[cvc121]

Hi,

I am having a lot of trouble trying to figure out the value of V2 in order to plug it into the formula C1V1=C2V2 to solve for the unknown, C2. Initially, I have 0.2M and 0.1L of copper (II) sulfate pentahydrate (CuS04x5H20). This gives me C1 and V1. Now, I transfer 0.01L of the 0.2M solution into another container and add water until it reaches 0.1L. What would the value of V2 be in order to calculate the concentration of the solution after the dilution?
Also, would the original solution or the diluted solution react with zinc metal at a faster rate? Why is this?

Thanks. All help is very much appreciated!

[Hunter2]

You calculate the moles in the 0,01 l (V1) of the 0,2 M(C1) solution. This amount  you will find in the new solution of 0,1 l(V2).

New concentration is: C2 = V1*C1/V2

[morales613]

You can find the original moles (.1 L *.2M) = moles Then you can add both volumes. Then divide original moles by total volume. Think about it. I know why you were confused because you got the same concentration, right? 

[Babcock_Hall]

Many problems in dilutions calculations can be avoided by picturing what is going on.  Let's assume that you transfer the solution with a 10-mL volumetric pipet of some sort.  The formula C₁V₁ = C₂V₂ really means that you cannot create or destroy matter, that the number of moles of solute you start with will not change after you dilute it.  The value of 0.1 L for the starting volume of your solution is not part of this problem, but using it is a common mistake.

I would think about the mathematics of the rates of reactions to answer your section question.  Show us your attempt, and we can critique it.

[Hunter2]

You can find the original moles (.1 L *.2M) = moles Then you can add both volumes. Then divide original moles by total volume. Think about it. I know why you were confused because you got the same concentration, right?

That is a mistake. You have to consider the 10 ml of this solution immersed in a 100 ml flask and filled with water.  The amount of 100 ml before doesn't matter. It can be also an ocean of this solution. Fact is 10 ml was taken and diluted.

And 10 ml put in 100 ml is a dilution 1:10. So what is the result.