Sciencemadness Discussion Board - Iodometry titrating for Fe3+ content

Iodometry titrating for Fe3+ content

CHRIS25 - 15-5-2014 at 08:49

Object: to accurately assess amount of Fe²⁺ and Fe³⁺ in a solution of Iron Chloride.

In all the information I read about adding Potassium Iodide to the unknown analyte I read only that an excess should be added. This is vague for two reasons:
1. How do you know when you have added excess to the Iron chloride solution?
2. What kind of molarity KI would generally be used?
and finally
3. What kind of molarity of sodium thiosulphate would generally be considered reasonable as a starter?

I need to have some idea about where to begin otherwise I could be wasting chemicals and repeating mistakes without knowing why. This sort of information would be given by an instructor and in the absence of that I only have PDF's that give excellent and thorough explanations about what is happening and why, but do not really help me to at least have a baseline from which to do this for the very first time. (and No I will not be standardizing the thiosulphate solution with iodate, and will not be using some carbonate to stabilize the thiosulphate at this time).
Thankyou.

blogfast25 - 15-5-2014 at 09:43

Quote: Originally posted by CHRIS25

1. How do you know when you have added excess to the Iron chloride solution?
2. What kind of molarity KI would generally be used?
and finally
3. What kind of molarity of sodium thiosulphate would generally be considered reasonable as a starter?

A.1. You need to have some rough idea of the Fe3+ concentration in your sample. It should be somewhere between 0.05 M and 0.15 M, ideally. If the concentration is much higher you will have to dilute the original sample to fall between the limits of 0.05 and 0.15 M, using a KNOWN dilution factor. Then use 20 ml of this diluted solution as the sample to be titrated.

A.2. The exact molarity of the iodide is not important but assuming you'll be titrating an Fe3+ solution of about 0.1 M and that you add the iodide as a 20 ml aliquot, it will have to be about 0.2 M to ensure there's an excess iodide, in accordance with 1.

A.3. The concentration of thiosulphate I would recommend here is 0.1 M (some may use 0.05 M or even 0.01 M). If the concentration of Fe3+ in 20 ml of sample was about 0.1 M, you will need about 20 ml of titrant solution to reach end point.

But imagine the Fe3+ concentration was 0.5 M? The analysis would not work at all.

The approx. initial Fe3+ concentration can usually be worked out from the origin of the solution. Otherwise it becomes a case of trial and error.

[Edited on 15-5-2014 by blogfast25]

CHRIS25 - 15-5-2014 at 10:03

A1 --- That is something I would have missed, but I presume this is because titrating iodine doe snot work at all in strongly acidic solns? But I would never have guessed to have diluted it so much so Thankyou.

A2 --- All in all then Something that was completely absent from every doc I ever read, keep the Molarity of Fe soln low.

A3 --- Thankyou.

I know my present soln is around 3M to 4M and I know the amount of grams of iron, but I want to learn this method of titrating. Also since Iodide is oxidized by the Fe³⁺ and is itself then reduced to Fe²⁺; it seems that any Fe²⁺ that happen to be in that soln already would not be able to be calculated?

blogfast25 - 15-5-2014 at 10:16

The Fe3+ concentration needs to be in the right area because otherwise your burette (25 ml) can't dispense enough titrant to react with all the iodine. Since as you're using 0.1 M thiosulphate, the Fe3+ concentration in your sample should be in the same order of magnitude.

To dilute the 3 - 4 M original solution, take 5.0 ml and dilute it accurately to 250.0 ml (really you should use a suitable pipette and volumetric flask for that task!). Or 10.0 ml to 500.0 ml. The dilution factor is then 50. Suppose your analysis came out as (e.g.) 0.0775 M on the diluted sample, then original sample was 50 x 0.0775 M = 3.875 M.

Fe2+ cannot be detected with iodometry, unless you oxidise it to Fe3+, prior to the titration.

But for now stick to the Fe3+. Then we can work something out for the Fe2+... For instance titration of the Fe2+ with potassium permanganate.

[Edited on 15-5-2014 by blogfast25]

Could someone Double check for me please

CHRIS25 - 23-5-2014 at 10:06

Method:

From a theoretical/suspected 3M soln of FeCl₃ I extracted 0.1M. This was 4mLs.
Added 4mLs of FeCl₃ to 116mLs water = Theoretical 0.012moles

Prepared KI 0.2M (0.004moles in 20mLs water)
Prepared S₂O₄ 0.1M (0.0035moles in 35mLs water)
Prepared 0.75g corn flour in 100mLs boiling water, let cool, decanted the supernatant fluid

Titration Results:

When soln turned very light yellow corn flour was added and soln went deep blue. titration continued until deep blue disappeared at 17.5mLs. However, although opaque at this stage the solution was not entirely colourless, continued until 26mLs had been added and then the soln turned truly colourless with the clarity of water. I assume it is this mark that I should be calculating?

If so then I fall flat on my face with these calculations;
26mLs x 0.1M S₂O₄ / 4mLs FeCl₃ = 0.65M
0.65M x 30 dilution factor = a very very concentrated FeCl₃ soln of 19.5M???

Ok, I have really done my best to ensure complete accuracy, something is wrong and I simply can not see it.
(Assuming a 3M concentration in 125mLs that I have then the maximum can not be more than: 0.125 x 3 = 0.375moles of Fe which is close to the 0.5 moles that I originally started with; 19.5M gives me 2.4moles which is not possible).

[Edited on 23-5-2014 by CHRIS25]

[Edited on 23-5-2014 by CHRIS25]

[Edited on 23-5-2014 by CHRIS25]

DraconicAcid - 23-5-2014 at 11:05

Quote: Originally posted by CHRIS25

Method:

From a theoretical/suspected 3M soln of FeCl₃ I extracted 0.1M. This was 4mLs.
Added 4mLs of FeCl₃ to 116mLs water = Theoretical 0.012moles

Prepared KI 0.2M (0.004moles in 20mLs water)
Prepared S₂O₄ 0.1M (0.0035moles in 35mLs water)
Prepared 0.75g corn flour in 100mLs boiling water, let cool, decanted the supernatant fluid

Titration Results:

When soln turned very light yellow corn flour was added and soln went deep blue. titration continued until deep blue disappeared at 17.5mLs. However, although opaque at this stage the solution was not entirely colourless, continued until 26mLs had been added and then the soln turned truly colourless with the clarity of water. I assume it is this mark that I should be calculating?

