Volume of Tris Needed to Prepare Buffer Solution

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Discussion Overview

The discussion revolves around calculating the volumes of 1.0 M Tris base, 1.0 N HCl or 0.2 NaOH, and water needed to prepare 50 ml of a 0.1 M Tris buffer at pH 9.0. Participants explore the theoretical aspects of buffer preparation, including stoichiometry and the Henderson-Hasselbalch equation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant expresses uncertainty about how to start the calculations for preparing the buffer solution.
  • Another participant suggests calculating the amount of Tris needed for a 50 mL of 0.1 M solution and emphasizes the need to add strong acid or base to convert part of the Tris to its conjugate form.
  • A participant calculates that 0.005 moles of Tris are needed and attempts to determine the volume of 1.0 M Tris required to achieve this concentration.
  • There is a discussion about the relationship between the concentrations and volumes of the solutions involved in the dilution process.
  • Participants clarify that Tris acts as a base and discuss the role of HCl in generating the conjugate acid (Tris+).
  • One participant mentions the importance of stoichiometry in the context of buffer preparation and the completion of the protonation reaction.

Areas of Agreement / Disagreement

Participants generally agree on the need for a conjugate acid-base pair in buffer preparation and the role of Tris as a base. However, there is some uncertainty regarding the specific stoichiometric calculations and the exact volumes of HCl or NaOH needed, indicating that the discussion remains unresolved.

Contextual Notes

Some participants express confusion about the stoichiometric relationships and the calculations required for the buffer preparation, indicating potential limitations in their understanding of the underlying principles.

MissSpice
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Homework Statement


Calculate the theoretical volumes of 1.0 M Tris base, 1.0 N HCL or 0.2 NaOH, and water needed to prepare 50 ml of a 0.1 M Tris buffer pH 9.0. pKa of Tris: 8.21

Homework Equations


Henderson-Hasselbalch: pH=pKa + log[A-]/[HA]

The Attempt at a Solution


Honestly, I don't even know how to start.

I know how to calculate moles of each, ratio of [Tris]/[Tris+] (I got 6.16 by 10^(9-8.21)), etc, but I don't what I actually need to use for the calculation... Help?
 
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First of all - how much TRIS do you need to prepare 50 mL of 0.1 M solution?

Then you need to add strong acid (or strong base) to convert part of the TRIS to conjugate acid (or base) - this is a simple stoichiometry.

Finally you need to fill up to 50 mL.
 
Borek said:
First of all - how much TRIS do you need to prepare 50 mL of 0.1 M solution?

I would need .005 mols, (0.1M Tris)x(0.05L)=.005. But then how do I go back to ml from here? And where does the 1.0M Tris base come in, since that is the solution I am diluting to get the 0.1M solution?

EDIT: I think I figured it out. Would it be (1.0M Tris)(X L)=(0.1M Tris)(.05L), so I would need 5mL of 1.0M Tris?

I still don't understand the HCl and NaOH part though; what do you mean by simple stoichiometry?
 
Last edited:
MissSpice said:
Would it be (1.0M Tris)(X L)=(0.1M Tris)(.05L), so I would need 5mL of 1.0M Tris?

Yes, it is just an application of molar concentration definition (##C = \frac n V##) and mass conservation (amount of substance doesn't change during dilution).

I still don't understand the HCl and NaOH part though; what do you mean by simple stoichiometry?

First things first - for buffer solution you need a pair of conjugate acid and base. Is TRIS a base, or an acid? How can you prepare other substance from the pair?
 
Borek said:
First things first - for buffer solution you need a pair of conjugate acid and base. Is TRIS a base, or an acid? How can you prepare other substance from the pair?

It's a base. So I would use the 1.0 N HCl to generate the acid, Tris+?
 
TRISH+ to be exact, but yes, you are no the right track.

Assume protonation reaction went to completion - that's where the stoichiometry comes into play.
 

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