This isn't really a homework problem. In lab we had an unknown buffer solution and we had to to titrate it with NaOH and HCl to try to identify what the buffer is. In the end I got my pka value equal to 4.52 I can't identify the buffer. My Ka is 3E-5 and I can't match any Ka values to this. Did I just totally screw up my lab??
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 Admin Is there a list of buffers that you have to select from? Or can it be anything?
 Nope, we weren't given a list to choose from so I'm guessing it could be anything?

Probably. Although some compounds are much more likely than others.

How did you get your pKa?
 I required 19.92 mL of NaOH to neutralize a 10 mL portion of my buffer. I did the same thing using another 10 mL portion of the buffer but this time with HCl and it took me 17 mL. I then used the henderson-hasselbalch equation to find my pKa. I measured my pH of my buffer with a pH meter and got it to be 4.40. I had already standardized my acid and base previously and got 0.1472 M NaOH and 0.1325 M HCl. Also I made sure to change the ml to L for the following calculations. For the concentration of my base and acid I did this: [A] = (19.92 ml NaOH x 0.1472 M NaOH)/10 mL = 0.293 [B] = (17ml HCl x 0.1325 M HCl)/10 ml =0.225 4.40= pka + log [0.225]/[0.293] pka= 4.52
 Posted something before thinking it was right... still wrong.

 Quote by kooombaya I required 19.92 mL of NaOH to neutralize a 10 mL portion of my buffer. I did the same thing using another 10 mL portion of the buffer but this time with HCl and it took me 17 mL.
What do you mean by "neutralize the buffer"?

 Quote by Borek What do you mean by "neutralize the buffer"?
This was how the question was stated in the lab book.
I found out what I did wrong by the way. Thanks for your time.

 Quote by kooombaya This was how the question was stated in the lab book.
Can you explain what they meant? I have never seen something like that, even if it is wrong, I have nothing against knowing.

 Quote by Borek Can you explain what they meant? I have never seen something like that, even if it is wrong, I have nothing against knowing.
From what I understood it goes something like this:
Say we added 10 mL of 0.1 M NaOH. Then this is the amount of acid in the buffer solution that reacted with the NaOH to reach a new equivalence point.
It's the same for HCl, except the HCl acts with the base in the buffer solution to reach a new equivalence point.
 Admin So it was just shifting pH of the buffer by addition of strong acid or base. There was an acid base reaction involved (which can be technically called neutralization), but it didn't end with neutral solution. Not the best wording if you ask me. -- buffer calculator, concentration calculator pH calculator, stoichiometry calculator

 Quote by Borek So it was just shifting pH of the buffer by addition of strong acid or base. There was an acid base reaction involved (which can be technically called neutralization), but it didn't end with neutral solution. Not the best wording if you ask me.
Yup exactly. Thanks again.