If so then I fall flat on my face with these calculations;
26mLs x 0.1M S₂O₄ / 4mLs FeCl₃ = 0.65M
0.65M x 30 dilution factor = a very very concentrated FeCl₃ soln of 19.5M???

Ok, I have really done my best to ensure complete accuracy, something is wrong and I simply can not see it.
(Assuming a 3M concentration in 125mLs that I have then the maximum can not be more than: 0.125 x 3 = 0.375moles of Fe which is close to the 0.5 moles that I originally started with; 19.5M gives me 2.4moles which is not possible).

[Edited on 23-5-2014 by CHRIS25]

[Edited on 23-5-2014 by CHRIS25]

[Edited on 23-5-2014 by CHRIS25]

Ignore the dilution factor, since you titrated the whole thing anyway.

Okay, if the blue colour vanished at 17.5 mL, and your thiosulphate was 0.1 M, then you used 1.75 mmol thiosulphate. A 1:1 ratio means that you had 1.75 mmol Fe³⁺, which came from 4 mL solution, so that's 0.44 M Fe³⁺.

[Edited on 23-5-2014 by DraconicAcid]

CHRIS25 - 23-5-2014 at 11:36

So my maths, as is usual, came adrift. I get it thankyou, so waiting till the blue solution has gone and is clear, as is what I have read, is a little misleading, one waits untill the blue disappears, regardless of the clarity. Dilution? That was stupid of me. Thankyou sincerely, Draconic.

Except this means my titration is off somewhere. 0.44M means only 0.05moles of Iron - this is way off. Or not all the ferrous has been oxidized? But even with this factor considered and the colour of the solution all indicating oxidation of ferrous is more than 80% there.

Another titration tomorrow....

[Edited on 23-5-2014 by CHRIS25]

blogfast25 - 23-5-2014 at 13:50

Quote: Originally posted by CHRIS25

Method:

From a theoretical/suspected 3M soln of FeCl₃ I extracted 0.1M. This was 4mLs.
Added 4mLs of FeCl₃ to 116mLs water = Theoretical 0.012moles

Unfortunately both last sentences already don't make any sense. You can't 'extract 0.1M'.

Did you take 4 ml (four millilitres) of 3 M FeCl3 solution? If so what did you do with that?

Did you dilute that to 120 ml, i.e. '4mLs of FeCl₃ added to 116mLs water'? Is that what you did?

General notational advice: ALWAYS, ALWAYS, ALWAYS separate a number from its unit of measurement by a space: e.g. 3 M, 5.6 g, 12 ml, 5.2 mol/L, 3.2 moles etc.

Quote: Originally posted by CHRIS25

Prepared S₂O₄ 0.1M (0.0035moles in 35mLs water)

S₂O₄ 0.1M ?? Did you mean 'Na2S₂O₃ 0.1 M'? Preparing only 35 ml of that would be folly: far too inaccurate.

[Edited on 23-5-2014 by blogfast25]

CHRIS25 - 23-5-2014 at 16:13

====Did you take 4 ml (four millilitres) of 3 M FeCl3 solution? If so what did you do with that?

====Did you dilute that to 120 ml, i.e. '4mLs of FeCl3 added to 116mLs water'? Is that what you did?

Yes this is what I did.

==== ALWAYS, ALWAYS, ALWAYS ok point absorbed.

====S2O4 0.1M ?? Did you mean 'Na2S2O3 0.1 M'? Preparing only 35 ml of that would be folly: far too inaccurate.

woa I have no idea why I made such a bloody ignorant mistake (my actual notes have the correct formula); but yes, obviously I used sodium thiosulphate pentahydrate

well two changes then to tomorrows titration; first I will make a 100 mL thiosulphate soln, and second, I will add the iodide, directly, as anhydrous into the ferric soln, and not make a solution of potassium iodide and then add this.

[Edited on 24-5-2014 by CHRIS25]

blogfast25 - 24-5-2014 at 03:42

Chris:

From the dilution (dilution factor = 120 / 4 = 30), which amount did you then titrate?

Definitely not 4 ml, right? This where your error lies: in the calculation you need to use that number:

vthiosulphate x Cthiosulphate = vdiluted sample x Cdiluted sample with vdiluted sample the volume of diluted sample you actually subjected to titration.

Then multiply Cdiluted sample with the dilution factor to get the concentration of the original, undiluted sample.

What you need is a couple of decent (second hand will do) 250.0 ml Class A volumetric flasks and ditto 10.0 and 20.0 ml bulb pipettes and you'll soon be up there with the pros!

CHRIS25 - 24-5-2014 at 04:59

Now I am confused.

1. I do not know the concentration of the diluted sample since I do not know the number of moles in the 4 mL sample taken. It was 4 mLs + 116 mLs water. And since I titrated into this whole solution, (not a part of it) I thought that what Draconic said was right - I do not need to bother with concentration. In fact when I titrate sodium hydroxide into a diluted copper chloride solution I never calculate the dilution factor into it, and luckily I saw many demonstrations that did not calculate concentration in this instance (copper chloride etching process).

2. I do not know what you mean by the word "with" in (with v diluted sample); which mathematical notation is the 'with' referring to?

3. I will deal with extra lab stuff later, but for now I have to use syringes for complete accuracy, (they are so much cheaper than bulb pipettes), and since they are from a veterinary source I can hardly doubt their accuracy for my needs at this moment in time.

blogfast25 - 24-5-2014 at 08:31

Chris:

Normally one would never titrate the entire dilution but only a known part of it, typically 20.0 ml. Otherwise you risk having to use far, far too much titrant solution to be practical.

Anyway, you titrated the whole 120 ml (containing the entire 4 ml of sample), so lets work with that.

You used 26 ml of 0.1 M Na2S2O3, so that you calculate the number of mol thiosulphate used being (0.026 L x 0.1 M) = 0.0026 mol.

But already there's a snake in the grass.

The titration reaction is:

S2O3(2-) + I2 === > 1/2 S4O6(2-) + 2 I-

And the original reaction:

Fe3+ + I- === > Fe2+ + 1/2 I2

In other words, each mol of thiosulphate titrates for TWO (2) mol of Fe3+ !!

