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### Topic: Chemical Equilibrium in Solution: Lab help:  (Read 19359 times)

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#### arcanum

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##### Chemical Equilibrium in Solution: Lab help:
« on: February 27, 2009, 03:30:26 AM »
Thanks for reading this, just stumbled upon this community looking for ideas how to solve my problem and thought I give you guys a try. Ill try to make my question easy as possible to understand do not be intimated by the size of the post, I just try to add everything possible, the question should be short, since I just don't understand what I'm suppose to be doing and grabbing from my data collected.

If you still have your old books perhaps, the experiment that I'm writing my lab report on is Experiment 12: From "Experiments in Physical Chemistry, by Carl W. Garland, Joseph W. Nibler, and David P. Shoemaker Eighth Edition.) I seen 7th edition, which its practically the same thing, so if you have that book you can follow along or it makes it a world easier to understand what I'm trying to ask.

My question, I don't know if I'm making this harder than it should be, but what I'm trying to find is the equilibrium concentrations of I_2,   I_3- , and I-  each in aqueous solutions. Which I have figured out its distribution constant k, but I have no idea what is needed to be done to find the equilibrium concentrations of I_2,   I_3- , and I-

I listed my collected data down below which I hope its legible. I already solved for the Distribution constant which I believe its part of the requirement to solve for the rest. What the book is telling me how to solve for it, is by...

(I_3- ) = T - (I_2 ) and (I- ) = C - ( I_3- )

To be fair, here is the whole Section verbatim.

"If, then one starts with a solution of KI whose concentration C is known accurately and dissolves iodine in it, the iodine will be present at equilibrium partly as neutral I_2 molecules and partly as I_3 - ions (formed when I_2  reacts with some of the I- ions initially present.) The total iodine concentration, T = (I_2) + (I_3-), can be found by titration with a thiosulfate solution. If there is an independent way to measure (I_2) at equilibrium, then (I-) and (I_3-) can be obtained from

(I_3- ) = T - (I_2 ) [eq 4]

and

(I- ) = C - ( I_3- )  [eq 5]

We can measure (I_2) at equilibrium by taking advantage of the fact that CCl_4 (We used Hexane instead in this experiment works the same, but safer), which is immiscible with aqueous solutions, dissolves molecular I_2 but not any of the ionic species involved.

-------------------------------------------(I already found the distribution constant k, so you can disregard runs 1-3, which are in blue. The results 4-6 are in green. I just added all the data just to make sure nothing gets left out.)

Average k:     Run 1: = 0.137      Run 2 = 0.112      Run 3 =  0.122

Warning, I did a preview before, and it looks like all my tables/results are off center but in the editing part they're center. My apologies in advance for it looking strange.

Table 1: Nominal initial concentrations and Volumes of the aqueous KI solutions and of the solutions of I_2 used in Hexane.

Run #   Hexane (50 mL) MolarIty I_2                       Molarity KI
1                    0.080                                               0
2                    0.040                                               0
3                    0.020                                               0
4                    0.080                                             0.15
5                    0.040                                             0.15
6                    0.080                                             0.03

Results from runs (1-3)
Group 1

Sample 1    Sample 2        Sample 3

Aq. Layer             3.5        1.2           0.6
Hexane Layer    17.5      10.3           6.6

Group 2

Sample 1      Sample 2                Sample 3

Aq. Layer         1.4      0.92                          0.72
Hexane Layer   23.6      8.56                    4.13

Group 3

Sample 1    Sample 2    Sample 3

Aq. Layer                2.1               0.95       0.47
Hexane Layer            13.9               8.57       4.65

Results from runs (4-6)
Group 1

Sample    Hexane #1    Hexane #2    Aqueous #1     Aqueous #2
4                 2.7       3.2      18.2               17.6
5                 1.5      0.95      10.1                10
6                 11      9.85      11.3               10.7

Group 2

Sample    Hexane #1    Hexane #2    Aqueous #1     Aqueous #2
4                 2.11          3.9       21.88               21.85
5                 3.98         4.02       11.87                 12
6                 9.2        8.55       11.51                11.2

So with all that, I'm just trying to figure out how to calculate the equilibrium concentrations of the different I-. I_2 and I_3-.

If there is anything that I could do to make this question easier to understand please tell me, still new to asking question on forums, but I hope this is enough information where someone could point me in the right direction.

#### Borek

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##### Re: Chemical Equilibrium in Solution: Lab help:
« Reply #1 on: February 27, 2009, 03:43:06 AM »
Honestly, after reading everything I still have no idea what your problem is.
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#### arcanum

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##### Re: Chemical Equilibrium in Solution: Lab help:
« Reply #2 on: February 27, 2009, 04:47:43 PM »
Honestly, after reading everything I still have no idea what your problem is.

