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Henderson-Hasselbalch buffer calculations - B + A Question

  • Thread starter jibjab
  • Start date
  • #1
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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! :)
 

Answers and Replies

  • #2
Borek
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2,801
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?
 
  • #3
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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.
 
  • #4
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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?
 
  • #5
Borek
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28,402
2,801
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?
 
  • #6
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A proton donor/proton acceptor
 
  • #7
Borek
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And what happens to Tris when you add a strong acid, like HCl?

(think what would happen to ammonia in the same situation)
 
  • #8
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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?
 
  • #9
Borek
Mentor
28,402
2,801
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.
 

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