# Creating buffer of pH 0.5

1. Feb 26, 2012

### RCN

1. The problem statement, all variables and given/known data
The assignment is to create a pH buffer in lab.
My assignment is pH 0.5.

2. Relevant equations
[H+] = (Ka)([HA]/[A-])
log [HA] = pKa - pH + log [A-]
The second equation was provided by the instructor, the first is from the textbook.

3. The attempt at a solution

This seems to apply easily for most pH values, such as pH 5.0, where weak acids are available with useful Ka values.

Example:
Make 100mL pH 5.0 buffer
Acetic Acid, HAc and Ac- with Ka 1.8 E -5
Say we have .1M HAc and .1M Ac-

[HAc]/[Ac-] = [H+]/Ka
= (10-5.0)/(1.8E-5)
= .556

If x = volume of HAc added,
.556 = (moles HAc)/(moles Ac-)
= ( (.1M)(x) )/( (.1M)(.100-x) )
= 36 mL HAc and 64 mL Ac-

The problem here is:
Weak acids don't have Ka values within a useful range, obviously.
A list of suggestions from the instructor lists STRONG acid HCl and SALT KCl. I have tried extensively to research this online and have found these listed together as buffer components but never with any calculations or explanation of how to use them or how they can work instead of weak acid-conjugate base (or vice-versa) combinations. I realize that techinically they do not completely ionize even though we often treat them like they do, but I also cannot find consistent values for Ka-- every value I find is very different, between 1000 and 10,000,000. For that matter, with these high Ka values, wouldn't it take an enormous ratio of salt to acid to create a usable buffer? I am certainly stumped (and exhausted) and would appreciate whatever help you can offer.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 27, 2012

### Staff: Mentor

Please read about a buffer capacity. The buffer is a solution that resist pH changes. Check the plot. Do you see how solutions of just a strong acid (or a strong base) are keeping pH constant?

3. Feb 27, 2012

### RCN

Thanks for the link, the concept makes quite a bit more sense to me now.
Also, I am now in good shape and have been able to solve the problem.

Thanks again!