Question

$\underset{1}{C{H}_{3}OH}+\underset{2}{(C{H}_3{)}_{2}NH}\rightleftharpoons C{H}_3-O^{-}+C{H}_3-\underset{H}{\underset{|}{\stackrel{+}{NH}}}-C{H}_3$
Identify each species in the following equilibrium according to the code:
SA = stronger acid, SB = stronger base, WA = weaker acid, WB = weaker base.
The $p{K}_a$ of $(C{H}_3{)}_{2}NH$ is 36; the $p{K}_a$ of $p{K}_a$ of $C{H}_{3}OH$ is 15.2.

• A. 1 - WA, 2 - WB
• B. 1 - WB, 2 - WA
• C. 1 - SA, 2 - SB
• D. 1 - SB, 2 - SA
• E. 1 - WA, 2 - WA

Answer 1

The correct option is A 1 - WA, 2 - WB

Here, $C{H}_{3}OH$ donates proton and hence acts as an acid while $(C{H}_3{)}_{2}NH$ is acting as a proton acceptor i.e. it is a base. Judging by their $p{K}_a$ values we can say that Option A is correct.