So you need to double that 0.0026 mol of thiosulphate to 0.0052 mol of Fe3+. Now divide that by the volume in which they were contained: 4 ml or 0.004 L: 0.0052 mol / 0.004 L = 1.3 M Fe3+

Still quite a distance from the suspected 3 - 4 M. I'll later look at the amount of KI added, to see if it is correct. How much of the 0.2 M KI solution did you add to your titrated sample?

Two general remarks: the doubling you overlooked is why experienced titrators use NORMALITY instead of MOLARITY.

Secondly, using syringes and such like is fine but difficult. Experienced titrators use a simple protocol which involves Normal solutions (of correct strength), a 10 or 25 ml burette, volumetric flasks and calibrated pipettes, because small errors soon mount to quite a lot and it can be difficult finding out what went wrong. QED.

Take this for example: "Prepared S2O4 0.1M (0.0035moles in 35mLs water)". That's 0.86863 g of sodium thiosulphate pentahydrate, difficult to accurately weigh without using 0.1 mg scales.

I just prepared a 0.1 N Na2S2O3 solution, 250.0 ml of it. That's about 3.1 g of sodium thiosulphate pentahydrate, which I can accurately enough weigh with my 1 mg scales (measuring error is about 0.3 %)

[Edited on 24-5-2014 by blogfast25]

CHRIS25 - 24-5-2014 at 08:56

I stared at the balanced equation for this reaction and saw that one mol of FeCl₃ resulted in half the amount of iodine but I assumed something else wrongly in my thinking.
Yes you are right, I weighed out 0.9g of thiosulphate because my scales do not go above one decimal place. (cheap ebay), anyway I realise now that measuring in such small quantities magnifies errors more than measuring in large quantities. (a case of economics at home); but I will do the titration again with larger amounts and with probably 10g of ferric solution and use 0.15 M of thiosulphate extracted from 100 mLs I suppose.
I poured all 20 mLs of the 0.2 M KI into the ferric solution, this I measured out at 0.66g but put in 0.7g and then pinched a bit out of the container in an attempt to lower it closer to 0.6 while still on the scales, (scales again yes I know new scales would be better). The mistakes you have highlighted though - thanks.

[Edited on 24-5-2014 by CHRIS25]

blogfast25 - 24-5-2014 at 09:09

Ouch: 1 decimal place! Too little. The only way to use that quasi accurately is by doing everything on a 10 x scale factor. E.g. 9 g of thio in 10 x the amount of water. A totally false economy.

2 digits isn't great but it's workable, the ones I sell cost just under £10. Can't go wrong with that. Better: 0.001 g (about £50, I think).

Not sure what you mean by: "[...] and use 0.15 M of thiosulphate extracted from 100 mLs I suppose."

[Edited on 24-5-2014 by blogfast25]

CHRIS25 - 24-5-2014 at 09:14

Actually I just measured 0.15 M ready for 200 mLs of water; extracted? I meant merely that I am making a beaker with this amount ready to use, 7.44g in 200 mLs = 0.03 m. 10 pounds? Mmm.

[Edited on 24-5-2014 by CHRIS25]
on a side note yes it is nice to have good tools for the job, but I suppose for me at the moment the learning and doing is so much more important, even if accuracy is temporarily sacrificed - the practise with what I have is important.

[Edited on 24-5-2014 by CHRIS25]

blogfast25 - 24-5-2014 at 09:17

0.15 M, so you mean 0.15 mol thio per litre. Why are you increasing it from 0.1 to 0.15 M? It would be slightly more to weigh, so a little more accurate. Is that the purpose?

CHRIS25 - 24-5-2014 at 09:22

Quote: Originally posted by blogfast25

0.15 M, so you mean 0.15 mol thio per litre. Why are you increasing it from 0.1 to 0.15 M? It would be slightly more to weigh, so a little more accurate. Is that the purpose?

Yes I thought a little more leads to just a little more accuracy?
0.15 M x 0.2L = 0.03 mol. 0.03 mol x 248 g/mol = 7.44g

blogfast25 - 24-5-2014 at 09:30

Quote: Originally posted by CHRIS25

Quote: Originally posted by blogfast25

0.15 M, so you mean 0.15 mol thio per litre. Why are you increasing it from 0.1 to 0.15 M? It would be slightly more to weigh, so a little more accurate. Is that the purpose?

Yes I thought a little more leads to just a little more accuracy?
0.15 M x 0.2L = 0.03 mol. 0.03 mol x 248 g/mol = 7.44g

Yes, it's a little better: 0.1 / 7.4 x 100 % = 1.8 % relative measuring error. Much better than 0.1 / 0.9 x 100 % > 10 %.

Spare a thought for the men and women who determined atomic masses to 5 decimal places!

[Edited on 24-5-2014 by blogfast25]

CHRIS25 - 24-5-2014 at 09:36

Relative measuring error is something totally new to me. Is it a standard that chemists use? And the way you demonstrated above is the maths for this?

blogfast25 - 24-5-2014 at 09:40

Quote: Originally posted by CHRIS25

Relative measuring error is something totally new to me. Is it a standard that chemists use? And the way you demonstrated above is the maths for this?

The more I do error analysis, the more I realise just how damn hard it is to measure ANYTHING with known and sufficient accuracy.

The relative measuring error is just a basic comparative tool.

For more details on error analysis, consult some basic texts on the Tinkerwebs. It ranges from fairly basic algebra to full blown 'Monte Carlo' statistical analysis.

[Edited on 24-5-2014 by blogfast25]

CHRIS25 - 24-5-2014 at 09:41

ok, thanks , but 0.1 / 7.4 = 0.0135, 0.0135 x 100 = 1.35?

blogfast25 - 24-5-2014 at 09:43

Quote: Originally posted by CHRIS25

ok, thanks , but 0.1 / 7.4 = 0.0135, 0.0135 x 100 = 1.35?

Yup. Typo on my calculator. Another rocket that would go after its creator! ;-)

[Edited on 24-5-2014 by blogfast25]

aga - 24-5-2014 at 12:07

Quote: Originally posted by blogfast25

Spare a thought for the men and women who determined atomic masses to 5 decimal places!

How was that done ?

I have also searched for exactly HOW chemical forumulae and compound composition were deduced in the 1800s and have drawn a blank.

Any pointers gratefully received.

[Edited on 24-5-2014 by aga]

blogfast25 - 24-5-2014 at 12:27

Quote: Originally posted by aga

Quote: Originally posted by blogfast25

Spare a thought for the men and women who determined atomic masses to 5 decimal places!

