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Equilibrium Titration? 
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#1
Dec409, 12:35 PM

P: 20

1. The problem statement, all variables and given/known data
Well we have this lab on equilibrium using titration. I found the lab here http://wikieducator.org/Chemistry/Equilibrium_Titration So I'm kinda stuck on number 3c, where it asks what is the starting concentration of Fe^{3+}. I don't know if I did number 3b right either... 2. Relevant equations Everything is on that webpage. 3. The attempt at a solution 1a) 0.05 M b)0.05 M 2)0.0013 mol ( volume of KSCN used was 13mL) 3a) n_{Ag}=n_{KSCN} = 0.0013 mol For 3b) what I did was take the starting moles and subtracted the end number of moles. So I calculated that there were 0.003mol in the starting solution of 0.05M Ag^{+} and then 0.00013mol of Ag^{+} at equilbrium. I then got 0.0017mol as the moles at equilibrium. It is the concentration that I'm not really sure to what I'm doing. So what I did was take 0.05M (concentration of diluted Ag) and then subtracted the concentration of Ag at equilibrium. For the equilibrium concentration, I put 0.0013mol / (0.05L + 0.05L + 0.013L) but I dont think this is right.. What should I do here? For 3c, would the starting amount of Fe^{3+} be zero? Any help will be appreciated. Thanks. 


#2
Dec409, 04:59 PM

P: 37

All right...I've never helped anyone on here before (usually I'm the one asking for help!), but let's see here:
3C) Yes, all products are [0.00] at the very start of a reaction. All right, let's go through all of problem 3. The first question of 3A is correct: (.10moles/1000ml) x 13ml = .0013mols of SCN and thus Ag+; I think you have an extra zero in the second sentence's # of mols. But what about the second part of 3A? The concentration of Ag+? And the volume would be of the actual solution titrated, not the volume of your first solution. Remember how you only took .025L of the solution? The solution was at equilibrium BEFORE you titrated. Now you're just measuring the mols of SCN = mols of Ag+ @ equilibrium. You mixed the two solutions, let it sit (time for it to reach equilibrium), and now you have the solution at equlibrium. Thus any number concentration number gained AFTER you titrated is an equilibrium concentration. Hope this helps? I tried not to solve it for you, but maybe point you in the right direction? ~Ibrahim~ 


#3
Dec409, 05:20 PM

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P: 23,369

Could be I am missing something, but there is something strange here.
Your solution after being mixed was 0.05 M in Ag^{+}. Of that you took 25 mL for titration  that means at most 0.025L*0.05M=0.00125 moles of Ag^{+} (1.25 mmole). Titration used 13 mL of 0.1M KSCN, that means 0.013L*0.1M=0.0013 moles (1.3 mmole of KSCN). That means you have titrated more Ag^{+} that was initially present in the solution! That's impossible even before some of Ag^{+} was consumed by the reaction with Fe^{2+}. Edit: sorry, I was distracted by some other things and I see ikjadoon has posted in the meantime  seems like he has addressed Fe^{3+} concentration earlier.  


#4
Dec409, 05:32 PM

P: 37

Equilibrium Titration?
Oh, no, it's definitely OK. Another set of eyes always helps. :)
I think it took 25ml of a solution of a .1M Ag+/.1M Fe2+. So, .100 mols/1000ml x 25ml = .0025mols of Ag+ initially, .0013mols @ equilibrium. ~Ibrahim~ 


#5
Dec409, 06:31 PM

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#6
Dec409, 09:25 PM

P: 37

I guess you titrated it a bit too much because, even though they're close, I'm guessing that it has a K>1 so the equilibrium concentration would have been even lower.... Thanks for the heads up. ~Ibrahim~ 


#7
Dec409, 11:14 PM

P: 20




#8
Dec509, 04:08 AM

Admin
P: 23,369

First of all  please pay attention to the problem I have pointed, seems like you will be getting negative concentrations.
Concentration is number of moles IN SOLUTION divided by the VOLUME OF THAT SOLUTION. You took 25 mL of the solution for titration, so number of moles determined were in THAT 25 mL.  methods 


#9
Dec509, 11:59 AM

P: 20




#10
Dec509, 12:40 PM

P: 20

Haha, except now my values for Ag+ at eqilibrium are negative because we overtitrated the solution with excess KSCN... I guess I'll be taking the absolute value?



#11
Dec509, 12:44 PM

Admin
P: 23,369

No, you can't take absolute value  it makes about as much sense as using random numbers generator.



#12
Dec509, 12:44 PM

P: 20




#13
Dec509, 12:56 PM

Admin
P: 23,369

There is no value you can use, all you can do is to explain why you can't calculate equilibrium constant using your experimental results.
You may later add something like "assuming volume of the titrant was xx procedure of calculating equilibrium constant would be...".  chemical calculators  , www.titrations.info  all about titration methods 


#14
Dec509, 12:58 PM

P: 20




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