Expert: Dr. Jeffery Raymond - 9/20/2011

Question
QUESTION: Hello

I need to make 100L of solution containing the following CaCl2 and NaCl
The Amount of Na added is not limited to a specific concentration

10ppm Ca
100ppm Cl

I don't understand this

thank you.

ANSWER: Basically, the question is asking you to add enough CaCl2 to get 10 ppm Ca ion into solution.  When this happens, you also have 20 ppm of Cl (2 Cls for every Ca).  Then add enough NaCl to generate 80 ppm of Na and Cl.  This will give you a total Ca concentration of 10 ppm, 100 ppm of Cl and 80 ppm Na (which the problem states is not important).

I hope this helps.

---------- FOLLOW-UP ----------

QUESTION: Yes a bit but is there a calculation I need to do to figure out how much?

ANSWER: Sorry, I had assumed that you had the equations and just needed help understanding the phrasing of the question.  Here is how to work a ppm problem:

ppm is usually calculated by taking the mass of the solute in units mg and dividing it by the mass of the solvent in kg.  So, with n= mg solute and m= kg solvent, n/m=x... where x is concentration in ppm. For this problem, you will want to switch your calculation around to n = m*x, since you have your mass of solvent (m, here 0.1 kg) and you have your ppm and need mass of the salts added.

---------- FOLLOW-UP ----------

QUESTION: Thank you so far for your help still having trouble. So n = how much I want to add so n = 10ppm (NaCl) * .1kg (I assume this is my 100L) = 1mg of CaCl2 added? (10mg for NaCl)

I think I am wrong or is it that simple.

Answer
It is pretty easy once you know what you are up to... however your division is off. You also need to account for atomic mass in relation to the whole molecule.  Here is an example:

A,B,C are all elements.  The atomic weights for each are 10, 20 and 15 respectively.

AC and BC3 are both salts.  I need 100 ppm A and 200 ppm C and I don't care how much B I have in solution.  How much of each should I add to 1 kg or water?

1.  Molc Wt of AC is 25.  So for every 25 grams of AC I add, I will be getting 10 grams of A.
2.  so to get 100 ppm A I use 100 ppm/1 kg = 100 mg A.
3.  Convert A to AC from the conversion factor in 1.  100 mg A * (25 g AC/10 g A) = 250 mg AC
4.  How much C does give us in solutions?  We just added 250 mg of AC, 100 mg of which was A, so we have 150 mg C in solution.
5.  How many mg of C do we need in total? 200 ppm C/1 kg = 200 mg of C. We have 150 in solution already, so we need 50 more mg of C.
6.  So now we have to look at the BC3 salt.  Molc Wt is then 65.  So for every 55 grams of AC I add, I will be getting 45 grams of C.
7.  We need 50 mg of C, so we convert again:  50 mg C * (65 g BC3/45 g C) = 72.2 mg of BC3
8.  All together we have to add 250 mg of AC and 72.2 mg BC3 to get 100 ppm A and 200 ppm C.

Use this method with your molecules and atomic/molecular weights and you should be just fine.

Take care