Textbook 6D.5

[Jiayi_Cao_3E]

Hello!

I tried to calculate for the answer multiple times but I can't seem to obtain pOH = 3.00, pH = 11.00, and % protonation = 1.8%
The question is as follows:
Calculate the pH, pOH, and percentage protonation of solute in each of the following aqueous solutions: (a) 0.057 M NH3 given that the pKa of its conjugate acid is 8.21.

The way I did it was I calculated for
pKb = 14 - pKa = 5.79.
Kb = 10^-5.79 = 1.62 x 10^-6
The expression for Kb = [OH-][NH4+]/[NH3] = x^2/(0.057 - x). the -x is ignored since 0.057 is more than 10000 larger than Kb.
x = 3.05 x 10^-4
Since x = [OH], pOH = -log(3.04 x 10^-4) = 3.517, but this is different from the answer pOH = 3.00

Anything helps!

[Hayley Vu]

For this question, you want to refer to the Basicity Constants table to get the Kb of NH3. The 8.21 pKa value only applies to part d of this problem. From the table, we know that the Kb of ammonia (NH3) is 1.8*10^-5. You can do the usual calculations from there. Hope this helps!

[Jiayi_Cao_3E]

Oh that makes so much more sense. Thank you!

[Tony Chen 1F]

I didn't know we had to use the table. I hope the textbook could be more clear on this :(