How was that done ?

I have also searched for exactly HOW chemical forumulae and compound composition was deduced in the 1800s and have drawn a blank.

Any pointers gratefully received.

[Edited on 24-5-2014 by aga]

When French chemist Louis-Nicolas Vauquelin in 1798 discovered beryllium (in the mineral Beryl), he determined the atomic mass to be about 9. He was right but today we know this with astonishing accuracy: 9.0121831 g/mol. How did we get from simply "9" to 9.0121831? That is a very long story of technological evolution and ever refining of available analytical and measurement techniques and not something that can be easily summarised here.

Similarly the story of stoichiometry, that goes back almost to the ancient Greeks who were the first to propose the idea of an atom, after which the idea went dormant for several centuries.

Gradually and over several hundreds of years, competing 'philosophies' (as they were then called) vied to provide the best explanations for the bewildering amount of compounds and materials we see in the world. Ultimately the so-called 'atomists' won and the notion that chemical elements combine into compounds in fixed ratios became established.

That part of the story, intertwined with the first is a picture so large that you can't even fit it onto a single canvass (so to speak).

The history of science is almost as compelling as the science itself. Today there's no great shortage of great books that popularise the history of physics and chemistry. I'll see if I can rustle up some great titles.

In terms of popular science history shows, I like Jim al Khalili's 'Chemistry: a Volatile History' a lot. Regularly repeated on Beeb 4.

[Edited on 24-5-2014 by blogfast25]

aga - 24-5-2014 at 12:52

I have just searched for Louis-Nicolas Vauquelin and again learnt nothing as to HOW the chemical compositions or atomic weights were deduced.

Am i being thick ?

Can everyone else pick up a lump of charcoal and say :-

'Hmm. Thats Carbon that is. Atomic weight's about twelve i reckon. Dunt feel like 14 isotope to me, more like 12' ?

blogfast25 - 24-5-2014 at 13:06

Quote: Originally posted by aga

Am i being thick ?

No, but as I tried to explain, it's a difficult question because there's no single answer. Historically, just to get to the notion that there is an atomic weight is a serious chunk of scientific history. It's not a simple 'trick' that can be revealed here in 5 minutes.

Here's someone who bravely tries to summarise the history of atomic mass: 'The mess with the mass':

http://h2g2.com/entry/A2911259

[Edited on 24-5-2014 by blogfast25]

Good News

CHRIS25 - 25-5-2014 at 01:21

Lessons Learned: Double check your notes!!!!! I had 70 mL of ferric chloride not 125. My last entry in my notes, but alas I continued to evaporate the solution and forgot to note new volume before storing.

Anyway I titrated with amounts previously written, I added iodide as anhydrous not in a solution.

Results: 56 mLs thiosulphate (I took the measurement the second the blue disappeared, I then added a further 55 mLs of thiosulphate just to see if the soln would go to clarity and it did not).
So 56 x 0.15 M = 0.0084 (x2) = 0.0168 mol thiosulphate titrated. Then 0.0168 x 0.010 mLs chloride soln = 1.68 M Fe³⁺ in 70 mLs of chloride solution.

But this begs the question that this means: 1.68 M x 0.07 L = 0.12 mols of Ferric (not ferrous) in solution. I added 29g 0.5 mol originally. Am I missing something here or is my titration still way up the creek? The ferric is well and truly oxidized without peroxide, it is a deep yellow against the sides of the glass but otherwise to look at a very deep brown/black. So there can not be much ferrous left surely?

[Edited on 25-5-2014 by CHRIS25]

aga - 25-5-2014 at 03:01

That helps enormously.
thankyou.

blogfast25 - 25-5-2014 at 05:48

Chris:

If I understand well you now took 10 ml (not 'mLs', which could stand for 'minor Luddite syndrome' but not much else ; -) ) and titrated that directly w/o any dilution? A minor problem with that is that the solution may be a bit too acidic for a thiosulphate titration but we'll assume it isn't. But it is the drawback of not diluting before titration. Maybe you did add some water to the 10 ml original sample?

So you titrated in 56 ml of thiosulphate 0.15 M to end-point (yes, the first disappearance of the blue is your end-point), which is 0.056 L x 0.15 mol/L = 0.0084 mol thiosulphate, equivalent to 0.0168 mol Fe3+. Contained in 10 ml that gives 0.0168 mol / 0.010 L = 1.68 M.

I think your titration is correct (but w/o any error analysis). The missing Fe remains a bit of a mystery. Considering the huge error margin with the thiosulphate solution in the first attempt, both results are roughly in line: your ferric chloride solution is definitely NOT 3 - 4 M.

Now here's an important point: you complain about turbidity (non-clarity). When does this first appear? Was the original sample clear? Does the turbidity occur during titration and what does it look like?

Another point is the use of 56 ml of titrant solution: that is a HUGE volume by most titrators standards. That is why we usually dilute the original sample and then only titrate a smaller, known volume of it. Are you using syringes for the injection of whales or elephants? ;-) Multiple additions of full syringes also introduces error...

[Edited on 25-5-2014 by blogfast25]

CHRIS25 - 25-5-2014 at 06:03

No, I diluted with 100 mLs of water (mLs was watson fawkes correction of my very first postings where I used Mls back then). Anyway I added the titrant to the whole lot. Also I used the burette now that I have one. The missing iron is a mystery, really who pinched it? So, the turbidity, it was opaque, you could quite clearly see through the titrated soln, it was simply not the clarity of water, it had an extremely subtle blue silvery tint, now this has to be something to do with taking too long? After completion, I left the titration soln to stand just to observe a thick milky white turbidity develop within about 5 minutes. I have to read up on this because I know I have read an explanation somewhere but can not remember. I do know that the whole titration has to be done quickly, but this time I took more time with the whole process, probably about 10 minutes.

[Edited on 25-5-2014 by CHRIS25]
Sorry - to answer another question, the original sample was a nice bright yellow as expected with dilution and the colour of ferric ions in this.

[Edited on 25-5-2014 by CHRIS25]
Darn it, that 3 M approximation was in actual fact acid reserve as I go through my notes, not Molarity of Fe. I do admit confusion here, but I originally began with 1.9 M HCl (and increased it) and 0.5 mol Fe filings. What exactly am I measuring? it is the amount of oxidised Fe in soln. So if it is 1.68M = 0.12 mol of Fe³⁺ then that surely must mean that the soln has a long way to go before the ferrous oxidises? I can not logically deduce it any other way based on the fact that Iron simply can not evaporate.

