Help in making molar solutions

Bruce Micales bmicales at facstaff.wisc.edu
Mon Apr 24 20:24:06 EST 1995

```Gary,

and then dilute it the stock.  For instance, prepare a 1M Tris -HCl, pH 7.5
stock solution.  Now determine the dilution factor needed to bring a 1M
solution to 40 mM..in this case it is 25.  Divide the final volume by this
dilution factor and the answer is the number of mls needed of the concentrated
solution.  Do this for each of the components, pH the solution (if called for)
and bring it to the final volume.  Each of your components will be at the
required molar concentration.

For example:

Stock solutions:
1 M Tris pH 7.5
500 mM EDTA pH 8.0
1.5 M NaCl

Dilution Factors:

1M (1000 mM) Tris ---> 40 mM = 25
500 mM EDTA --> 1mM = 500
1.5M NaCl --> 0.15M NaCl = 10

Therefore for 1 liter (1000ml) of solution use:

1000/25 = 40  ml 1M Tris
1000/500 = 2 ml 500 mM EDTA
1000/10 = 100 ml 1.5M NaCl

This may appear the "long way around", but it works and is nice if you have
stock solutions.  In general,the molar concentrations of one compound is not
going to effect the molar concentration of another.

I hope this helps.

Bruce Micales

>Help!

>I am a microbiologist, and we recently started making our own reagents for
>a molecular biology project (pulsed-field gel electrophoresis used for
>epidemiological purposes).  I am having problems figuring out how to make
>certain buffers and solutions with some of the published protocols.  For
>example, one of the solutions we use is called TEN solution and has the
>following formula:

>40 mM Tris-Cl, pH 7.5
>1 mM EDTA, pH 8.0
>0.15 M NaCl

>I remember enough of chemistry to make up separate molar solutions of
>each, but what's the secret to combining them?  If I mixed them in equal
>amounts, wouldn't their overall molarity change?  I am missing something
>fundamental here, and could really use the help.

>Thanks,

>-gary cage
> AZ State Laboratory
> Dept of Microbiology

Bruce Micales
bmicales at facstaff.wisc.edu
Department of Anatomy