Preparing 40mM TES pH 7.4 - Calculating A-/HA

  • Thread starter Thread starter vande060
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on preparing a 40mM TES buffer at pH 7.4, utilizing the molecular weight (MW) of TES, which is 229.25 g/mol, and its pKa of 7.5. The calculation reveals that to achieve the desired buffer composition, 56% of the solution should be in the base form (A-) and 44% in the acid form (HA). The correct approach involves using 0.917 grams of TES and determining the amount of strong base needed to convert 44% of the acid to the base form, ensuring the pH remains stable.

PREREQUISITES
  • Understanding of buffer solutions and their components
  • Knowledge of pH calculations using the Henderson-Hasselbalch equation
  • Familiarity with molecular weight and its application in solution preparation
  • Basic skills in stoichiometry for calculating moles and grams
NEXT STEPS
  • Learn about the Henderson-Hasselbalch equation in detail
  • Research the properties and applications of TES buffer
  • Explore methods for preparing and adjusting buffer solutions
  • Investigate the effects of strong bases on buffer pH
USEFUL FOR

Chemists, laboratory technicians, and students involved in biochemistry or molecular biology who require knowledge in buffer preparation and pH management.

vande060
Messages
180
Reaction score
0

Homework Statement



prepare 40mM TES buffer pH 7.4
prepare 100mls of it
mw TES = 229.25
pka = 7.5



Homework Equations



ph = pka +log(A-/HA)



The Attempt at a Solution



My problem is that I don't know what the MW of the acid or base is for this buffer. Is the given MW the acid, and I subract a proton weight for the base or something? I can do the calculation though:



7.4 = 7.5 +log(A-/HA)

-.1 = log(A-/HA)

A-/HA = .79

56% base
44% acid

.56* 40mM * MW of base *.1L = moles of base
.44*40mM *MW of acid *.1L = moles of acid

So how should I figure out the MW of acid and base?
 
Physics news on Phys.org
http://en.wikipedia.org/wiki/TES_(buffer )

Given molar mass refers to the acid.
 
Last edited by a moderator:
Borek said:
http://en.wikipedia.org/wiki/TES_(buffer )

Given molar mass refers to the acid.

Thank you,

Ok, that's kind of what I thought. So the acid MW is 229.25, and the base would be (229.25 - 1.008) is that correct?
 
Last edited by a moderator:
You don't need molar mass of the conjugate base. Use enough acid to prepare 40 mM solution, neutralize it with any strong base.
 
so I would go:


229.25g/mol * .04moles/liter *.1L = .917g TES added

then to calulate strong base needed to convert 56% of acid to base

.04moles/liter *.1L = .004 moles TES * .56 = .00224 moles of strong base needed
 
Check your math. If more than half of the acid is converted to base pH is higher than pKa.

But in general you are on the right track.
 
Borek said:
Check your math. If more than half of the acid is converted to base pH is higher than pKa.

But in general you are on the right track.

whoops it should be 44% converted to base
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 29 ·
Replies
29
Views
5K