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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: fonclbrish2010 on October 04, 2010, 01:50:51 AM

Title: How to make a solution of chemicals with specific molarities
Post by: fonclbrish2010 on October 04, 2010, 01:50:51 AM

The PROBLEM:

You are given 30 ml of a 1 M KCL solution, 30 ml of a 0.5 M NaCl solution and a bottle of Tris buffer (MW: 121.14). How would you make 200 ml of a single solution containing 0.02 M NaCl, 0.02 M KCl, and 0.1 M Tris?
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My ATTEMPT:
I think I use the equation C1*V1=C2*V2, with the liquids, but I dont know how I go about doing this with the MW of the Tris buffer. This is what I figured out so far:

V1*(1 M KCL)= 200 ml * (.02 M KCL)
 So, 4 ml of KCL
V1* (.5 M NaCl)= 200 ml *(.02 M NaCl)
So, 8 ml NaCL
Does that mean I add 188 ml of Tris buffer? If I'm doing it completely wrong, can you show me how do it right step by step.

Title: Re: How to make a solution of chemicals with specific molarities
Post by: DrCMS on October 04, 2010, 06:10:57 AM

Quote from: fonclbrish2010 on October 04, 2010, 01:50:51 AM

The PROBLEM:

You are given 30 ml of a 1 M KCL solution, 30 ml of a 0.5 M NaCl solution and a bottle of Tris buffer (MW: 121.14). How would you make 200 ml of a single solution containing 0.02 M NaCl, 0.02 M KCl, and 0.1 M Tris?
--------------------------------------------------
My ATTEMPT:
I think I use the equation C1*V1=C2*V2, with the liquids, but I dont know how I go about doing this with the MW of the Tris buffer. This is what I figured out so far:

V1*(1 M KCL)= 200 ml * (.02 M KCL)
 So, 4 ml of KCL
V1* (.5 M NaCl)= 200 ml *(.02 M NaCl)
So, 8 ml NaCL
Does that mean I add 188 ml of Tris buffer? If I'm doing it completely wrong, can you show me how do it right step by step.

No you do not Tris buffer is a solid http://en.wikipedia.org/wiki/Tris

The first 2 parts are correct.

How many g of tris buffer are needed to make 200ml of  0.1M solution?