The equilibrium constant expression for the ionization of HCN is as follows: The corresponding expression for the reaction of cyanide with water is as follows: If we add two of the Equations above, we obtain the following (recall that the equilibrium constant for the sum of two reactions is the product of the equilibrium constants for the individual reactions): In this case, the sum of the reactions described by Ka and Kb is the equation for the autoionization of water, and the product of the two equilibrium constants is Kw: Thus if we know either Ka for an acid or Kb for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair. Just as with pH, pOH, and pKw, we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, by defining pKa as follows: and pKb as
(BH+), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH. The Common Ion Effect To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of H+). Le Châtelier’s principle can be used to predict the effect on the equilibrium position of the solution. A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. Recall that the dissociation reaction of acetic acid is as follows: and the equilibrium constant expression is as follows: Sodium acetate (CH3CO2Na) is a strong electrolyte that ionizes completely in aqueous solution to produce Na+ and CH3CO2− ions. If sodium acetate is added to a solution of acetic acid, Le Châtelier’s principle predicts that the equilibrium in the second to last Equation above will shift to the left, consuming some of the added CH3CO2− and some of the H+ ions originally present in solution: Because Na+ is a spectator ion, it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which [H+] is less than the initial value. Because [H+] has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding any base to a solution of a weak acid should increase the pH.
The two Equations are both forms of the Henderson-Hasselbalch equationA rearranged version of the equilibrium constant expression that provides a direct way to calculate the pH of a buffer solution: pH = pK a + log([base]/[acid])., named after the two early-20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch equation may be limited to solutions whose concentrations are at least 100 times greater than their Ka values. There are three special cases where the Henderson-Hasselbalch equation is easily interpreted without the need for calculations: [base] = [acid]. these conditions, [base]/[acid] = 1 in the Equation above. Because log 1 = 0, pH = pKa, regardless of the actual concentrations of the acid and base. Recall that this corresponds to the midpoint in the titration of a weak acid or a weak base. [base]/[acid] = 10. In the Equation above, because log 10 = 1, pH = pKa + 1. [base]/[acid] = 100. In the Equation above, because log 100 = 2, pH = pKa + 2. Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = pKa − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit. Note the Pattern If [base] = [acid] for a buffer, then pH = pKa. Changing this ratio by a factor of 10 either way changes the pH by ±1 unit. The Henderson-Hasselbalch equation can also be used to calculate the pH of a buffer solution adding a given amount of strong acid or strong base. Note the Pattern Only the amounts (in or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch equation, not their concentrations.