Ok how about I take it a little slower and tell you guys step by step. I think I'm giving information backwards which may make it confusing, ill give you the experiment set up what we did and maybe that may help a little bit. Thanks for being patient :+)

What we did was Add 0.080M of I_2 [For the first run, I_2 M was the changing factor] as a solute and the solvent was 200 mL of DI water and 50 mL of Hexane added to that as well. (Reasoning for adding the hexane to the water and I_2 I'm not sure.) Then we put a stopper in the flask and shook it up for near five minutes and let it rest in a water bath at (25-26 C) for about 10 minutes, then shook again and repeat for an hour. Then we let the solution rest until the layers separated.

We extracted the aqueous layer out and place into another flask filled with 10 mL of 0.1M of KI. We then pulled the hexane layer out and place it into another flask filled with 10 mL of 0.1 M of KI.

Then we titrated both of the layers separately with Thiosulfate solution with a starch indicator, until the solution turned clear.

And we repeat these steps for the rest of the runs 1-3 which the only thing changes is the Molarity of I_2
Runs 4-6 two factors change where in addition to the molarity of I_2 changing, the experiment asked for 200mL of KI to be added before titration, each with different molarities.

Run #   Hexane (50 mL) MolarIty I_2                       Molarity KI
1                    0.080                                               0
2                    0.040                                               0
3                    0.020                                               0
4                    0.080                                             0.15
5                    0.040                                             0.15
6                    0.080                                             0.03

--------------------------

So that was the required experiment that was performed.

From my understanding the experiment is about equilibrium and layer separation and how well they separate with different molarity of I_2.

That's why we added the hexane layer and shook up the flasks really well for 5 minutes, so that the layers mix up properly.

When we extract the layers we placed the different layers each into a 10mL 0.1 M KI which is suppose to reduce the loss of I_2 from the aqueous solution by evaporation while trying titrate, which in the ends forms a less volatile I_3-
(Did I think that out properly there? Not sure there but the numbers look that way-)

Then we titrated the samples with thiosulfate solution which the reaction that should be taking place should be...

2 S_2 O_3^(-2) + I_3- ---> S_4 O_6 ^(-2) + 3 I -

That means the I_3- formed when we added the 10 mL of KI gets converted to  I- when we titrated with thiosulfate
So the experiment went from I_2 to I_3- to I- .

Overall this experiment was suppose to show the distribution of molecular iodine I_2 between two immiscible solvents, water and hexane. (Or I think at at least, by looking at the changes.)
--------------------------------------

[Question]
[color= red]So, where I'm stuck at is the idea of how am I suppose to find the equilibrium concentrations of I_2 , I_3- and I- in aqueous solutions. (Based on results of runs 4-6) [/color]

The equations the book gives me is:

(I_3- ) = T - (I_2 ) [eq 4]

and

(I- ) = C - ( I_3- )  [eq 5]

Where T = (I_2) + (I_3-) and C is suppose to be the initial KI solution concentration.

The real problem I have identifying which is suppose to be which at equilibrium concentrations.

Because initial I start out with 0.080 M of I_2 but that gets converted over to I_3- by the 10mL of 0.1 M KI then gets converted over by titration with thiosulfate to I-. Which how am I suppose to identify my I_2, and I_3- if they don't really exist anymore.

I am assuming that the endpoint of my titration with Thiosulfate tells me that all of the I_3- have been converted over to I-.

I hope this makes my question a little bit clearer. Thanks for taking the time to read all this :+)

#### Borek

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##### Re: Chemical Equilibrium in Solution: Lab help:
« Reply #3 on: February 27, 2009, 05:51:53 PM »
From my understanding the experiment is about equilibrium and layer separation and how well they separate with different molarity of I_2.

My bet is that it is about determination of two equilibrium constants - one is separation equilibrium of iodine between two phases, other is equilibrium I2 + I-  I3-. Your system is described by three equations - mass balance (iodine in two phases and iodine in I3-) and two equilbrium equations. Write all three equations, it is relatively simple algebra to find out all concentrations and all constants from your data.
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#### arcanum

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##### Re: Chemical Equilibrium in Solution: Lab help:
« Reply #4 on: February 27, 2009, 06:43:00 PM »
Ok, let me give it a try and see if I'm on the right mind track or way off base.

I2 = my initial molarity which is like 0.080 M or whichever run I'm on.
I3- = KI molarity that is added. which is like 0.15 or 0.03 depending on sample.
I- = is what I need to find by using the equations.

