Chapter 16 - Acid-Base Equilibria - Exercises - Page 681: 16.8

Please see the work below.

a. We know that $K_b=\frac{K_w}{K_a}$ $K_b=\frac{1.0\times 10^{-4}}{6.8\times 10^{-14}}=1.47\times 10^{-11}$ b. We know that $K_a=\frac{K_w}{K_b}$ $K_a=\frac{1.0\times 10^{-12}}{4.2\times 10^{-10}}=2.38\times 10^{-5}$