The Henderson-Hasselbalch equation relates the concentrations of a protonated form of a weak acid (base) to the deprotonated form. It can be used to easily calculate (predict) the pH of a buffer where we have substantial concentrations of both the protonated and deprotonated forms of the compound. For simplicity, we'll write it here for the acid, but it could just as easily be written for the base.

By taking the log of the equilibrium expression we find

\[\rm{pH = pK_a + \log\left({[A^-] \over [HA]}\right)}\]

where the pKa = -log(K_{a}). Thus we can see that in a situation in which we have equal amounts of the protonated and deprotonated forms of a conjugate pair, the pH = pKa. Alternatively, you could say that at a pH that is equal to the pKa, these concentrations are equal. This means that for any compound, you can determine if the majority will be in the protonated or deprotonated form at a given pH by comparing the pH of the solution to the pKa. If the pH is less than the pKa, the solution is "too acidic" and the majority of the compound will be in the protonated (acidic) form. If the pH is greater than the pKa, the majority will be in the deprotonated (basic) form. The exact ratio can be obtained from the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation can also be approximated to determine the pH of a buffer from the initial concentrations that I use to make the solution. If one makes a solution with both the protonated and deprotonated forms of a compound, the differences in concentration at equilibrium compared to the initial concentrations will be very small. Thus one can approximate the concentrations at equilibrium as the initial concentrations, in this case you can predict the pH is

\[\rm{pH \approx pK_a + \log\left({[A^-]_0 \over [HA]_0}\right)}\]

where the concentrations of [HA]_{0} and [A^{-}]_{0} are the initial concentrations which we have placed into the equilibrium expression (assuming the change is small). Sometimes we show the initial or mixed concentrations as \(C_{\rm HA}\) and \(C_{\rm A-}\) which would give the following H-H equation:

\[\rm{pH \approx pK_a + \log\left({C_{\rm A-} \over C_{\rm HA}}\right)}\]

The same idea can be written out for K_{b} and OH^{-} with the same approximations.

\[\rm{pOH \approx pK_b + \log\left({[BH^+]_0 \over [B]_0}\right)}\]

Calculating the pH of an acidic buffer

Calculation of the effect of adding a strong base to a buffered solution

How to prepare a buffer with a particular pH

© 2013 mccord/vandenbout/labrake