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

In summary, to make a 1, 3, 5, or 7ppm solution from a 100ppm solution, you can use the formula C1V1 = C2V2 to calculate the amount needed. For example, for a 5ppm solution, you would need to take 5mL of the stock 100ppm solution and dilute it to a final volume of 100mL. This method can be used for any desired ppm concentration.
  • #1
higherme
129
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. =)
 
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  • #2
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
 
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  • #3
but I am taking from a solution that is already 100 ppm...
 
  • #4
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
 
  • #5
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.
 

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

1. What is flame atomic emission spectroscopy?

Flame atomic emission spectroscopy is an analytical technique used to determine the elemental composition of a sample by measuring the wavelengths of light emitted when the sample is vaporized in a flame.

2. How does flame atomic emission spectroscopy work?

In flame atomic emission spectroscopy, a small amount of the sample is introduced into a flame, causing the atoms to become excited and emit light at specific wavelengths. The emitted light is then passed through a spectrometer, which separates and measures the different wavelengths to identify the elements present in the sample.

3. What is the difference between flame atomic emission spectroscopy and flame atomic absorption spectroscopy?

Flame atomic emission spectroscopy measures the light emitted by excited atoms, while flame atomic absorption spectroscopy measures the amount of light absorbed by unexcited atoms. Both techniques are used to determine the elemental composition of a sample, but they differ in the way the light is measured.

4. What are the advantages of using flame atomic emission spectroscopy?

One of the main advantages of flame atomic emission spectroscopy is its ability to detect a wide range of elements, including metals and non-metals, with high sensitivity. It is also a relatively simple and cost-effective technique compared to other analytical methods.

5. What are the limitations of flame atomic emission spectroscopy?

One limitation of flame atomic emission spectroscopy is that it requires a liquid sample, which can be challenging if the sample is not soluble in water or other solvents. Additionally, the technique cannot differentiate between different oxidation states of an element, and it is not suitable for analyzing complex mixtures of elements.

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