[Edited on 25-5-2014 by CHRIS25]

[Edited on 25-5-2014 by CHRIS25]

blogfast25 - 25-5-2014 at 07:11

To be entirely SC ('scientifically correct') a ml is one cm3 ('one cubic centimeter') and 1 L should be one dm3 ('one cubic decimeter') but unfortunately that hasn't caught on so much and old habits die hard.

The milky white turbidity is elemental sulphur but it can only arise from excess thiosulphate in acid conditions, unless the S4O6(2-) reaction product also shows that behaviour. Must look into that...

If you started from 0.5 mol Fe filings, and assuming no loss of iron and all iron as Fe3+ then estimating the Fe3+ concentration is only a mater of knowing the final volume of the solution after all your manipulations: [Fe3+] = mol Fe3+ / Final Volume (in L), so go back to your notes.

You're using the burette: serious progress!

But 10 min is quite a long time, there's a risk of some of the iodide being oxidised by air, back to iodine. 2 - 3 mins per titration, tops.

[Edited on 25-5-2014 by blogfast25]

CHRIS25 - 25-5-2014 at 07:33

Yes I did that this morning = 7 M. So what does that say about this titration then? Where exactly is the point in this, What does it say? Why the huge discrepancy, I do not understand - times like this when you wish you were in college.
unless the S4O6(2-) reaction product also shows that behaviour.
will look into this also

[Edited on 25-5-2014 by CHRIS25]

blogfast25 - 25-5-2014 at 08:19

Quote: Originally posted by CHRIS25

Yes I did that this morning = 7 M. So what does that say about this titration then? Where exactly is the point in this, What does it say? Why the huge discrepancy, I do not understand - times like this when you wish you were in college.

7 M would mean a final volume of about 71 ml: 0.5 mol / 0.071 L = 7 M. That sounds waaaayyy too high to me. 7 M is 1096 g FeCl3 / L !!!! The solubility limit of FeCl3.6H2O is only 918 g/L. Such solutions would be very, very dark, almost black (I have a saturated solution of FeCl3 in my lab: it's like black syrup!)

Such high concentrations could only be achieved by painstaking evaporation of a more dilute solution: they CANNOT be achieved directly by dissolving iron in acid.

Your titrations are obviously a bit flawed but not that much.

Something is wrong at the level of your concentration estimate. Either you used much less iron than you seem to remember or something like that.

To be able to titrate a 7 M FeCl3 solution with a 0.15 M thiosulphate solution (via iodide, of course), you would have to dilute the sample 25 times, for instance 10 ml sample in 250 ml water. Then take 20.0 ml of that dilution, add iodide and titrate with 0.15 M sodium thiosulphate. That should give you an end point AROUND 20 ml.

[Edited on 25-5-2014 by blogfast25]

blogfast25 - 25-5-2014 at 09:20

Just looking at the amount of KI needed.

Assuming you're right about the approx. 7 M, you took 10 ml = 0.01 L of that, so it contains 0.07 mol Fe3+.

Acc. Fe3+ + I- === > Fe2+ + 1/2 I2

... you need to add AT LEAST 0.07 mol iodide or 0.07 mol x 166 g/mol = 11.6 g KI. Anything less and you'll get a faulty (too low) reading. But such a titration would require a whopping 230 - 240 ml of 0.15 M sodium thiosulphate to reach end point!!

[Edited on 25-5-2014 by blogfast25]

CHRIS25 - 25-5-2014 at 09:31

Right. Determination here to sort this out. My notes and my memory agree, 28 grams iron, then two filtrations and another 2 grams iron due to a little loss during both filterings so I added a gram each time. half a mol in 125 mLs HCl and then a further 40 mLs later on. This is just a summary. I know it was 28g because I need that amount for something else. So another titration, different and larger measurements - BIG ones and we will see what happens this time.

blogfast25 - 25-5-2014 at 09:43

Quote: Originally posted by CHRIS25

BIG ones and we will see what happens this time.

No, don't do BIG ones, it won't help much.

Try what I wrote above.

Take a 250 ml beaker and put in 10.0 ml of the concentrated solution. Fill up to the 250 ml mark and stir well.

Now transfer 20.0 ml of that dilution into your conical flask, adding another 50 ml or so of water. Add 1.2 g of KI and allow to dissolve and react for a few minutes, constantly swirling that flask.

Titrate this with 0.15 M thiosulphate from your burette, adding starch towards the end. The end point should be in the region of 17 to 22 ml of 0.15 M thiosulphate, based on 7 M Fe3+ in the original sample. See what you find and we'll calculate the result later.

Determination? That's what we like to hear!

[Edited on 25-5-2014 by blogfast25]

Three titrations done

CHRIS25 - 27-5-2014 at 04:48

Sorry took so long, had 11 hours and a 585 km round trip yesterday so too tired, this morning re-freshed...

Firstly Gert, I can not work out mathematically how to calculate the dilution factors into the stoichiometry so please show me (I mean only from no.1 titration, the last two I can do). I can do maths so long as there are examples but can not work out mathematical concepts to save my life. So here are the results from 3 titrations (complicated by dilutions):
1. Extracted 20 mLs Ferric Chloride from 250 mLs water (this was 20 mLs added to 240 so a dilution factor of 24); Then added 50 mLs to the 20 mL sample and titrated. endpoint at 3.5/4.0 mLs thiosulphate. (this reading is suspect because endpoint reached far too quickly and starch did not turn soln blue so I was passed the endpoint, I did not shake the flask quick enough as it dripped in).

Lesson learned carry on....

2. 30 mLs ferric chloride extracted from same diluted 250 mL solution. NO extra water added this time. 5.5 mLs thiosulphate reached endpoint.
3. Same again, 30 mLs ferric chloride extracted and 5.3 mLs thiosulphate to reach endpoint.

As I said I am thick at maths without first an example to follow when it gets to diluting now. So your explanation would be extremely appreciated.

