Predicting Acid–Base Reactions from pKa Values

Compilations of pK_a values like those in Table 2.3 and Appendix B are useful for predicting whether a given acid–base reaction will take place, because H⁺ will always go from the stronger acid to the stronger base. That is, an acid will donate a proton to the conjugate base of a weaker acid, and the conjugate base of a weaker acid will remove a proton from a stronger acid. Since water (pK_a = 15.74) is a weaker acid than acetic acid (pK_a = 4.76), for example, hydroxide ion holds a proton more tightly than acetate ion does. Hydroxide ion will therefore react to a large extent with acetic acid, CH₃CO₂H, to yield acetate ion and H₂O.

Another way to predict acid–base reactivity is to remember that the product conjugate acid in an acid–base reaction must be weaker and less reactive than the starting acid and the product conjugate base must be weaker and less reactive than the starting base. In the reaction of acetic acid with hydroxide ion, for example, the product conjugate acid (H₂O) is weaker than the starting acid (CH₃CO₂H), and the product conjugate base (CH₃CO₂^–) is weaker than the starting base (OH^–).

Worked Example 2.4

Predicting Acid Strengths from pK_a Values

Water has pK_a = 15.74, and acetylene has pK_a = 25. Which is the stronger acid? Does hydroxide ion react to a significant extent with acetylene?

Strategy

In comparing two acids, the one with the lower pK_a is stronger. Thus, water is a stronger acid than acetylene and gives up H⁺ more easily.

Solution

Because water is a stronger acid and gives up H⁺ more easily than acetylene, the HO^– ion must have less affinity for H⁺ than the HC≡C:−HC≡C:− ion. In other words, the anion of acetylene is a stronger base than hydroxide ion, and the reaction will not proceed significantly as written.

Worked Example 2.5

Calculating K_a from pK_a

According to the data in Table 2.3, acetic acid has pK_a = 4.76. What is its K_a?

Strategy

Since pK_a is the negative logarithm of K_a, it’s necessary to use a calculator with an ANTILOG or INV LOG function. Enter the value of the pK_a (4.76), change the sign (–4.76), and then find the antilog (1.74 × 10^–5).

Solution

K_a = 1.74 × 10^–5.