NovisHenderson-Hasselbalch calculator
Find a buffer's pH or pKa, the base-to-acid ratio you need for a target pH, how much of a group is ionized, or the new pH after adding HCl or NaOH, with the stoichiometry written out. Every step is shown, the way you would write it on an exam.
How the Henderson-Hasselbalch equation works
pH = pKa + log10( [A-] / [HA] )
The pKa is the pH where the acid is exactly half given up: [A-] and [HA] are equal, the ratio is 1, and log10(1) is 0, so pH equals pKa. Every whole pH unit above the pKa means ten times more base form than acid form. Every unit below means ten times more acid form.
That is why a buffer works best within about one pH unit of its pKa: there is enough of both forms to soak up added acid or base.
Worked example: adding HCl to a buffer
100 mL of a buffer is 0.10 M acetic acid and 0.10 M acetate (pKa 4.76). You add 0.0020 mol of HCl. What is the new pH?
- Work in amounts: 0.10 M x 0.100 L = 0.010 mol of HA and 0.010 mol of A-.
- The H+ reacts completely with the base form: H+ + A- -> HA.
- A- after: 0.010 - 0.0020 = 0.0080 mol. HA after: 0.010 + 0.0020 = 0.012 mol.
- New ratio: 0.0080 / 0.012 = 0.667. The volume cancels, so amounts are enough.
- New pH: 4.76 + log10(0.667) = 4.76 - 0.176 = 4.58.
The pH fell by only 0.18. The same HCl in 100 mL of plain water would give 0.020 M H+ and a pH of 1.70. That difference is the whole point of a buffer.
When the equation stops working
If the acid or base you add is more than the buffer form it reacts with, one form is gone and the log of zero or of a negative number does not exist. The pH then comes from a concentration, so the calculator asks for the total volume after mixing:
- More than the buffer can take: the leftover strong acid or base sets the pH. For example, 0.015 mol HCl into the buffer above leaves 0.005 mol of H+ in 0.100 L, so [H+] = 0.050 M and the pH is 1.30.
- Exactly enough to use one form up: only the weak acid (or only its conjugate base) is left, so the pH comes from its own equilibrium, solved with the quadratic.
Mistakes that cost points
Practice problems
An acetic acid buffer (pKa 4.76) has 0.10 M acetic acid and 0.10 M acetate. What is the pH?
Same buffer, but with 0.20 M acetate and 0.10 M acetic acid. What is the pH?
A phosphate buffer has pKa 7.21. What ratio of HPO4 2- to H2PO4 - gives pH 7.40?
Blood bicarbonate: pKa 6.1, blood pH 7.4. What is the ratio of HCO3- to CO2?
A carboxylic acid group has pKa 4.0. What fraction is deprotonated at pH 7.4?
A lysine side chain has pKa 10.54. What fraction is protonated at pH 7.4?
100 mL of 0.10 M HA and 0.10 M A- (pKa 4.76) gets 0.0020 mol HCl. What is the new pH?
The same buffer gets 0.0020 mol NaOH instead. What is the new pH?
Practice it until it sticks
Novis turns problems like these into a spaced-repetition deck, checks your own written work one step at a time, and tells you which step went wrong. Novis opens October 20.