Which of the followings primarily exist as positively charged at pH7 (pKa for NH2 is ~9 and for imidazole ring ~6, be careful with your calculation using pH=pKa+log(A/HA+):

We have to realize that both of these substances are weak bases, and they should be treated as such. This is the reason they give us the hint to be careful using the formula pH=pKa+log(A/HA+) which aplies to weak acids.

Instead, we are going to use a formula which looks similar but applies to weak bases:

B + H₂O ⇄ BH⁺ + OH⁻ Kb = [ BH⁺ ] [OH⁻ ]/ B

pOH = pKb + log [BH⁺]/[B]

By determining the ratio of conjugate acid concentration to weak base concentration we can answer this problem, after some data manipulation.

pH + pOH = 14 ⇒ pOH = 14 - 7 = 7

pKw = pKa + pKb where pKw is 14 from Kw =10⁻¹⁴

For NH₂:

pKb = 14 - pKa = 14 - 9 = 5

So at pOH = 7

7 = 5 + log [NH₂⁺]/[NH₂] ⇒ 7 - 5 = log [NH₂⁺]/[NH₂]

taking inverse log to both sides of the equation:

10² = [NH₂⁺]/[NH₂]

The ratio is 100 so the equiulibrium will favor primarily the product side, that is the protonated base.

For imidazole ring:

pkb = 14 - pka = 14 - 6 = 8

7 = 8 + log [ protonated Imidazole]/[Imidazole]

7 - 8 =log([ protonated Imidazole]/[Imidazole])

Taking inverse log to both sides of the equation:

10⁻¹ = [ protonated Imidazole]/[Imidazole]

In this case the equilibrium favors the left side sine K=0.10. Hence we wil find primarily more unprotanated imidazole.

A note of caution is that we are given the pKa as approximated, therefore the pkb are also approximations.