How Do You Prepare 1, 3, 5, and 7ppm Solutions from a 100ppm Stock?

  • Thread starter Thread starter higherme
  • Start date Start date
  • Tags Tags
    Atomic Emission
Click For Summary

Discussion Overview

The discussion revolves around the preparation of 1, 3, 5, and 7 ppm solutions from a 100 ppm stock solution. Participants explore the use of dilution formulas and the appropriate calculations for achieving the desired concentrations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant questions the method for preparing diluted solutions, suggesting the use of the formula C1V1 = C2V2 for calculations.
  • Another participant explains the concept of ppm, providing a detailed breakdown of how to achieve a 1 ppm solution from a 1 kg solvent, including examples of volume and mass ratios.
  • A participant emphasizes that they are starting from a 100 ppm solution, indicating a focus on dilution rather than initial preparation.
  • Repetitive posts confirm the correctness of the initial calculations, stating that using 5 mL of stock solution diluted to 100 mL yields a 5 ppm solution, and suggesting similar calculations for other concentrations.

Areas of Agreement / Disagreement

Participants generally agree on the method of dilution using the formula provided, with confirmations of the calculations. However, there is no explicit consensus on the broader implications of the ppm concept as discussed in the second post.

Contextual Notes

The discussion does not address potential limitations or assumptions in the calculations, such as the precision of measurements or the properties of the solutions involved.

higherme
Messages
126
Reaction score
0
my question is : How would you make a 1, 3, 5, 7ppm solution from a 100ppm solution?

do i just just this formula: C1V1 = C2V2 to calculate the amount i need?

for example, for 5ppm solution => (5ppm) V1 = (100ppm)V2

so for V1, I just choose my final volume to be whatever i like... like if i was to prepare it in a 100ml volumetric flask, then my V1 would be 100ml.
so I would need (5ppm*100ml) / 100ppm = 5 ml of the stock 100 ppm soln.

Is that correct?

Thank you in advance. =)
 
Physics news on Phys.org
If it's ppm then the solution has mgX/kgSol. So 100 mgX/kgSol = 100 ppm.

If one wants a 1 ppm X solution, take 0.01 (by volume) of 1 kgSol (or 10 gm, which could be 10 ml of aqueous soln) and add 990 gm of solvent (e.g H20), which would give 1 mgX/kgSoln = 1 ppm.

Or if one uses less volume, e.g. take 0.001 (by volume) of 1 kgSol (1 gm, or 1 ml aq), and add only 99 gm, to get 0.1 mg/0.1 kgSoln.

See - http://delloyd.50megs.com/photo.html

http://www.iun.edu/~cpanhd/C101webnotes/aqueoussolns/ppmppb.html

http://pages.towson.edu/ladon/concas.html

http://antoine.frostburg.edu/chem/senese/101/measurement/faq/ppm-as-unit-conversion.shtml
 
Last edited by a moderator:
but I am taking from a solution that is already 100 ppm...
 
higherme said:
my question is : How would you make a 1, 3, 5, 7ppm solution from a 100ppm solution?

do i just just this formula: C1V1 = C2V2 to calculate the amount i need?

for example, for 5ppm solution => (5ppm) V1 = (100ppm)V2

so for V1, I just choose my final volume to be whatever i like... like if i was to prepare it in a 100ml volumetric flask, then my V1 would be 100ml.
so I would need (5ppm*100ml) / 100ppm = 5 ml of the stock 100 ppm soln.

Is that correct?

Thank you in advance. =)


this sounds right to me
 
It is right, as long as the 5 mL of stock solution is diluted to 100 mL (yielding 5ppm solution). Use 7 mL for the 7 ppm and so forth.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
5K
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
39K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
12K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
3
Views
8K
Replies
5
Views
9K
  • · Replies 3 ·
Replies
3
Views
2K