If we let BH stand for ethylenediamine, it has two basic components; however, only the first one needs to be considered.
pKb = -log Kb. You know pKb, convert to Kb. Then
...............BH + HOH ==> BH^+ + OH^-
I............0.25......................0...........0
C.............-x........................x............x
E,,,,,,,,,,,,,,0.25-x................x.............x
Write the Kb expression, plug in the E line and solve for x = PH^-. Convert to pH.
Consider the titration of 10.00 mL of a 0.250 M aqueous solution of ethylenediamine (H2NCH2CH2NH2) with 0.250 M HCl(aq). pKb1 = 4.07 and pKb2 = 7.15.
What is the pH of the ethylenediamine solution prior to the start of the titration?
6 answers
Okay, so I got a pH of 11.7, is that correct?
i'm still confused though. Wouldn't you multiply L of ethylenediamine by its M before plugging that into the ICE table?
It's late and I'll calculate it tomorrow and see if I get that answer. The volume doesn't matter. The concentration is what you want. It's 0.25 M whether it's 1 mL or 1000 mL.
Yes, 11.7 is correct.
Thank you, I still have a few more parts related to this specific problem. Do you want me to add onto this forum or start a new question for each part?
7 answers