Thus, using equation 4

(I3-) = T - (I2)

(I3-) = (0.080 + 0.15) - 0.080 = [ 0.15]

That means using that into equation 5 which is...

(I-) = C - (I3-)

(I-) = 0.15 - 0.15 = 0

Theoretically. Or am I looking at this the wrong way, because I'm quite sure my data titration of thiosulfate has something to do with the calculations, but I can't figure that out.

The formula weight is suppose to be 112.132g but all we had was solution no samples, so I'm trying to figure out if there is a relationship between thiosulfate and Iodine, but I don't see how that works. What does the titration have to do with the experiment anyways other than making I- at the endpoint? I guess that's where I'm blind at the moment.

Sorry to cut this post short, but I must be going to work soon. Ill be back later tonight and thank you for your help so far, its helping me think about everything in whole when I write stuff down like this and try to explain it and solve. :+)

Just fixed one of your sub tags, that's all.  -Macman104
« Last Edit: February 27, 2009, 11:59:32 PM by macman104 »

#### arcanum

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##### Re: Chemical Equilibrium in Solution: Lab help:
« Reply #5 on: February 28, 2009, 03:39:44 AM »
Ok, I'm back from work and I've been working on this a little bit more.

Tell me if I'm in the correct direction:

For runs 1-3 we are suppose to solve for the k (distribution constant)

Which is just simply k = (I2aqueous / (I2hexane)

Example for sample 1 group 1 is...

k = 3.5mL/ 17.5mL = 0.2

And that is suppose to be my k (distribution constant)

With that I use that to solve Kc which I use an ice table to solve.
I2  +          KI                       I3-
I           0.08           0.15                        0
C          -0.2           -0.2                      +0.2
E          -0.12          -0.5                       0.2

Thus my I3- would be 0.2 and the other concentrations found. Now this isn't the average distribution constant, but I just did this for the sake of doing the calculations to see if I'm on the right track, or am I way off?

By the way, thank you Macman104 for fixing one of my error tags, I was in a rush to get to work didn't get a chance to fix that error. Thank you again :+)

#### Borek

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##### Re: Chemical Equilibrium in Solution: Lab help:
« Reply #6 on: February 28, 2009, 04:26:42 AM »
With that I use that to solve Kc which I use an ice table to solve.
I2  +          KI                       I3-
I           0.08           0.15                        0
C          -0.2           -0.2                      +0.2
E          -0.12          -0.5                       0.2

Where did you get 0.2 from?

Obviously it can't be OK, as you can't have negative concentrations at equilibrium.

(I3-) = T - (I2)

Quote
(I-) = C - (I3-)

What are T and C?
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#### arcanum

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##### Re: Chemical Equilibrium in Solution: Lab help:
« Reply #7 on: March 01, 2009, 01:02:52 AM »
With that I use that to solve Kc which I use an ice table to solve.
I2  +          KI                       I3-
I           0.08           0.15                        0
C          -0.2           -0.2                      +0.2
E          -0.12          -0.5                       0.2

Where did you get 0.2 from?

Obviously it can't be OK, as you can't have negative concentrations at equilibrium.

(I3-) = T - (I2)

Quote
(I-) = C - (I3-)

What are T and C?

I got 0.2 from my distribution constant from group 1 sample 1, which I did my example on the last post. I'm not even sure if that is a 100% correct. I'm trying to think which way I'm suppose to go on this. But I figure that my distribution constant would be the change in each of the factors of KI and I2

And that isn't the average of all the group, matter of fact I'd think that would be the data to throw out. However looking at all the other data the lowest distribution constant (k) I have is 0.059 and that's the only one that seem that would be working for that sort of way. So maybe I'm incorrect in my calculation in k.

Which I'm quite sure that its suppose to be k = aqueous layer mL titrated/ Hexane layer mL titrated.

As for what is T and C

T is suppose to be "The total iodine concentration, T = (I_2) + (I_3-), can be found by titration with a thiosulfate solution." Which I'm guessing that T will equal the I2 + amount of thiosulfate used in each titration?

C is suppose to be If, then one starts with a solution of KI whose concentration C is known accurately and dissolves iodine in it, the iodine will be present at equilibrium partly as neutral I_2 molecules and partly as I_3 - ions (formed when I_2  reacts with some of the I- ions initially present.)

I'm guessing that C is suppose to be just the initial amount of KI in the solution?

The stuff in italics is whats in the book that relative to the equation. I'm still not sure what is suppose to be what, think that's where I'm most lost.

Thanks for the replies. It's best to have someone to discuss this with :+)

• Mr. pH