C1V1 = C2V2

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dan167
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C1V1 = C2V2

Post by dan167 » Sat Oct 13, 2007 9:05 am

Hi, I am lost with this. I want to prepare a 10ml of a 20?g/mL working solution, by diluting 0.1% w/v stock solution. Using C1V1 = C2V2

Does any know how to do this? Or have a website where the method is shown? thanks

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Katy_Bobbles
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Re: C1V1 = C2V2

Post by Katy_Bobbles » Sat Oct 13, 2007 11:05 am

0.1% is equivalent to 0.1g/100ml which is equivalent to 0.001g/ml. To make it an easier number to work with I would convert the units to mg/ml so C1 = 1mg/ml

Well C2 is your final concentration value so C2 = 20 (g/ml). But you also have to convert this to mg/ml = 20000

V2 is your final volume so that would be 10 (ml)

You want to find out the V1 so:

V1 = C2 X V2/C1
= (1) x 10/20000
= 0.0005 ml = 0.5μL

Thats the way I'd have done it. Is that the kind of value your looking for?? It's tiny in a 10ml final volume. :? Or is the question mark here:


dan167 wrote:Hi, I am lost with this. I want to prepare a 10ml of a 20?g/mL working solution, by diluting 0.1% w/v stock solution. Using C1V1 = C2V2


supposed to be μg/ml?? Or am I just being stupid n have wrote a lot of rubbish?!? :lol:

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Post by blcr11 » Sat Oct 13, 2007 3:44 pm

I assume you mean to prepare 10 ml of a 20 microgram/ml solution from a 1 mg/ml (0.1% w/v) stock. You can’t make a 20 g/ml solution by diluting a stock solution of only 0.001 g/ml, so there’s something wrong with Katy’s reasoning, though the gist of the explanation is correct.

The idea of using V1 x C1 = V2 x C2 is simple algebra (well, that and the principle of mass balance), but the trick is to make sure all your units are consistent, as Katy was suggensting, otherwise you may end up numerically off by 3 or more orders of magnitude. Let V1 = the unknown volume of stock solution needed, C1 = 0.1%(w/v) = 1 mg/ml, as Katy said, but I’ll take it 3 orders of magnitude lower and restate it as 1000 microgram/ml. Then V2 is the final volume of 10 ml of a C2 = 20 microgram/ml solution. Solve the expression for the unknown volume, V1 and you should have:

V1 = (V2 x C2)/C1.

Plug in the numbers and you get:

V1 = (10 ml x 20 microgram/ml)/ 1000 microgram/ml = 0.2 ml

So dilute 0.2 ml of 0.1% (w/v) stock to a final volume of 10 ml.

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Post by jchanley » Tue Feb 03, 2009 3:26 am

C1V1=C2V2

Can someone please help me with this problem? If you could help in layman terms, I think I will be able to get the rest of my assignment. It's been a long time since I've done any type of chemistry. Thanks - here goes:

Given a stock solution of 5.0% sodium chloride (NaCl), how would you prepare 20 ml of the following solution?

2.0% sodiwm chloride solution.

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canalon
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Post by canalon » Tue Feb 03, 2009 3:34 am

C1=5%
V1= that is what you want to calculate
C2=2%
V2=20ml
I let you solve the equation and figure out the missing ingredient.
Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

Alexcology
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Re: C1V1 = C2V2

Post by Alexcology » Sat Apr 10, 2010 6:54 am

for the preparation of 2% NaCl (2g/100)from the 5% NaCl (5 g/100ml)
please apply the formul C1V1=C2V2
C1= 5 g
V1= requred to know
C2= 2 g
V2 = 20
put the values in the following formula
V1=C2 X V2 /C1
V1=2 X 20 / 5
V1= 8 ml
hence tkae the 8 ml solution from the stock solution of 5% NaCl and make up to 20 ml with distilled water.
by this way you can make 2% NaCl solution from 5% NaCl stock solution.

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