Why Should the Mole Ratio in Alkaline Buffers Be 5:1 Base to Acid?

  • Context: Chemistry 
  • Thread starter Thread starter stunner5000pt
  • Start date Start date
  • Tags Tags
    Buffer Chemistry
Click For Summary
SUMMARY

The mole ratio in alkaline buffers should be 5:1 base to acid, as established through stoichiometric principles. The discussion clarifies that using sulfuric acid, which is diprotic, necessitates a focus on the bisulfate ion rather than sulfuric acid itself. The correct application of the Henderson-Hasselbalch equation confirms that the concentration of the base must exceed that of the acid significantly to maintain the buffer's effectiveness. Participants emphasized the importance of understanding the stoichiometry involved in buffer solutions.

PREREQUISITES
  • Understanding of buffer solutions and their components
  • Knowledge of the Henderson-Hasselbalch equation
  • Familiarity with diprotic acids, specifically sulfuric acid
  • Basic principles of stoichiometry in chemical reactions
NEXT STEPS
  • Study the Henderson-Hasselbalch equation in detail
  • Explore the properties and behavior of diprotic acids
  • Research the stoichiometry of buffer solutions
  • Learn about the role of bisulfate in acid-base reactions
USEFUL FOR

Chemistry students, laboratory technicians, and anyone involved in preparing or studying alkaline buffer solutions will benefit from this discussion.

stunner5000pt
Messages
1,447
Reaction score
5
Homework Statement
Which combination will produce an alkaline buffer

a. 0.1 mol ammonia & 0.05 mol sulfuric acid
b. 0.5 mol ammonia & 0.10 mol sulfuric acid
c. 0.1 mol acetic acid & 0.05 mol NaOH
d. 0.1 mol acetic acid & 0.50 mol NaOH
Relevant Equations
Henderson Hasselbach equation
THe answer is supposed to be (B) but I'm not understanding properly

I initially chose A - to form an alkaline buffer, the # mol of acid should be half of this. But this doesn't seem to be the case

I don't understand why the mol of base in this case should be 5x the mole of acid

your help is greatly appreciated, as always!
 
Physics news on Phys.org
Hint: sulfuric acid is diprotic.
 
  • Like
Likes   Reactions: DeBangis21
Ok that does help - so we should be looking at bisulfate rather than sulfuric acid
but wouldn't the concentration of bisulfate be the same i.e. the second part of the henderson hasselbach would still be

for a. log (0.05/0.1) = log(0.5)

and for
b. log(0.1/0.5) = log(0.2)

I feel like i'm missing something obvious ....
 
stunner5000pt said:
a. 0.1 mol ammonia & 0.05 mol sulfuric acid

stunner5000pt said:
bisulfate

Nope, not bisulfate. Just look at the reaction stoichiometry and amounts of substances involved.
 
  • Like
Likes   Reactions: DeBangis21

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
9K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
12K