Henderson-Hasselbalch Equation

Add to the pK_a of the acid the base-10 logarithm of the concentration ratio of conjugate base to acid: $pH = pK_{a} + \log{}(\frac{[A^{-}]}{[HA]})$. Insert the concentrations in $\frac{mol}{L}$; only their ratio matters, not their absolute value. Example acetic acid buffer with $pK_{a} = 4.75$, $[CH_{3}COO^{-}] =\frac{ 0.1 mol}{L}$ and $[CH_{3}COOH] =\frac{ 0.05 mol}{L}$: $pH = 4.75 + \log{}(\frac{0.1}{0.05}) = 4.75 + 0.30 = 5.05$. In the logarithm it is always base divided by acid, not the other way round. If both concentrations are equal, the logarithm is zero and the pH equals the pK_a.

Because then the concentration ratio $\frac{[A^{-}]}{[HA]}$ is exactly one and the logarithm of one gives zero. In $pH = pK_{a} + \log{}(\frac{[A^{-}]}{[HA]})$ the second term drops out, so $pH = pK_{a}$ remains. This point is called the half-equivalence point and is especially important: there the buffering effect is greatest, because small additions of acid or base barely shift the pH. For this reason the pK_a of an acid is determined experimentally by reading the pH at the half-equivalence point of a titration. A good buffer works within about one pH unit around its pK_a.

The equation is an approximation and works well near $pH \approx{} pK_{a}$, that is in the actual buffer range, about one pH unit around the pK_a. It fails for very dilute solutions and at the edges, when one component is almost completely consumed. For strong acids or bases that dissociate completely it is also not applicable, because there is no equilibrium between acid and conjugate base. The self-ionization of water and the use of activity instead of concentration are also neglected. For standard buffer problems in chemistry class, however, it gives very reliable results.

Because the logarithm contains the ratio $\frac{[A^{-}]}{[HA]}$, a factor of ten shifts the pH up by exactly one unit, since $\log{}(10) = 1$, so $pH = pK_{a} + 1$. If the ratio rises by a factor of 100, the pH rises by two units. Conversely a tenfold excess of acid lowers the pH by one unit below the pK_a. This logarithmic relationship means that large concentration changes cause only moderate pH changes, which explains the buffering effect. A buffer keeps the pH stable as long as both components are present in comparable amounts and neither dominates by more than a factor of ten.

Blood uses mainly the carbonic acid-bicarbonate system as a buffer. Here HA is carbonic acid H₂CO₃ and A⁻ is bicarbonate HCO₃⁻, with an effective pK_a of about 6.1. The physiological pH of 7.4 is reached because the ratio of bicarbonate to dissolved CO₂ is roughly 20 to 1: $pH = 6.1 + \log{}(20) \approx{} 6.1 + 1.3 = 7.4$. The body regulates this buffer dynamically: breathing controls the CO₂ content, the kidneys control the bicarbonate. As a result the blood pH stays within a very narrow, vital range despite metabolic fluctuations.