Posted 18 March 2012 - 06:43 AM
We need to find dilution, volume stock and volume buffer.
(2 mg/mL= 2000 μg /ml)
For 400 μg /ml, dilution= 1:5, vol stock= 10 mL, vol buffer= 40 mL
For 100 μg /ml, dilution= 6:25, vol stock= 12 mL, vol buffer= 38 mL
For 20 μg /ml, dilution= 1:5, vol stock= 13 mL, vol buffer= 52 mL
For 5 μg /ml, dilution= 1:20, vol stock= 2.5 mL, vol buffer= 47.5 mL
For 1 μg /ml, dilution= 1:50, vol stock= 0.5 mL, vol buffer= 49.5 mL
Are they correct or false? Someone can teach me?
Posted 18 March 2012 - 06:05 PM
so for 400 ug/ml, you have 2000 (c1) x V1 = 400 (C2) x 50 (V2).
therefore V1 = 400 x50/2000
V1=10 ml, + 40 ml buffer to make 50.
I don't see how you got the rest, unless you were making some sort of serial dilution...
Posted 18 March 2012 - 11:28 PM
Bob1's answer is another method...
What about this?
Question for 400 μg /ml,
Initial concentration : stock concentration;
400:2000 = 1:5 = 1/5 x final volume = 1/5 x 50
= 10 mL (volume stock)
Hence, dilution = 1:5, volume stock = 10 mL, volume buffer = (50 - 10)
= 40 mL
This method is applied to find all the final concentration...
Is my method acceptable?
Posted 19 March 2012 - 06:18 PM
For example, in the second one, you've said you are going to add 12 ml of the stock to 38 ml of buffer.
12 x 2000ug/ml (stock conc.)= 24000ug
24000ug in your 50ml final volume (12+38)= 480ug/ml
It does kind of appear that you are doing some kind of dilution series though, but without knowing which "stock" you use for each of the subsequent dilutions, we can't tell you if you are right or not.
Your method is probably ok, but I would suggest getting used to using the standard C1V1 = C2V2 method as not all dilutions you do in the future will be as simple as the one you have given us here.
Posted 20 March 2012 - 10:18 AM