Can strong acids and salts be used to create a pH 0.5 buffer?

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SUMMARY

The discussion centers on creating a pH 0.5 buffer using strong acids and salts, specifically hydrochloric acid (HCl) and potassium chloride (KCl). The participant initially struggled with the concept due to the lack of available Ka values for strong acids and the confusion surrounding their use in buffer solutions. However, they concluded that while strong acids do not ionize completely, they can still contribute to buffer capacity when combined with salts. The participant successfully solved their assignment after gaining a clearer understanding of buffer systems.

PREREQUISITES
  • Understanding of buffer solutions and their capacity
  • Familiarity with acid-base equilibrium and the Henderson-Hasselbalch equation
  • Knowledge of strong acids, specifically hydrochloric acid (HCl)
  • Basic concepts of ionic compounds, particularly potassium chloride (KCl)
NEXT STEPS
  • Research the role of strong acids in buffer solutions
  • Explore the Henderson-Hasselbalch equation in detail
  • Investigate buffer capacity and its practical applications in laboratory settings
  • Learn about the ionization of strong acids and their implications for pH stability
USEFUL FOR

Chemistry students, laboratory technicians, and educators looking to deepen their understanding of buffer systems and the application of strong acids and salts in pH control.

RCN
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Homework Statement


The assignment is to create a pH buffer in lab.
My assignment is pH 0.5.


Homework 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.

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.
 
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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?
 
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!
 

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