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Author Topic: BACK TITRATION  (Read 4342 times)

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troulla100

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BACK TITRATION
« on: April 15, 2016, 02:27:13 AM »

I have a problem in back-titration where the concentration of iron in the sample have to be found.
100cm3 of water convert the iron present into Fe+2. I know the volume and concentration of K2Cr2O7 and Fe+2.
The reaction is 6Fe+2  + Cr2O7 2-  + 14H+   :rarrow:  6 Fe+3  + 2Cr3+ + 7H2O

I calculated the moles of Fe+2 reacted, the moles of K2Cr2O7 added originally and that reacted but I am stuck and I don't know how to continue. Could you please help me with this?

Thank you
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Burner

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Re: BACK TITRATION
« Reply #1 on: April 15, 2016, 03:30:57 AM »

the concentration of iron in the sample have to be found.

Do you mean 'percentage by mass of iron in the sample'?
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troulla100

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Re: BACK TITRATION
« Reply #2 on: April 15, 2016, 03:36:12 AM »

it is a concentration of iron in sample in ppm that I have to find.
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Burner

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Re: BACK TITRATION
« Reply #3 on: April 15, 2016, 05:15:11 AM »

it is a concentration of iron in sample in ppm that I have to find.

Ok, that's simliar concept.

I calculated the moles of Fe+2 reacted, the moles of K2Cr2O7 added originally and that reacted but I am stuck and I don't know how to continue. Could you please help me with this?

You have found number of moles of Fe2+ reacted, so you can find the number of moles of iron in the sample(which should be equal to that of Fe2+). Than calculate the mass of iron in the sample.
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troulla100

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Re: BACK TITRATION
« Reply #4 on: April 15, 2016, 05:54:38 AM »

Thank you.
I do not have to find the moles of K2Cr2O7 reacted and the total?
How can I find the mass of iron in the sample from the Fe+2 moles reacted?
Thank you
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Borek

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Re: BACK TITRATION
« Reply #5 on: April 15, 2016, 08:04:25 AM »

Please give us all information you have, including complete procedure followed. At the moment there is not enough information to calculate anything, and there is nothing in your posts that explains why it is a back titration (as opposed to the normal titration).

Plus, ppm levels are typically way too low for titrimetric methods.
« Last Edit: April 15, 2016, 08:22:59 AM by Borek »
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troulla100

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Re: BACK TITRATION
« Reply #6 on: April 15, 2016, 08:27:30 AM »

100cm3 sample of water used to convert any iron present to Fe+2. Addition of K2Cr2O7 leads to the equation written to the previous post. Then the excess K2Cr2O7 was back-titrated with Fe+2 Solution.
and the concentration of iron in the sample need to be found.

Thank you!
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Borek

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Re: BACK TITRATION
« Reply #7 on: April 15, 2016, 11:17:02 AM »

That's the procedure, what is the data? It is still not clear where you got stuck and why.
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troulla100

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Re: BACK TITRATION
« Reply #8 on: April 15, 2016, 01:08:39 PM »

I found the number of moles of Fe+2 reacted in the titration. then, I used the ratio of reactants 6:1 and I found the number of moles of Cr2O72- which is the amount in excess. Then, I found the total number of moles of K2Cr2O7 originally added to the diluted solution. Then I found the amount of moles of K2Cr2O7 added and titrated. And then, I stucked there. How can I used the 100cm3 of water and how to proceed.
Thank you.
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Burner

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Re: BACK TITRATION
« Reply #9 on: April 15, 2016, 03:42:04 PM »

I found the number of moles of Fe+2 reacted in the titration. then, I used the ratio of reactants 6:1 and I found the number of moles of Cr2O72- which is the amount in excess. Then, I found the total number of moles of K2Cr2O7 originally added to the diluted solution. Then I found the amount of moles of K2Cr2O7 added and titrated. And then, I stucked there. How can I used the 100cm3 of water and how to proceed.
Thank you.

You have calculated the number of moles of Cr2O72- reacted with the Fe2+ in the sample . The next step is to calculate the number of moles of Fe2+ from the sample.

100cm3 water is just used to convert all the Fe in the sample to Fe2+. Since you are calculating the ppm of iron in the solid sample, you don't need the number 100cm3 in your calculation.
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troulla100

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Re: BACK TITRATION
« Reply #10 on: April 15, 2016, 11:07:52 PM »

Thank you so much about that.
To find the number of moles of Cr2O72- I used the equation n(K2Cr2O7 titrated) + n(K2Cr2O7 reacted) = n(K2Cr2O7 total added originally). So, I found the number of moles of K2Cr2O7 reacted. So, now to find the number of moles of Fe2+ from the sample I have to use the ratio 6:1 with the K2Cr2O7 reacted? or the number of moles of Fe2+ from the sample is the amount that I calculated in the very beginning which was the amount reacted in the titration?(from which I used the ratio 6:1 to find the moles of K2Cr2O7 titrated-in excess)?
I am sorry for being so confusing writing
Thank you
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Burner

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Re: BACK TITRATION
« Reply #11 on: April 15, 2016, 11:22:47 PM »

So, now to find the number of moles of Fe2+ from the sample I have to use the ratio 6:1 with the K2Cr2O7 reacted?

Yes
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troulla100

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Re: BACK TITRATION
« Reply #12 on: April 15, 2016, 11:24:52 PM »

So, I have to used again the 6:1 ratio with the moles of K2Cr2O7 reacted to find the Fe2+ in the sample?
Thank you
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troulla100

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Re: BACK TITRATION
« Reply #13 on: April 15, 2016, 11:52:37 PM »

Thank you so so much about the help. Now seems so clear to me. thank you.
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Amanyy

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Re: BACK TITRATION
« Reply #14 on: December 29, 2018, 01:27:10 AM »

Can you tell plz what you did to solve this question?!!
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