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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: screen on April 15, 2006, 09:43:29 PM

Title: Calculation problem with standard preparation
Post by: screen on April 15, 2006, 09:43:29 PM

I have 3 standard solutions. They are all 2000ppm. I want to prepare two mixed standard solutions (10ml) that contain these 3 standards. Solution One should have 100ppm of each standard and Solution Two should have 500ppb of each standard.

Question 1: I would like to ask if my calculation is correct
Solution One (each standard in this solution is 100ppm):
For each standard: M1V1= M2V2
                          2000ppm x V1 = 100ppm x 10ml
                                            V1 = 0.5ml

It means I should add 0.5ml of each standard solution to a 10ml volumetric flask, and then the final concentration of each standard become 100ppm.


Question 2: Solution Two (each standard in this solution is 500ppb):
Does it mean each standard in the mixed solution is 500ppb after I added 0.05ml of Solution One in another 10ml volumetric flask?

100ppm x V1 = 500ppb x 10ml
               V1 = 0.5 ppm x 10ml / 100ppm
               V1 = 0.05 ml

Title: Re: Calculation problem with standard preparation
Post by: Borek on April 16, 2006, 04:05:58 AM

You are OK for the solution 1, but you are obviously wrong for the solution 2 - you can't make solution more concentrated by diluting!

Title: Re: Calculation problem with standard preparation
Post by: screen on April 16, 2006, 12:28:35 PM

Thanks for your reply.

Actually the concentration for each standard in the Solution Two is 500ppb = 0.5ppm. The Solution One is 100ppm for the concentration of each standard.

Is the Solution Two preparation correct?

Or

For solution two that I have to prepare three individual 100ppm standard solution first,

and then I pipette 0.05ml of each 100ppm standard solution into a 10ml volumetric flask to get a mixed solution

that contains three standards with 500ppb concentration each?

Title: Re: Calculation problem with standard preparation
Post by: Borek on April 16, 2006, 12:55:15 PM

Sorry, you are right - i have missed that b somehow. I need new monitor or new glasses.

Looks OK with me. To make solution 2 in one step you will have to dilute 2000 ppm solution 4000 times - something like 0.025 mL diluted to 100 mL. Two steps looks more ogical.

Only thing that makes me wonder is that you used formula M₁V₁ = M₂V₂ - which suggests molar concentrations. ppm and ppb are not molar, they are usually weight/weight. But it doesn't matter as dilution calculation looks the same in this case.