Just so that you are clear I used 1.2 g iodide, 0.15M thiosulphate and the dilutions that you asked only on the first titration, second and third titrations were as you said, except without the extra 50 mLs water After that I had extracted the ferric sample from the 250mLs (10 mLs ferric chloride + 240 mLs water)

[Edited on 27-5-2014 by CHRIS25]

[Edited on 27-5-2014 by CHRIS25]

blogfast25 - 27-5-2014 at 12:31

Chris:

”1. Extracted 20 mLs Ferric Chloride from 250 mLs water (this was 20 mLs added to 240 so a dilution factor of 24); Then added 50 mLs to the 20 mL sample and titrated. endpoint at 3.5/4.0 mLs thiosulphate. (this reading is suspect because endpoint reached far too quickly and starch did not turn soln blue so I was passed the endpoint, I did not shake the flask quick enough as it dripped in).”

This is confusingly put or maybe I’m easily confused.

I take it to mean this. You took 20 ml of original sample and added it to 240 ml water. That would make 260 ml and a dilution factor of 260 / 20 = 13.

Of this dilution you took 20 ml and added 50 ml water and titrated this with 3.5 ml thiosulphate 0.15 M. That is 0.0035 L x 0.15 mol/L = 0.000525 mol thiosulphate or 0.00105 mol Fe3+, contained in 20 ml of DILUTION, so the dilution is 0.00105 mol / 0.02 L = 0.0525 M.

You diluted by a factor of 13, so the original solution is 13 x 0.0525 = 0.6825 M.

For titration no2 (30 ml of dilution, 5.5 ml thiosulphate used) we get: 0.0055 x 0.15 = 0.000825 mol thiosulphate = 0.00165 mol Fe3+, in 30 ml dilution or 0.00165 / 0.030 = 0.055 M. The original solution: 0.055 x 13 = 0.715 M.

The KI amounts are correct, yet nothing tallies!

Are you sure your sodium thiosulphate solution is 0.15 M and not 1.5 M? For 1.5 M things would work out at 7 M, as you suspect.

To prepare for instance 200 ml (that is 200/1000 = 0.2 L) of 0.15 M thiosulphate from solid sodium thiosulphate pentahydrate (molar mass 248.18 g/mol) you need to weigh:

0.2 L x 0.15 mol/l x 248.18 g/mol = 7.44 g and dissolve it in 200 ml of water. Are you sure this is what you did?

And then there's this bit:

"After that I had extracted the ferric sample from the 250mLs (10 mLs ferric chloride + 240 mLs water)"

... which hopelessly contradicts what you wrote above. Chris, which is it? 10 ml ferric chloride + 240 ml of water or 20 ml ferric chloride + 240 ml of water? BIG difference: dilution factors of respectively 25 and 13! But even that doesn't explain what is going on... For No2 with dilution factor 25 would give 1.37 M for the ferric chloride solution.

[Edited on 27-5-2014 by blogfast25]

CHRIS25 - 27-5-2014 at 14:22

Ok. I guess I am working at too many tasks at once. Sorry. please allow me to start again. The following is what I used, I did what you advised for the first titration. Yes I placed 7.44g of thiosulphate made up to 200 mLs water. 1.2g iodide each titration. The ferric chloride (my typo above - so sorry). I took 10 mLs from the original ferric chloride solution, put into beaker and added 240 mLs water.
Titration 1: Took 20 mLs from this beaker, placed it into the flask under the burette and added 50 mLs water and titrated it with 0.15 M thiosulphate.
Titration 2: Took 30 mLs from this same beaker, placed it into the flask (NO EXTRA WATER ADDED), titrated with 0.15 thiosulphate.
Titration 3: Exactly the same as Titration number 2.

I do apologize for the typo. The thiosulphate endpoints are as written all correct.
My calculations are now working out at between 12.6 M and 13.2 M giving me 0.82 and 0.86 mols of Iron respectively. Going from too little to too much.

Lesson learned, concentrated solutions are inacurate when titrating? These results appear a more accurate reflection, but can not account for the extra.

[Edited on 27-5-2014 by CHRIS25]

[Edited on 27-5-2014 by CHRIS25]

blogfast25 - 28-5-2014 at 04:56

Chris:

"My calculations are now working out at between 12.6 M and 13.2 M giving me 0.82 and 0.86 mols of Iron respectively. Going from too little to too much."

How do you arrive at these numbers? Because I don't. As I wrote above, I expected about 20 ml thiosulphate for 20 ml of dilution and you're barely getting 5 ml! Something is still very wrong, trust me...

CHRIS25 - 28-5-2014 at 07:47

It seems my calculator can't do maths either....
Ok, concentrating on titration number 2 only:

0.0055 L x 0.15 M = 0.000825
0.000825 x 2 = 0.00165
0.00165/0.030 L sample taken = 0.055
0.055 x 24 (dilution factor) = 1.32 M
1.32 M = 0.0858 mols Fe.

If this is all wrong then I am thickly stupid and will drink my ferric chloride for dinner.
Interestingly these figures mathch the very first titrations that I did, where I had a 1.3M. Something is wrong, and I will make another solution of FeCl.

[Edited on 28-5-2014 by CHRIS25]

[Edited on 28-5-2014 by CHRIS25]

[Edited on 28-5-2014 by CHRIS25]

blogfast25 - 28-5-2014 at 09:59

You are careless around numbers, young man. Line 4 of your calculation: the dilution factor is 25, NOT 24.

The similarity with your earlier results struck me too.

Would you mind sending me some of your CURRENT solution? I need to find some rest: by analysing your sample myself, I might restore my biorhythm! Seriously.

CHRIS25 - 28-5-2014 at 10:30

I used 24 on purpose, I thought that 10 mLs ferric soln with 24 times the water = 250 mLs total.

DraconicAcid - 28-5-2014 at 11:22

Now you're being thick. The dilution factor is the ratio of the final volume of the solution to the initial volume of the solution, not the amount of water you've added. If you had taken 1 mL of your solution, and diluted it with 1 mL of water, would you say that that the concentration should change by a factor of 2 (i.e., half of what it was), or a factor of 1 (i.e., stay the same)?

blogfast25 - 28-5-2014 at 11:23

Quote: Originally posted by CHRIS25

I used 24 on purpose, I thought that 10 mLs ferric soln with 24 times the water = 250 mLs total.

Yes, but that makes the dilution factor = Final Volume / Initial Volume = 250 / 10 = 25. Important, that...

You can also do a density measurement on the ferric chloride, that is a non-destructive measurement and accurate enough to distinguish between say 1.5 M and 7 M FeCl3.

