Science, Tech, Math › Science pH and pKa Relationship: The Henderson-Hasselbalch Equation Definition and Example Share Flipboard Email Print Nicola Tree / Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Anne Marie Helmenstine, Ph.D. Chemistry Expert Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels. our editorial process Facebook Facebook Twitter Twitter Anne Marie Helmenstine, Ph.D. Updated January 30, 2020 The pH is a measure of the concentration of hydrogen ions in an aqueous solution. pKa (acid dissociation constant) and pH are related, but pKa is more specific in that it helps you predict what a molecule will do at a specific pH. Essentially, pKa tells you what the pH needs to be in order for a chemical species to donate or accept a proton. The relationship between pH and pKa is described by the Henderson-Hasselbalch equation. pH, pKa, and Henderson-Hasselbalch Equation The pKa is the pH value at which a chemical species will accept or donate a proton.The lower the pKa, the stronger the acid and the greater the ability to donate a proton in aqueous solution.The Henderson-Hasselbalch equation relates pKa and pH. However, it is only an approximation and should not be used for concentrated solutions or for extremely low pH acids or high pH bases. pH and pKa Once you have pH or pKa values, you know certain things about a solution and how it compares with other solutions: The lower the pH, the higher the concentration of hydrogen ions [H+].The lower the pKa, the stronger the acid and the greater its ability to donate protons.pH depends on the concentration of the solution. This is important because it means a weak acid could actually have a lower pH than a diluted strong acid. For example, concentrated vinegar (acetic acid, which is a weak acid) could have a lower pH than a dilute solution of hydrochloric acid (a strong acid).On the other hand, the pKa value is constant for each type of molecule. It is unaffected by concentration.Even a chemical ordinarily considered a base can have a pKa value because the terms "acids" and "bases" simply refer to whether a species will give up protons (acid) or remove them (base). For example, if you have a base Y with a pKa of 13, it will accept protons and form YH, but when the pH exceeds 13, YH will be deprotonated and become Y. Because Y removes protons at a pH greater than the pH of neutral water (7), it is considered a base. Relating pH and pKa With the Henderson-Hasselbalch Equation If you know either pH or pKa, you can solve for the other value using an approximation called the Henderson-Hasselbalch equation: pH = pKa + log ([conjugate base]/[weak acid])pH = pka+log ([A-]/[HA]) pH is the sum of the pKa value and the log of the concentration of the conjugate base divided by the concentration of the weak acid. At half the equivalence point: pH = pKa It's worth noting sometimes this equation is written for the Ka value rather than pKa, so you should know the relationship: pKa = -logKa Assumptions for the Henderson-Hasselbalch Equation The reason the Henderson-Hasselbalch equation is an approximation is because it takes water chemistry out of the equation. This works when water is the solvent and is present in a very large proportion to the [H+] and acid/conjugate base. You shouldn't try to apply the approximation for concentrated solutions. Use the approximation only when the following conditions are met: −1 < log ([A−]/[HA]) < 1Molarity of buffers should be 100x greater than that of the acid ionization constant Ka.Only use strong acids or strong bases if the pKa values fall between 5 and 9. Example pKa and pH Problem Find [H+] for a solution of 0.225 M NaNO2 and 1.0 M HNO2. The Ka value (from a table) of HNO2 is 5.6 x 10-4. pKa = −log Ka = −log(7.4×10−4) = 3.14 pH = pka + log ([A-]/[HA]) pH = pKa + log([NO2-]/[HNO2]) pH = 3.14 + log(1/0.225) pH = 3.14 + 0.648 = 3.788 [H+] = 10−pH = 10−3.788 = 1.6×10−4 Sources de Levie, Robert. “The Henderson-Hasselbalch Equation: Its History and Limitations.” Journal of Chemical Education, 2003.Hasselbalch, K. A. "Die Berechnung der Wasserstoffzahl des Blutes aus der freien und gebundenen Kohlensäure desselben, und die Sauerstoffbindung des Blutes als Funktion der Wasserstoffzahl." Biochemische Zeitschrift, 1917, pp.112–144.Henderson , Lawrence J. "Concerning the relationship between the strength of acids and their capacity to preserve neutrality." American Journal of Physiology-Legacy Content, vol. 21, no. 2, Feb. 1908, pp. 173–179.Po, Henry N., and N. M. Senozan. “The Henderson-Hasselbalch Equation: Its History and Limitations.” Journal of Chemical Education, vol. 78, no. 11, 2001, p. 1499.