[danika.frederick]

How many mLs of 12 M HCl must be added to 100 mLs of 0.1 M tris buffer (pKa = 8.1) at pH 8.8 to bring it to a pH of 6.8?

I understand you must use the Henderson-hasselbalch equation. I think I have the first part of the question right but I'm really not sure, and I'm totally lost. Here is what I have so far but I have no idea if it's correct or not:

100 mL of 1 M tris = 0.1 mols, therefore the total amount of conjugate base and acid will be 0.1 mols (I think...)

Using H-H, 8.8 = 8.1 + log [0.1 - x]/

• . After solving this and using antilog etc. I have determined x = 0.017, and therefore mols HCl = 0.017 and mols CB = 0.983 (0.1-0.017).

It is from this point (assuming that all of the above was correct) that I'm not sure what to do. If someone can please EXPLAIN to me (rather than give me the answer) what to do next, I would really appreciate it.

Edit: Also, apparently the answer is 6.55 mLs of HCL

[Borek]

See these examples, they should help:

http://www.chembuddy.com/?left=buffers&right=pH-change