Method, see here:

http://www.sciencemadness.org/talk/viewthread.php?tid=30274&...

[Edited on 28-5-2014 by blogfast25]

[Edited on 28-5-2014 by blogfast25]

CHRIS25 - 28-5-2014 at 11:37

Ok, (156.5 - 95.3)(142.5 - 95.3) = 61.2 x 47.2
I understand density and SG, but have no idea what I am supposed to do with these figures. Sorry. Have not learned this technique yet. Sorry forgot - 50 mLs this was.

[Edited on 28-5-2014 by CHRIS25]Hold on, is tis 1.29 SG then?

[Edited on 28-5-2014 by CHRIS25]

blogfast25 - 28-5-2014 at 11:54

Quote: Originally posted by CHRIS25

Ok, (156.5 - 95.3)(142.5 - 95.3) = 61.2 x 47.2
I understand density and SG, but have no idea what I am supposed to do with these figures. Sorry. Have not learned this technique yet. Sorry forgot - 50 mLs this was.

[Edited on 28-5-2014 by CHRIS25]Hold on, is tis 1.29 SG then?

[Edited on 28-5-2014 by CHRIS25]

Error alert: (156.5 - 95.3) / (142.5 - 95.3) = 61.2 / 47.2 = 1.30 (1.2966 = 1.30)

Hold on while I calculate the theor. density for FeCl3 7 M.

blogfast25 - 28-5-2014 at 12:10

Ok, making a few reasonable (but not necessarily entirely correct) assumptions, I get a theoretical density for 7 M FeCl3 of 1.74 g/cm3, quite far removed from 1.30 g/cm3, confirming my suspicion that this solution is quite concentrated but not nearly 7 mol/L.

Since as FeCk3 has a much higher density (2.898) than water (1), the higher the concentration of FeCl3, the higher the density of the solution.

[Edited on 28-5-2014 by blogfast25]

CHRIS25 - 28-5-2014 at 12:12

I have just been searching the web, I have no notes on this one. How do you convert this density into the Molarity?

DraconicAcid - 28-5-2014 at 12:21

The CRC Handbook gives a density of 1.32 g/mL for 32% FeCl3, which is 2.59 M.

blogfast25 - 28-5-2014 at 12:22

7 mol/L means 7 mol/L x 162.2 g/mol = 1135.4 g FeCl3 / L.

Now we assume that when FeCl3 and water are mixed no contraction or expansion takes place. This is certainly true for dilute solutions (< 1 M).

That one litre is made up of 1135.4 g / 2.898 g/cm3 = 392 cm3 (or ml, if you prefer) of FeCl3. The rest of that 1 L is made up of water, i.e. 1000 - 392 = 608 ml.

Since as the density of water is 1, 608 ml of water = 608 g of water.

So we have (1135 g FeCl3 + 608 g water) / (392 ml FeCl3 + 608 ml water) = 1.74 g/ml (g/cm3

[Edited on 28-5-2014 by blogfast25]

CHRIS25 - 28-5-2014 at 12:27

I throw my hands in the air. I throw my hands in the air.
I did not mean to the above, I meant to Draconic's revelation about the molarity with a 1.32 density and a 2.59 M

[Edited on 28-5-2014 by CHRIS25]

blogfast25 - 28-5-2014 at 12:35

Yes, that one's a bit tricky when one is not math minded.

CHRIS25 - 28-5-2014 at 12:42

Quote: Originally posted by blogfast25

Yes, that one's a bit tricky when one is not math minded.

No, but the good news is that I am just intelligent enough to reverse engineer everything I come across. So I always see an example and an answer then I take things apart to see how it is done and then that's it. So I understand all the reasoning now. I just can't reason maths out from scratch - a weakness of mine.

blogfast25 - 28-5-2014 at 13:07

Quote: Originally posted by DraconicAcid

The CRC Handbook gives a density of 1.32 g/mL for 32% FeCl3, which is 2.59 M.

That kind of tallies: I get 1.16 g/ml for 1.5 M.

[Edited on 28-5-2014 by blogfast25]

CHRIS25 - 28-5-2014 at 13:15

Quote: Originally posted by blogfast25

Quote: Originally posted by DraconicAcid

The CRC Handbook gives a density of 1.32 g/mL for 32% FeCl3, which is 2.59 M.

That kind of tallies: I get 1.16 g/ml for 1.5 M.

[Edited on 28-5-2014 by blogfast25]

all this wonderful bad news means that my titration must be wrong, or there is ferrous still in the solution. Anyway we will see.

blogfast25 - 29-5-2014 at 05:04

I will test the sample for ferrous ions, so we'll know. It's highly unlikely though...

blogfast25 - 6-6-2014 at 09:36

On Chris’ sample, for density relative to water, I got 1.36, which isn’t too far from Chris’ own value.

But here’s a surprise; when testing for ferrous iron:

Left: 2 drops of the ferric solution with about 1 mm of water.

Middle: same as left but with 0.5 ml of K3Fe(CN)6 reagent solution added.

Right: same as left but with a few grains of FeSO4 added and with 0.5 ml of K3Fe(CN)6 reagent solution added.

Middle and left show strong formation of Prussian Blue (Fe7(CN)18, idealised) due to presence of Fe2+. So the ferric chloride solution still contains ferrous ions, but this test cannot tell me how much.

Tomorrow the Fe3+ will be titrated.

[Edited on 6-6-2014 by blogfast25]

CHRIS25 - 6-6-2014 at 10:00

Absolutely great news, about my measurements, was dreading the moment you tested it to be honest thinking he's bound to find my horrendous embarrassing mistake. That there are still ferrous ions in solution is a surprise, yet paradoxically also good news since my 0.5 mol iron can not simply vanish. This gives me new insight into length of time needed without pumping air, without peroxide, to obtain ferric from ferrous. It has been 3 weeks at least now.

blogfast25 - 6-6-2014 at 12:30

It did surprise me a little to find such a strong positive for Fe2+. It's a shame so little of the original solution is left, otherwise I might try and complete the oxidation for a second titration. Let me think about that...

Loss of peroxide during the oxidation may be partly responsible here. The reduction of the peroxide is supposed to go according:

H2O2 + 2 H+ 2 e- === 2 H2O

.. but often oxygen effervescence can be observed during oxidations with peroxide:

H2O2 === > O2 + 2 H+ + 2 e-

This side reaction is OXIDATION of H2O2 to oxygen and doesn't contribute anything to the oxidation of Fe2+ to Fe3+. So too much effervescence may leave one short of the desired stoichiometry.

