Well Dr. Brian Bower has already explained it well. I would be adding up just a little touchup.
1. Always remeber the basics for Molar (M) solutions
1M = 1mole/L
To state in word form: A one molar (M) solution contains 1 mole (gram molecular weight) of solute in 1 liter of solution.
2. Determination of concentration molar (M) of stock solution starting with the density mentioned on the stock bottle (Dr. Brian has already explained that well so i wont repeat it here).
3. The amount of volume that needs to be drawn from the stock bottle this is the next step.
Here we generally use the equation C1 x V1 = C2 x V2
C1 and V1 (for stock bottle)
C2 and V2 (for what concentration of solution you want to make)
So as in your case, in the first step you found out the (M) concentration of the stock bottle so C1 is known (i.e. 17.595 M), but V1 (?) is the amount you want to calculate out.
C2 is the concentration (M) of solution which you want to generate (i.e. 20 mM or 0.02 M)
V2 is what volume of final solution you want to make of this C2 concentration (It is also a known aspect of calculation, say 50 ml)
So, to sum up you will get volume to be drawn from stock by calculating out V1(?) from the equation. Like Dr. Brian described in the calculation V1 = 56.83uL and water must be upto 50ml to form the solution (Final of 20mM concentration).
The practical aspect is you take 50ml of water, pipette out of it 57ul and add to it 57uL of volume from the stock bottle of H2O2 (50% W/W).
I do hope my math isn't wrong. That would be embarrassing.
First, determine the effective concentration of the stock reagent:
50% H2O2, Density: 1.197 g/mL at 20*C (per Sigma, but you may want to confirm gravimetrically)
[H2O2] (g/mL): 0.5985 g/mL at 20*C
H2O2 MW: 34.0147 g/mol
[H2O2]: 0.017595 mol/ml
[H2O2]: 17.595mol/L (M)
Then determine the desired volume and concentration:
[H2O2] Needed: 0.02mol/L (M) or 20mmol/L (mM)
Volume Needed: 50 mL (0.05L)
Then Calculate the volume of stock needed to prepare the desired volume:
0.05L * 0.02mol/L = 0.001mol / 17.595mol/L = 5.68332E-05 L * 1e6 ul/L = 56.83324979 uL
Stock to Add: 56.83325 uL
Water to Add: Add to 50mL (~49.94ml)
You might get a more accurate final concentration by performing several serial dilutions from the stock solution (e.g. make a 5% stock instead of a 50% stock). It would reduce your % error in the final concentration.
Well Dr. Brian Bower has already explained it well. I would be adding up just a little touchup.
1. Always remeber the basics for Molar (M) solutions
1M = 1mole/L
To state in word form: A one molar (M) solution contains 1 mole (gram molecular weight) of solute in 1 liter of solution.
2. Determination of concentration molar (M) of stock solution starting with the density mentioned on the stock bottle (Dr. Brian has already explained that well so i wont repeat it here).
3. The amount of volume that needs to be drawn from the stock bottle this is the next step.
Here we generally use the equation C1 x V1 = C2 x V2
C1 and V1 (for stock bottle)
C2 and V2 (for what concentration of solution you want to make)
So as in your case, in the first step you found out the (M) concentration of the stock bottle so C1 is known (i.e. 17.595 M), but V1 (?) is the amount you want to calculate out.
C2 is the concentration (M) of solution which you want to generate (i.e. 20 mM or 0.02 M)
V2 is what volume of final solution you want to make of this C2 concentration (It is also a known aspect of calculation, say 50 ml)
So, to sum up you will get volume to be drawn from stock by calculating out V1(?) from the equation. Like Dr. Brian described in the calculation V1 = 56.83uL and water must be upto 50ml to form the solution (Final of 20mM concentration).
The practical aspect is you take 50ml of water, pipette out of it 57ul and add to it 57uL of volume from the stock bottle of H2O2 (50% W/W).
- Badges
- Science topic
- Similar topics
- Cell Biology
- Methods