The Henderson-Hasselbalch equation relates the pH of a solution of a weak acid and its salt to the relative concentrations of the acid and its salt. The equation allows one to predict the pH, given a certain ratio of acid to salt, or to calculate the ratio of acid to salt, given the measured pH and the known pKa. Its usefulness in the laboratory is that it allows one to prepare a buffer of the desired pH by combining the acid and its salt in the appropriate ratio.
Consider the case of acetic acid (pKa = 4.76). Substitute this value for pKa into the living equation, and calculate the pH when the salt (say, sodium acetate) and the free acid are combined in the ratios 1/100, 1/10, 1/1, 10/1, and 100/1. When the salt and acid concentrations differ by an order of magnitude, what is the relationship between the pH and the pK? When they differ by two orders of magnitude? When the ratio is 1:1?
These relationships hold regardless of the pKa, as you can confirm by substiting another pKa and recalculating the pH at various ratios of salt to acid.
The pKa of lactic acid is 3.86. In what ratio would you combine sodium lactate and lactic acid to prepare a buffer of pH 4.86?
To use the equation below to solve problems enter any three values and click the Calculate button. The Clear button will remove all values. When entering values you may use "e" as a power-of-ten exponent. For example, 5.8e3 = 5.8 x 103, 5e-7 = 5.0 x 10-7.
Student hints: There is a good way to check many answers when doing problems with the Henderson-Hasselbalch equation. If the concentration of A- is greater than the concentration of HA (more than half of the ionizable groups are unprotonated), then the pH must be above the pKa. Similarly, if the concentration of HA is greater than the concentration of A-, then the pH is below the pKa. Of course, if the concentration of HA and A- are the same, pH = pKa.
Original material from Lehninger Principles of Biochemistry, 5a edición, de D. Nelson and M. Cox; 2009. ISBN: 0-7167-7108-X.
Dr. José Antonio Encinar. (IBMC-UMH)