HELP! absolute salt concentration

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HELP! absolute salt concentration

Postby sciencemom » Sun Nov 29, 2009 12:10 pm

i am looking for a simple answer as to how to calculate the absolute salt concentration for my stock solution and then the dilutions. Please make this as simple as possible i have not had chemistry and this is my first science fair. Please try to get back to me as quickly as possible. Its due pretty soon! :(
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Re: HELP! absolute salt concentration

Postby heatherL » Sun Nov 29, 2009 1:22 pm

Hi there,

Calculating the absolute salt concentration of your stock solution will depend on your starting materials. Are you starting with a stock solution that is already made? If so, the concentration of the solution should be indicated on the bottle. If you are starting with the salt in solid form and will be making your stock solution yourself, you can determine the concentration by the amounts that you mix. If you tell me a little more about your starting materials, I can help you with your specific calculations.

In the mean time, here is a website that helps you understand calculations associated with concentrations:
http://chemistry.about.com/od/lectureno ... ration.htm

Once you have a stock solution of known concentration and volume, it is fairly easy to determine the volume you need to make a dilution. There is a simple equation that only works for dilutions:

M1*V1 = M2*V2

M1 is the concentration in molarity (moles of solute / Liters of solution) of your starting solution.
V1 is the volume of your starting solution.
M2 is the concentration in molarity of your diluted solution.
V2 is the volume of your diluted solution.

To determine how much to dilute a solution to get a particular volume, let's do a simple example.
If I have a 1M solution of sodium chloride (NaCl), this means I have 1 mole of NaCl per liter of solution. Let's say I start with a liter of this solution, and I want to know how much water I need to add to make it a 0.5M solution.

M1*V1 = M2*V2
(1M)*(1L) = (0.5M)*V2
V2 = (1M)*(1L)/(0.5M) = 2L

So I need to add 1L of water, which will give me a total of 2L of solution. That will dilute my stock solution to 0.5M.

I hope this helps you get started. Please let me know what kind of solutions you are working with, and I will try to help you with the specific calculations.

Cheers,
Heather
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Re: HELP! absolute salt concentration

Postby sciencemom » Sun Nov 29, 2009 2:03 pm

Hi-
We made our stock solution like this... 3 cups of water + 1 cup of table salt. We then mixed, mixed, mixed and added enough water at that point to make it a quart (per the procedure in the How Salty Does Water Have to Be for an Egg to Float). We then made a serial dilution by adding 3/4 cup of tap water to 3/4 cup of the solution. We repeated with 3/4 cup of tap water and 3/4 of each solution there after. We are not science experts and this is our first science fair. The most simple explanation and help you can give us is best. We are also struggling with creating a serial dilution that is more precise, now that we know the egg floated in cup 2. ANY help is truly appreciated!!
Thanks!
sciencemom
 
Posts: 2
Joined: Sun Nov 29, 2009 12:01 pm
Occupation: mother of 6th grade science student
Project Question: how salty does water have to be for an egg to float?
Project Due Date: 12/3/09
Project Status: I am conducting my experiment

Re: HELP! absolute salt concentration

Postby donnahardy2 » Sun Nov 29, 2009 3:22 pm

Hi,

Here are the numbers you need for your calculation. I am using metric measurements because this is preferred for science fair projects. I'll go step-by-step:

1 cup of salt weighs 272 grams

4 cups of water is .944 liters

Salt, or sodium chloride, or NaCl weighs 58.4 grams per mole


272 grams NaCl /.944 liters x 1 Mole/58.4 grams = 5 Moles per liter (your starting concentration)

You diluted the salt solution by 1/2 each time you diluted it so the second cup contained 2.5 M (moles per liter) NaCl. The 3rd cup contained 1.25 M, the 4th .625 M, etc.

You could test a 2 M solution by mixing 2 parts of the 5 M NaCl solution with 3 parts of water, e.g. 2/3 cup plus 1and 1/3 cup water. You could make a 3 M solution by mixing 3 parts of the 5 M NaCl with 2 parts of water (1 and 1/3 cup NaCl plus 2/3 cup water).

I hope this helps.

Donna Hardy
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