# Volume of Tris Needed to Prepare Buffer Solution

1. Apr 4, 2012

### MissSpice

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Henderson-Hasselbalch: pH=pKa + log[A-]/[HA]

3. 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?

2. Apr 4, 2012

### Staff: Mentor

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.

3. Apr 4, 2012

### MissSpice

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: Apr 4, 2012
4. Apr 4, 2012

### Staff: Mentor

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).

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?

5. Apr 4, 2012

### MissSpice

It's a base. So I would use the 1.0 N HCl to generate the acid, Tris+?

6. Apr 4, 2012

### Staff: Mentor

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.