Textbook Problem Focus 6D.5

[Meghna Sehrawat 3A]

Hi! I was confused about how the textbook got the Kb for 6.D.5

The problem states that the we are "given that the pKa of its [the weak base's] conjugate acid is 8.2". So, when I was trying to find the Kb of the weak base, I used pKa + pKb = 14, to get pKb = 5.79. Then, I found the Kb by doing 10^-5.79 = 1.62 * 10^-6 = Kb. However, the book is saying that Kb = 1.8 * 10^-5. I was confused what I was doing wrong to get Kb from the pKa.

[Charlie 2H]

The problems say to use the tables in 6C to get the Kb or Ka and in this situation the Kb of NH3 is 1.8x10^-5. You then plug this value into the Kb equation to get the pKb and then subtract the pKb from 14 to get pKa.