6C.13 Comparing strength of bases

[Layla Qumsieh 3C]

6C.13 Arrange the following bases in order of increasing strength on the basis of the pKa
values of their conjugate acids, which are given in parentheses: (a) ammonia (9.26); (b) methylamine (10.56); (c) ethylamine (10.81); (d) aniline (4.63; see Exercise 6C.12). Is there a simple pattern of strengths?

The answer includes arylamines. Where is that coming from? I found each pKb by doing 14 minus each given pKa, and ranked them by highest pKb to lowest pKb. Is that the correct way to find the answer?

[Lynette Andreasyan 33H]

When trying to arrange bases in order of increasing strength you need either Kb or pKb. The greater the Kb, the stronger the base. Therefore, the smaller the pKb, the stronger the base. This is because Kb=10^-pKb. Because of the negative sign ,the greater the value of pKb, the smaller the overall value of Kb. So when given the pka, use the formula pKa+pKb=14 to find to find pKb. Then rank the pKb values you get from greatest to weakest. This would place the bases in order of smallest to greatest strength (increasing strength).

[Dashrit_Pandher_1J]

The question asks you to rank the following bases based on pKa. Theres not one single way to solve this, so the way you did is absolutely a valid way to solve the problem.

pKa and pKb are part of a conjugate seesaw, so as pKa gets larger, pKb gets smaller. The smaller pKa is, the stronger the acid is.

Since it's asking you to rank in order of increasing strength of the conjugate acids, a strong conjugate acid would be a weak base. A weak base would have a large pKb value, and thus a small pKa value.

Now knowing this, the smaller the pKa the weaker the base. That means it goes: aniline < ammonia < methylamine < ethylamine