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Posted

Hello again, everyone! Today, I present you all with a question about dilution and it's formula of C1V1=C2V2. I believe it's also common to see "M" in place of "C". 

My question is, "How many mL is needed to dilute a 100 mL solution of NaOH with a concentration of 10M to a concentration of 2M."

I believe the answer is 20 mL: 

C1V1=C2V2

C1 = 10M (initial concentration) 

V1 = ? (this is the amount of solution that must be transferred, right?)

C2 = 2M (desired concentration)

V2 = 100 mL (not entirely sure why this value is 100 mL, but I think it is because that's how much I want to end up with?)

(10M)(V1) = (2M)(0.1 L)

V1 = [(2M)(0.1 L)] / 10M

V1 = 0.02 L = 20 mL

Now, I believe this means that 20 mL of NaOH should be diluted in 80 mL of water? This will give the NaOH a concentration of 2M and a final volume of 100 mL. Do I understand this correctly? Thanks a lot for the help :) 

 

Posted
9 hours ago, Kagi98JP said:

My question is, "How many mL is needed to dilute a 100 mL solution of NaOH with a concentration of 10M to a concentration of 2M."

 

I doubt if that is the question as written.

How many mL are needed to dilute 100 mL of NaOH ?

How many mL of what wine ?, whisky ?   ....?

 

Secondly ask yourself "How much original solution are you trying to dilute ?

Well it appears to be 100 mL.

In which case you haven't finished the calculation since you have only only diluted 20mL of the stuff.

Since this is homework I will let you finish but just observe that you are on track to obtain the right answer.

 

Here is a comment.

The equation C1V1 = C2V2 is quick and dirty and much bandied about, but does not always work directly.

And you need a volume.

I have seen many Pharmacist students come to grief trying to rush a calculation with this formula.

Here is an alternative that always works, and works whether you have volumes, % or weights and concentrations can be used in any form,

so long as consistent (ie the same) units are used.

It works for mixing creams and so callled 'alligation'  questions. It works when the % of one component are 0% or 100%.

In short it works for all forms of mixture.

dilution1.thumb.jpg.7e6b22f6d506e69e6e463ace8de36278.jpg

If you like this we can work through how easy it is to get the right answer to your question with this method, once you have finished yours.

 

 

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