December 23, 2024, 06:37:16 AM
Forum Rules: Read This Before Posting


Topic: Chemical Equilibrium lab report. I am so beyond lost....  (Read 6797 times)

0 Members and 1 Guest are viewing this topic.

Offline Lhg07

  • New Member
  • **
  • Posts: 8
  • Mole Snacks: +0/-0
Chemical Equilibrium lab report. I am so beyond lost....
« on: February 26, 2014, 10:54:00 PM »
I am sure it's been seen before, but I'm working on my formal lab report for Chemical Equilibrium.

I have a reaction between:

Fe+3 (aq) +SCN- (aq)  ::equil:: FeSCN+2 (aq)


As part of the pre lab assignment, we were to plot in excel:

Concentration of FeSCN+2 vs. absorbance. The values were given:

.00004  vs.  0.213
.00008  vs.  0.348
.00012  vs.  0.524
.00016  vs.  0.719
.00020  vs.  0.879
.00024  vs.  1.046
.00028  vs.  1.215
.00032  vs.  1.387

I graphed it and got an R2 value of 0.99828 and y=0.1747x

Is this the calibration curve I use to get concentration of the FeSCN+2 ion?

I know that's significant but I don't know how or why.

Next, I was to combine 10.00 mL each of 0.1 M KSCN, 0.5 M HNO3, and 0.1 M Fe(NO3)3

It asks me to calculate the concentration of Fe+3 ion using the M1V1=M2V2 dilution equation and that this concentration will be equal to the FeSCN+2 ion concentration.

WHY??  ??? How do you isolate the Fe +3 ion?

I tried and got 0.0333 M for solution 1 by doing:

[0.1M][10.00 mL]=[M2][30.00 mL]

...but I don't know if that's correct. If it is, does that become the new concentration for solution 1 when I solve for solution 2?

Then it asks me to do 3 more solutions using serial dilution and calculate the concentration of Fe +3 ions again for each.

1. solution 2: add 10 ml of solution 1 + 40 ml HNO3
2. Solution 3: add 10 ml of solution 2 + 40 ml HNO3
3. Solution 4: add 10 ml of solution 3 + 40 ml HNO3

I am so beyond lost and I know this isn't even the hard part yet.

In lab, solution #1-3 had absorbance of 2.5 and solution #4 had absorbance of .250.

What do I do with that? Does that even sound correct?

Then the last part had me make 3 solutions  with:

-0.002 M Fe(NO3)3
-0.002 M KSCN
-0.05 M HNO3

solution #1: 10 mL of Fe(NO3)3, 4 mL KSCN, 6 mL of HNO3

solution #2: 10 mL of Fe(NO3)3, 6 mL KSCN, 4 mL of HNO3

solution #3: 10 mL of Fe(NO3)3, 8 mL KSCN, 2mL of HNO3


-Solution #1 had transmittance of 35.73, absorbance: 0.45
-Solution #2 had transmittance of 24.18, absorbance 0.62
-Solution #3 had transmittance of 14.27, absorbance of 0.85

It wants me to find
-[FeSCN+2] (equilibrium concentration)
-[Fe+3] initial concentration
-[SCN] initial concentration
-[Fe+3] eq. concentration
-[SCN] eq. concentration
-K constant
-average K


I am not asking for anyone to do this for me, I just need really dumbed down help on what to do with all this information. I don't understand what it's asking me to do and I'm so terribly frustrated  :'(

I am begging someone to help....PLEASE. This lab report is due tomorrow and I have no idea where to start. I'm going to be up all night trying to do this so anything would be helpful.
« Last Edit: February 27, 2014, 02:21:10 AM by Lhg07 »

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3551
  • Mole Snacks: +546/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Chemical Equilibrium lab report. I am so beyond lost....
« Reply #1 on: February 27, 2014, 10:09:32 AM »
As a general rule - and I'm not just being a smartass here - you shouldn't wait until the night before a lab report is due before you seek help. There are people here willing to assist you, but as you can see sometimes it takes a little while to get a response.

Your post is quite long, and to be honest it's hard to follow what you're supposed to be doing. Seeing the actual procedure and questions would be very helpful.

Here are some brief answers, based on my best guesses.

Quote
Is this the calibration curve I use to get concentration of the FeSCN+2 ion?
Yes.

Quote
WHY??   How do you isolate the Fe +3 ion?
It seems they want you to calculate the initial concentration of ferric ion. You are adding a solution of ferric nitrate to some other solutions. What will be the final volume of ferric ion after you combine these solutions? You would use the dilution equation to figure this out. Your answer of 0.033 M seems correct.

Before you move on, you should try to understand the chemistry of what is happening.  So basically what is happening is that you are adding some ferric ion and some thiocyanate ion and you are forming a colored complex of iron and thiocyanate according to your reaction. As you know, though, the reaction isn't 100%, meaning you have some relative concentrations of thiocyanate, iron and iron thiocyanate at equilibrium, and that proportion is governed by the equilibrium constant. If you know how much of the two reactants you start with and how much thiocyanate you make in equilibrium, you can calculate what the equilibrium constant is. The reason you are doing this at several different concentrations is to calculate an average of sorts, because any single measurement is prone to error. This is what you're doing with your serial dilutions. You're going to calculate your k value at several different concentrations and take an average.  If you don't know how to calculate an equilibrium constant, you can ask for help here or consult your textbook.

At the end, you're asked to prepare some other solutions with different starting amounts of ferric and thiocyanate ions and calculate the various concentrations of everything and the equilibrium constants.  Do you expect the equilibrium constant to vary depending on your input conditions? Make a hypothesis!! This is how you determine whether you really understand what is going on.



What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Sponsored Links