Henderson-Hasselbalch Equation Calculator

What the Equation Says

A buffer is a weak acid mixed with its conjugate base, or a weak base mixed with its conjugate acid. The Henderson-Hasselbalch equation rewrites the acid's $K_a$ expression in log form, so the pH of the mixture takes one line to find:

$$\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$$

In this equation, $\mathrm{p}K_a = -\log K_a$, $[\mathrm{A^-}]$ is the concentration of the conjugate base, and $[\mathrm{HA}]$ is the concentration of the weak acid. Because the acid and base are in the same solution, the volume cancels in the ratio, so you can plug in moles instead of molarities and get the same pH.

The log term is the part students mix up. The base goes on top. If you put the acid on top, the sign of the log flips, and the answer comes out on the wrong side of the pKa.

Worked Example: An Acetate Buffer

OpenStax works through a buffer made of 0.10 M acetic acid and 0.10 M sodium acetate, with $K_a = 1.8 \times 10^{-5}$ for acetic acid.

1. Convert to pKa: $-\log(1.8 \times 10^{-5}) = 4.74$.
2. Form the ratio: $0.10 / 0.10 = 1$, and $\log 1 = 0$.
3. Add: $\mathrm{pH} = 4.74 + 0 = 4.74$.

The same source then adds 1.0 mL of 0.10 M NaOH to 100 mL of this buffer. The hydroxide converts a little acetic acid into acetate, leaving 0.098 M acid and 0.100 M base, and the pH moves only to 4.75. Adding that same NaOH to 100 mL of an unbuffered solution at pH 4.74 sends the pH to 10.99. That gap of more than six pH units is what a buffer is for.

Reading the Ratio Without a Calculator

Each factor of 10 in the ratio moves the pH by one unit:

For ratios between those, remember that $\log 2 = 0.30$ and $\log 5 = 0.70$. A 2:1 base-to-acid buffer has a pH 0.30 above its pKa, and a 1:2 buffer has a pH 0.30 below it. These shortcuts cover most of the buffer math you will see without a calculator. The MCAT math without a calculator guide has more log tricks.

When the Equation Stops Working

• Strong acids and bases. They dissociate completely, so there is no equilibrium to describe. Use $\mathrm{pH} = -\log[\mathrm{H^+}]$ directly.
• A lopsided ratio. OpenStax notes that a buffer stops being useful once one component drops below about 10% of the other. The calculator flags ratios outside 0.1 to 10.
• Very dilute solutions. The equation assumes the starting concentrations barely change when the acid dissociates (the "x is small" approximation). When the acid and base are very dilute, solve the equilibrium with an ICE table instead.

How It Shows Up on the MCAT^®

The MCAT^® rarely asks for a bare buffer pH. It asks what form a molecule takes at a given pH. A side chain with a pKa near 4 is mostly deprotonated at pH 7.4, because the pH is more than three units above the pKa. That one comparison tells you the charge of an amino acid, which drives questions on protein folding, electrophoresis and isoelectric points. The Amino Acids reference lists the side-chain pKa values to practice with, and the ratio mode above turns any pH and pKa into a percent deprotonated.

Blood is the other classic setting. The carbonic acid and bicarbonate pair buffers plasma, so a passage about breathing that raises carbonic acid is asking you to see the base-to-acid ratio fall and the pH drop with it.