Clarissa V.

asked • 10/01/20

Determination of the order of Hydroxide

First of all, I really appreciate the time taken to help me figure this out. I included my questions below with the details to hopefully help clear up my questions. As context, this is part of a data analysis for an experiment that we did on Kinetics.

1.     Create a new spreadsheet in your workbook named ‘Rate Law Calculations’. Set up the following table in this new spreadsheet.


 A B C D E F
1 Trial # [CV+] (M) [OH−] (M) slope kobs k
2 1




3 2




4 3




5 4





2.     Enter the experimental values for the concentration of crystal violet (CV) and hydroxide (OH−) for each trial (not the stock concentrations!).

  1. The CV solution used was as following:
  2. Trial 1= 3.0x10^-5M
  3. Trail 2= 3.0x10^-5M
  4. Trial 3= 3.0x10^-5M
  5. Trial 4= 1.5x10^-5M

-These were the actual measurements given for the experiment, however, would these be the stock concentrations, and if so, would I need to adjust those values and make them half of the stock concentrations?


-My second question comes from trying to find the values for the hydroxide. In the experiment we used NaOH, so my question is would I have to find the amount of OH present in the NaOH solution?

-To better understand me, these are the following experimental values for the NaOH used:

  1. Trail 1= 0.1M NaOH in 0.1M NaCl
  2. Trial 2= 0.1M NaOH in 0.1M NaCl
  3. Trial 3= 0.2M NaOH
  4. Trial 4= 0.1M NaOH in 0.1M NaCl


Thank you!


1 Expert Answer

By:

J.R. S. answered • 10/01/20

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To find the concentrations of CV and OH- actually present during the kinetic analysis, you need to know (provide) the volume of each stock solution used and the final volume of the reaction mixture. For example if you used 1 ml of the stock CV and 1 ml of the stock NaOH, then the volume of the reaction mixture would be 2 ml and the concentration of CV and OH- would be 1/2 of the stock concentration for each (because you diluted each in half). So, if you can submit a comment to this answer and tell me exactly how you did the experiment (how much of each reagent you used and final volume of reaction), I can better help you answer this.

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Clarissa V.

These were the measurements that we were given for lab: Control Trial- CV solution: 3.0 x 10-5 M, NaOH solution:Deionized water, Temperature: Room Temp ; Trial 1- CV solution: 3.0 x 10-5 M, NaOH solution: 0.1 M NaOH in 0.1 M NaCl, Temp: Room Temp; Trial 2- CV solution: 3.0 x 10-5 M, NaOH solution: 0.1 M NaOH in 0.1 M NaCl, Temp:Room Temp; Trial 3- CV solution: 3.0 x 10-5 M, NaOH solution: 0.2 M NaOH, Temp: Room Temp; Trial 4- CV solution: 1.5 x 10-5 M, 0.1 M NaOH in 0.1 M NaCl, Temp: Room Temp
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10/02/20

J.R. S.

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First off, I don't see any difference between trial 1 and trial 2. They appear the same. Second point, what is being measured? Did you measure the time to observe some color change? We need more data so we can plot the graph and find the order of OH- from a kinetic analysis. From the data provided, I would say that these are the final concentrations of CV and OH- so no further calculations are needed for that part. But we need a measured variable to do the calculations.
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10/02/20

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