1 M tris pH 6.8 buffer preparation using tris and Tris HCL (No conc HCL) - (Sep/28/2016 )

for my sds page i require 1M tris pH 6.8. we prepare it using tris base and adjust pH with conc HCL. is there a way to prepare this buffer using tris and Tris HCl together (and not using conc HCL). there is a way to prepare pH 8.8 buffer in that manner (Sigma website) but not of pH 6.8. can anyone guide. thanks and regards.

Yes, this is possible... you can use the Henderson-Hasselbalch equation. There are also a number of online calculators that will do this for you. Roche Lab FAQs also has tables for this sort of thing.

i tried few calculator, but they give it for making tris buffer upto pH 7.00 only. i want to make pH 6.8. if anyone can give me precise amount of tris and tris Hcl to add to make 1M tris buffer pH 6.8, it would be really helpful.

thanks

OK... henderson-hasselbalch pH=pKa + log10(conc base/ conc acid)

pH =6.8, pKa =8.08.

substitution gives us that the log10(base/acid)=-1.28... antilog gives us ratio between the two as 0.05248. This means that to balance the pH you will need 19x more tris-HCL than tris base to get pH 6.8. Can you work from that?

Henderson hasselbalch is a great intellectual exercise but rarely works in real practice when attempting to make buffers with precise pH.

The practical way of making a buffer using acid and conjugate base is to make stocks of both at equal concentration, then add one to the other while stirring with a pH meter until desired pH is met.

For example, if you want 1 M Tris pH 6.8, make a 1 M Tris base solution and a 1 M Tris-HCl solution. Add the Tris base solution dropwise with stirring and a pH meter to the Tris HCl solution until pH is raised to 6.8. Because both Tris stocks are 1 M, the final solution is also 1 M Tris.  It may be helpful to do a rough HH calculation beforehand so you know about how much of each 1 M stock solution to make so you don't have an excessive amount of one left over as waste at the end (which bob1 did very nicely for you already).