Register to reply

Equilibrium Titration?

by fishfish
Tags: equilibrium, titration
Share this thread:
fishfish
#1
Dec4-09, 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 Fe3+. 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) nAg=nKSCN
= 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 Fe3+ be zero?

Any help will be appreciated. Thanks.
Phys.Org News Partner Science news on Phys.org
Sapphire talk enlivens guesswork over iPhone 6
Geneticists offer clues to better rice, tomato crops
UConn makes 3-D copies of antique instrument parts
ikjadoon
#2
Dec4-09, 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~
Borek
#3
Dec4-09, 05:20 PM
Admin
Borek's Avatar
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 Fe2+.

Quote Quote by fishfish View Post
For 3c, would the starting amount of Fe3+ be zero?
Yes.

Edit: sorry, I was distracted by some other things and I see ikjadoon has posted in the meantime - seems like he has addressed Fe3+ concentration earlier.

--

ikjadoon
#4
Dec4-09, 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~
Borek
#5
Dec4-09, 06:31 PM
Admin
Borek's Avatar
P: 23,369
Quote Quote by ikjadoon View Post
I think it took 25ml of a solution of a .1M Ag+/.1M Fe2+.
Solution was prepared by mixing 50 mL of 0.1 M Ag+ and 50 mL of 0.1M Fe2+ - final volume 100 mL, initial concentration of both Ag+ and Fe2+ 0.05M, they were both effectively diluted twice.

--
ikjadoon
#6
Dec4-09, 09:25 PM
P: 37
Quote Quote by Borek View Post
Solution was prepared by mixing 50 mL of 0.1 M Ag+ and 50 mL of 0.1M Fe2+ - final volume 100 mL, initial concentration of both Ag+ and Fe2+ 0.05M, they were both effectively diluted twice.
Whoops, you're definitely right. It's a .05M solution that they took 25ml, not a .1M solution. Good call.

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~
fishfish
#7
Dec4-09, 11:14 PM
P: 20
Quote Quote by ikjadoon View Post
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~
So for the concentration of Ag+, I'm assuming that it's the number of moles of Ag+ (0.0013mol) divided by the ENTIRE volume of the solution? Which is 0.05L of Ag + 0.05L of Fe + 0.013L of SCN ?
Borek
#8
Dec5-09, 04:08 AM
Admin
Borek's Avatar
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
fishfish
#9
Dec5-09, 11:59 AM
P: 20
Quote Quote by Borek View Post
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.

--
chemical calculators - buffer calculator, concentration calculator
www.titrations.info - all about titration methods
OH! Now I see! Thanks !! :D
fishfish
#10
Dec5-09, 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?
Borek
#11
Dec5-09, 12:44 PM
Admin
Borek's Avatar
P: 23,369
No, you can't take absolute value - it makes about as much sense as using random numbers generator.
fishfish
#12
Dec5-09, 12:44 PM
P: 20
Quote Quote by Borek View Post
No, you can't take absolute value - it makes about as much sense as using random numbers generator.
Then what value should I take?
Borek
#13
Dec5-09, 12:56 PM
Admin
Borek's Avatar
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
fishfish
#14
Dec5-09, 12:58 PM
P: 20
Quote Quote by Borek View Post
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...".
Alrighty, I'll do that then :) Thanks so much!


Register to reply

Related Discussions
Ingredient in aspirin is acetylsalicylic acid Biology, Chemistry & Other Homework 3
I figured out the problem myself. Thanks anyways Biology, Chemistry & Other Homework 1
Chemical equilibrium: Enthalpies and equilibrium contsants Biology, Chemistry & Other Homework 0
Titrate sulphuirc acid with sodium hydroxide Chemistry 28
Titration chemistry homework Introductory Physics Homework 0