in comparable amounts.
- The ratio [A⁻]/[HA] sits between about 0.1 and 10, so the log term stays between -1 and +1. Outside that window the approximation degrades.
- The equation uses the equilibrium concentrations, approximated by the amounts mixed in. That approximation holds because the weak acid ionizes very little in the presence of its own conjugate base.
- It does not describe a strong acid or strong base, and it cannot be stretched to a solution containing only one member of the pair.
Worked example
Problem. A buffer is made 0.20 M in acetic acid and 0.30 M in sodium acetate. Acetic acid has K_a = 1.8 × 10⁻⁵. What is the pH?
- Convert K_a to pKa: pKa = -log(1.8 × 10⁻⁵) = 4.74.
- Identify the pair: acetic acid is HA at 0.20 M, and acetate ion from the salt is A⁻ at 0.30 M.
- Take the ratio and its logarithm: [A⁻]/[HA] = 0.30 / 0.20 = 1.5, and log(1.5) = 0.176.
- Add the two terms: pH = 4.74 + 0.176 = 4.92.
- Sanity check: there is more base than acid, so the pH should sit just above the pKa of 4.74, and it does.
Answer. pH 4.92, a little above the acid's pKa because the conjugate base is present in excess.
Common mistakes
- Inverting the ratio. The conjugate base goes on top, so more base raises the pH and putting the acid on top moves the answer the wrong way.
- Using K_a directly instead of pKa. The equation needs the logarithmic form, and substituting 1.8 × 10⁻⁵ for pKa gives a meaningless result.
- Converting both concentrations to moles in different volumes. Because only their ratio matters, moles work as well as molarity, but both have to be measured the same way.
- Applying it to a solution of a weak acid on its own. Without the conjugate base there is no buffer, and the pH has to come from a K_a equilibrium calculation instead.
Related formulas
- pH formula:
pH = -log[H₃O⁺] - Titration:
n(titrant) = M(titrant) × V(titrant)