4
$\begingroup$

We prepare buffer solutions with concentration in terms of x, for example 50x TAE buffer. How do we come up with x? Is x related to molarity in some sense? And how much can we concentrate the solution. We have 50x TAE but usually 10x TBE buffer, why is it so?

$\endgroup$
3
  • 2
    $\begingroup$ X simply stands for "times" like 50 times concentrated. And there are limits of how much of a certain substance can be dissolved in water. This limits stock concentrations. $\endgroup$
    – Chris
    Jul 11 '16 at 7:28
  • $\begingroup$ Thanks for the comment @Chris. If X stands for "times" as you say, how much is 1 time, i.e, 1x, is it 1M, 0.05M as given here link? And is it same for all solutions? $\endgroup$
    – risingStar
    Jul 11 '16 at 9:13
  • $\begingroup$ That depends on your final concentration. If it is 100mM, then 10x is 1M (1000mM), 100x would be 10M. $\endgroup$
    – Chris
    Jul 11 '16 at 9:57
9
$\begingroup$

1x is the final working concentration of the solution (it could be anything depending on the type of the solution). Stock solutions are made at a higher concentration; if it is 10 times more concentrated than the working solution then it is 10x.

Why 50x TAE but 10x TBE?

The salts in TBE precipitate at higher concentrations. In other words, the salts in TBE have lower solubility compared to those of TAE. Therefore stock solutions of TBE are usually 10x or even 5x.

$\endgroup$
3
  • $\begingroup$ Thanks @WYSIWYG. Can you tell me how do we decide what should be 1x concentration by citing few examples? $\endgroup$
    – risingStar
    Jul 11 '16 at 14:18
  • 2
    $\begingroup$ @Biginer How to decide the 1x concentration: the first time when someone does that, then they 1. use rational guess 2. use trial and error, and settle on a concentration that is most optimal for the task. Subsequently, others follow it and tweak it when necessary. $\endgroup$
    – WYSIWYG
    Jul 11 '16 at 15:13
  • $\begingroup$ Thanks @WYSIWYG. I came to the same conclusion when I discussed it with my peers. Thanks a lot! $\endgroup$
    – risingStar
    Jul 11 '16 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.