David513

5+ Year Member
Dec 25, 2010
105
35
Status
Pre-Medical
Hello all,

I have no idea if this is an appropriate post (and if it is whether it's in the right area), but I was hoping to get some help with a biochemistry question! Thank you in advance for your time.

A compound is known to have a free amino group with a pKa of 8.8, and one other ionizable group with a pKa between 5 and 7. To 100 mL of a 0.2 M solution of this compound at pH 8.2 was added 40 mL of a solution of 0.2 M hydrochloric acid. The pH changed to 6.2. The pKa of the second ionizable group is...?

I know the answer involves use of the Henderson-Hasselbalch equation and probably involves calculating the change in the moles of acid and base, but the second ionizable group thing is throwing me off. I appreciate the assistance ! Here is my work so far...

1)
1a) 0.1L*0.2mol/L => 0.02mol unknown compound
1b) 0.04L*0.2mol/L => 0.008mol HCl

2)
pH=pKa+log(base/acid)
8.2=8.8+log(base/acid)
10^-0.6=base/acid
0.25=base/acid
Therefore, there are 0.004mol of the base and 0.016mol of the acid ORIGINALLY.

3)
Change is from 1b) above, i.e. 0.008 mol acid added, 0.008 mol base subtracted, but there is a limiting reagent. The final amounts of base and acid are 0.000 mol and 0.020 mol, respectively.

4)
6.2=pKa+log(0/0.020)
6.2=pKa+0
6.2=pKa

I know this is wrong so please tell me how I can fix my work! Also, if you would care to explain in words the concepts behind what this question is asking, I would appreciate it, but it's not necessary. Thanks again!