6C.7 Textbook Question

[Holiday Parsons 2D]

Hi I attached a picture of question 6C.7 from the textbook. I'm just wondering why we were able to apply the pKb value for NH2OH and (CH3)2NH on the table (6C.2, attached that also) for the bases that the problem gave us: +NH3OH and (CH3)2NH2+. I notice that the problem gave us the versions with an additional proton. So if those are the conjugate acids, are their pKb values equal to its corresponding base?

[Arnav Ranade 2K]

Yes. The last two are conjugate acids of weak bases, so to find their Ka you have to divide Kw by the Kb of their conjugate weak bases, which are given in the table.