Lecture 6 summary

In today's lecture, we continue our treatment of acid-base chemistry within the Brønsted-Lowry framework. After finishing with definitions and relationships that apply among them in aqueous systems, we address two general types of calculations: those involving strong acids or strong bases, and those used to treat weak acids and weak bases.

Since we begin the laboratory part of the course this week with Experiment 1, in which we investigate the properties of buffers, we devoted a few minutes at the outset to talk a bit about buffers. See the next lecture for a fuller discussion of buffers, including a derivation of the Henderson-Hasselbalch equation.

pKa Table: I will hopefully post a pKa table later with a dozen or so of the most important pKa values for this class. For now, refer to the table below which is for CHEM230. See also the Calbiochem booklet on Buffers.

Definition of K_b and pK_b

K_b is known as the base hydrolysis constant, or simply "base constant". It is defined by the chemical equation for abstraction by a base of a proton from water, called the base hydrolysis equation.

Relationship between K_a, K_b, and K_w

If we combine the general equations defining K_a and K_b, and use the rule for combining equilibrium constants when reactions are summed (the equilibrium constants are multiplied), we obtain a simple and useful relationship between K_a, K_b, and K_w:

HA  =  H⁺  +  A^–       K_a  =  [H⁺][A^– ] / [HA].

A^–  +  H₂O  =  HA  +  OH ^– .      K_b  =  [HA][OH ^–] / [A^–].

sum: H₂O  =  H⁺  +  OH ^–      K_eq  =  K_a· K_b  =  [H⁺][OH^– ]  =  K_w.

We have derived the relation  K_w  =  K_a· K_b, which hold for aqueous solutions of acids and bases. The logarithmic form of the relation is  pK_w  =  pK_a +  pK_b. When the temperature is 25°C, K_w. = 1.0 x 10^–14, so that

K_a· K_b  = 1.0 x 10^–14    (at 25°C), and

pK_a +  pK_b. = 14.00    (at 25°C).

Strong acid and strong base calculations

A typical problem is to calculate the pH of a solution of a strong acid or a strong base of a given (formal) concentration. This type of problem (see the example on p.161 of Harris) is relatively simple and straightforward, since we can assume that any compound that is a strong acid or strong base in water is completely dissociated.

Weak acid and weak base calculations

This situation is more complicated than the strong acid / strong base one because the formal concentration of the weak acidic or weak basic species does not tell us directly the resulting pH. For a weak acid HA, there is an equilibrium established between HA and its conjugate base A^–, and the extent of dissociation depends on both the formal concentration and the value of K_a. A typical problem would be to calculate the pH of a solution of a weak acid or a weak base of a given (formal) concentration and a K_a or a K_b value. We also need to be able to calculate the fraction of dissociation, denoted as a, and defined as

a  =  [A^–] / { [A^–] + [HA] }

for a weak acid HA. For a base, we would refer to a similar quantity as fraction of association or fraction hydrolyzed

a  =  [BH⁺] / { [BH⁺] + [B] }

for the example of a neutral base B in water, some of which is converted to its conjugate acid BH⁺ according to the base hydrolysis equation,

B  +  H₂O  = BH⁺  +  OH ^– .

We are also asked in some cases to calculate a K_a or a K_b value given a formal concentration and a measured pH or fraction of dissociation. Examples of all these types of problems can be found in the examples in the text, the example problems for today's class, and the other problems in the list for today's lecture selected from the end of Ch.8 of Harris.