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General : Rate of Reaction and Concentration
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 Message 1 of 2 in Discussion 
From: MSN NicknameN_2006  (Original Message)Sent: 3/9/2008 6:39 PM
Hey Steve,
In  class we did an experiment about iodine clock reactions.
We were asked at the end to fill out the initial concentrations of [I-] and [S2O8^2-]
 
So I wanted to know if I have to calculate those initial concentrations for th Reactants and how much (volume) we used in each experiment
 
Here was the chart that I used for the experiment:
 
Reactant                      Volume of Reactant (mL)
                                    EX 1      EX2     EX 3
0.200 M KI                 1.00       2.00       2.00
 
0.100 M K2S2O8       2.00      2.00       1.00
 
So what I am trying to figure out is, for the initial concentration would it be 0.200? Or I know we have to incorporate the different volumes we used.
Do I take the M (which is the molarity) and mulitply it by the volume in Litres to get the moles, then take the moles and divided that by the volume to get the mol/L?
That just seems like you would just leave it the way it is, because you multiply then divide, so that would cancel out.
 
Hopefully you can help!!!
THanks!


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 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 3/10/2008 7:11 AM
Right idea, but you need to divide by the total volume.  For example, in Exp 1, you use 1.00 mL of 0.200 M KI.  The total volume of the solution after adding the K2S2O8 solution is 3.00 mL.  So the mole of KI is 0.200 mol/L  X  0.00100 L  =  0.000200 mol.  The molarity is 0.000200 mol / 0.00300 L  =  0.0667 M.

Or, use the dilution formula M1V1  =  M2V2.

M1  =  0.200 M
V1  =  1.00 mL  (you do not have to convert to liters)
M2  =  ?
V2  =  3.00 mL

Plugging in and solving for M2 gives the same answer, 0.0667 M. 

Steve