Henderson-Hasselbalch buffer calculations - B + A Question

Click For Summary

Discussion Overview

The discussion revolves around the preparation of a buffer solution using Tris, HCl, NaCl, and sucrose, focusing on the calculations involved in determining the amounts of each component based on the Henderson-Hasselbalch equation. Participants explore the roles of acids and bases in the buffer system and the implications for stoichiometry in the calculations.

Discussion Character

  • Homework-related
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how to determine the amount of Tris base to add to the buffer, questioning the appropriate method to use for calculations involving the Henderson-Hasselbalch equation.
  • Another participant suggests that the confusion stems from misunderstanding the sources of acid and conjugate base, prompting a discussion on the definitions within the Bronsted-Lowry theory.
  • Multiple methods for calculating the amounts of acid and base are presented, with one participant outlining two different approaches but expressing uncertainty about when to apply each method.
  • There is a debate about whether HCl or Tris base should be considered the acid in this context, with one participant asserting that HCl is not the acid, which leads to further exploration of the roles of Tris in the buffer system.
  • Participants discuss the behavior of Tris when combined with a strong acid, leading to the conclusion that protonated Tris acts as the conjugate acid, which adds complexity to the stoichiometric calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to calculating the amounts of components in the buffer solution. There are competing views on the identification of the acid and base in the context of the Bronsted-Lowry theory, and the discussion remains unresolved regarding the appropriate method for calculations.

Contextual Notes

Participants express uncertainty about the definitions of acid and base in the Bronsted-Lowry context, and there are unresolved questions about the stoichiometry involved in the protonation of Tris. The discussion highlights the complexity of buffer preparation when multiple components are involved.

jibjab
Messages
13
Reaction score
0

Homework Statement


To purify osprase you must use 750 mL of an extraction buffer consisting of 0.500 M Tris, pH 7.8, 100 mM NaCl,
20% sucrose. You have the following chemicals available:
Tris base (FW = 121.1 g/mol; pKa = 8.1)
1.00 M HCl
NaCl (FW = 58.4 g/mol)
Sucrose (FW = 342 g/mol)
Outline how you would prepare your solution, show all calculations.


Homework Equations


pH=pKa+log(B)/(A)


The Attempt at a Solution


I plugged in the pH and pKa in the equation and got 0.5012A=B.
I multiplied 0.75L x 0.5 mol/L = 0.375 mol
I solved for A = 0.2498 mol then multiplied that by 1000mL/1mol = 249.8 mL HCl
I solved for B = 0.1252 mol x 121.1 g/mol = 15.16 g Tris base

^The biggest thing I need to know is how do you know what to do with your B value? What I mean is how do you know when you have to multiply the formula weight of the tris base with the B+A value(0.375 mol in this case), or when you should solve your A value, subtract the B+A value from the A value to solve for B, then multiply the FW with the B value for the amount of base added?

(Feeling around in the dark for this section)

I multiplied 0.75 L x .01 mol/L = 0.075 mol NaCl
0.075 mol NaCl x 58.4 g/mol = 4.38 g NaCl
0.375 mol x 342 g/mol Sucrose = 128.25g x 0.2 = 25.65 g Sucrose.

^I haven't solved one that had other stuff in the buffer before, I'm not sure what to do with that bit.

Any explanations provided are highly appreciated! :)
 
Physics news on Phys.org
If I understand you correctly, your problem is with determining what are sources of the acid an conjugate base. That in turn means you probably have problems with the Bronsted-Lowry theory.

What is the acid in this problem?

What is the conjugate base?

Where do each of them come from?
 
I understand that the acid is HCl and the conjugate base would be Tris-base, but what I don't understand is which method you use to determine the amount of base added to the buffer. For example, I've seen it solved one of two ways for different problems:

1) After solving for A, B would be found by subtracting A from the total mol of the buffer. Then you would multiply B mol by the FW of whatever B is to find out how much of B you would need in the buffer.

2) The FW of the base would be multiplied by the A+B = total mol buffer <---- that amt
and that is the B value. A is found in the usual way, by plugging it into A+B=total.

I don't understand how you know which method to use.
 
After reviewing the other problems I worked, it looks like you multiply the acid or the base FW with the total buffer mol when one or the other is given in aqueous form? I don't understand why?
 
jibjab said:
I understand that the acid is HCl and the conjugate base would be Tris-base

And you are wrong. HCl is not the acid here! That's why you don't understand what to do and when.

What is a definition of the acid in the Bronsted-Lowry theory? What is a definition of the base?
 
A proton donor/proton acceptor
 
And what happens to Tris when you add a strong acid, like HCl?

(think what would happen to ammonia in the same situation)
 
This is where I'm completely lost. It looks to me like the tris base is acting as a base because it bonds with the Hydrogen on HCl. I guess this would make it more acidic afterward, so it would behave like a weak acid?
 
You are on the right track - it is protonated Tris that becomes the conjugate acid. Now that you know what is the acid and what is the conjugate base, think about stoichiometry of the protonation, and perhaps it will become clear how to deal with concentrations.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
10K
Replies
3
Views
3K