Volume of Tris Needed to Prepare Buffer Solution

AI Thread Summary
To prepare a 50 mL solution of 0.1 M Tris buffer at pH 9.0, 5 mL of 1.0 M Tris base is required. The stoichiometry involves using strong acid, such as 1.0 N HCl, to convert part of the Tris into its conjugate acid form (Tris+). The calculation for the amount of Tris needed is based on the molarity and volume, applying the formula C = n/V. Understanding the relationship between the base and its conjugate acid is crucial for achieving the desired pH. The discussion emphasizes the importance of stoichiometry in buffer preparation.
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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