[Edited on 6-6-2014 by blogfast25]

CHRIS25 - 7-6-2014 at 04:35

Quote: Originally posted by blogfast25

It did surprise me a little to find such a strong positive for Fe2+. It's a shame so little of the original solution is left, otherwise I might try and complete the oxidation for a second titration. Let me think about that...

Loss of peroxide during the oxidation may be partly responsible here. The reduction of the peroxide is supposed to go according:

H2O2 + 2 H+ 2 e- === 2 H2O

.. but often oxygen effervescence can be observed during oxidations with peroxide:

H2O2 === > O2 + 2 H+ + 2 e-

This side reaction is OXIDATION of H2O2 to oxygen and doesn't contribute anything to the oxidation of Fe2+ to Fe3+. So too much effervescence may leave one short of the desired stoichiometry.

[Edited on 6-6-2014 by blogfast25]

Hi, I'm trying to interpret, did you add peroxide to the solution before the prussian blue precipitate in order to ensure complete oxidation?

blogfast25 - 7-6-2014 at 04:48

Quote: Originally posted by CHRIS25

Hi, I'm trying to interpret, did you add peroxide to the solution before the prussian blue precipitate in order to ensure complete oxidation?

No, no, not at all. The test was conducted on the sample as I received it (diluted 2 drops to 1 ml water). It clearly contains ferrous ions. The third tube is simply a reference test.

Had I completed the oxidation by adding peroxide I wouldn't have found any Prussian Blue. Which I hope to demonstrate later on today.

I was trying to provide an explanation for why your ferric chloride still contained ferrous chloride: possibly some of the peroxide you used got consumed into this side reaction.

[Edited on 7-6-2014 by blogfast25]

CHRIS25 - 7-6-2014 at 05:11

I feared that: I never use peroxide ever for oxidizing as a short cut. This sample has been oxidized by natural exposure to air over nearly 4 weeks

blogfast25 - 7-6-2014 at 05:30

Quote: Originally posted by CHRIS25

I feared that: I never use peroxide ever for oxidizing as a short cut. This sample has been oxidized by natural exposure to air over nearly 4 weeks

So was that the only oxidising action? Because that would explain a thing or two: air (oxygen) oxidation of Fe2+ in strongly acidic conditions at RT is notoriously slow, despite what many of the usual suspects say.

[Edited on 7-6-2014 by blogfast25]

CHRIS25 - 7-6-2014 at 05:48

On sunny days it went outside with the covering off. But I knew it was slow, just not this slow.

blogfast25 - 7-6-2014 at 09:05

Ok, mystery finally solved: your ‘ferric chloride solution’ is mainly ferrous chloride.

Here’s what I did.

1) 2.0 ml of the original sample (OS) was pipetted into a 100.0 ml volumetric flask, then diluted to the mark (dilution factor 100/2=50). Call this solution 1.

2) Another 2.0 ml of the original sample (OS) was pipetted into a small conical flask, slightly diluted and put in the fridge. Call this solution 2.

3) Solution 1 was titrated with KI and 0.1 N Na2S2O3. The end points were very variable. I suspect some of the Fe2+ is oxidising (then oxidises more iodide) during the titration. Very tentatively I got a value of 1.4 M for Fe3+

4) To Solution 2 was added about 10 ml H2O2 32 %, very slowly, almost drop by drop. The solution changed colour quite dramatically and much heat evolved, all signs of ferrous ions being oxidised by the peroxide.

5) To get rid of the excess peroxide (which would also oxidise iodide, thus creating false readings) the oxidised solution 2 was boiled for about ½ hour. This caused predictably the Fe3+ to precipitate as Fe(OH)3. After cooling 10 ml 98 % H2SO4 was added, which dissolved the Fe(OH)3 immediately.

6) This oxidised solution 2 tested negative for Fe2+ with K3Fe(CN)6.

Here solution 2 (left) after oxidation, compared to solution 1 (right). It’s much more coloured:

7) Solution 2 was quantitatively transferred into a 100.0 ml volumetric flask and diluted to the mark (dilution factor 50), then titrated as above. I found what is now TOTAL Fe (titrated as Fe3+

to be 3.9 M.

8) Another quick test showed that when the OS is slightly diluted and then precipitated with strong ammonia, black magnetite formed (Fe3O3, the mixed oxidation state oxide). This is typical of solutions that contain ferrous and ferric ions both in significant quantities.

So roughly over half of the iron had not been oxidised from the ferrous to the ferric form. Considering what a poor oxidiser air oxygen is in acid conditions (for this oxidation), this is hardly a big surprise.

[Edited on 7-6-2014 by blogfast25]

CHRIS25 - 7-6-2014 at 10:02

Well then, thanks to all your hard work Gert, a lot learned, a lot solved, and more techniques learned through this process. I have K ferricyanide and did not use it to test for ferrous ions because I was always led to believe, during my darkroom days, that this stuff should never come into contact with acids of any sort - so is it really safe to use below a certain acidity strength?

The production of magnetite is a dead give away. But I am so surprised that half the solution was still in the ferrous stage. I think now I will get out the air pump and pump away, (I only have 5% peroxide and it is ridulous adding so much water just to oxidize); so air pump from now on.

In your point 7, you say 3.9 M. But this does not tell me how many moles ferric in the solution. If I take my 50 mLs (from which the sample was taken) then this is only 0.195 moles ?????

blogfast25 - 7-6-2014 at 10:58

Ferro and ferrihexacyanates are very safe, even ay low pH. One of the two is even used as a food additive!

These substances are very strong complexes, so that the CN groups linked to the central Fe atom are bonded very, very, strongly to it.

To get HCN evolution you'd need strong acid + heating. Don't confuse complex cyanides with simple cyanides like NaCN or KCN: these simple cyanides evolve HCN just by showing them the acid bottle! But that's not the case here...

3.9 M applies to all of the solution you had. I seem to remember 70 ml. So that would be 3.9 M x 0.07 L = 0.273 mol. I know, that's a lot of Fe lost but I'll bet you your next attempt will be much better. But you will need some oxidiser, other than air oxygen. Have you any nitric acid?

[Edited on 7-6-2014 by blogfast25]

CHRIS25 - 7-6-2014 at 11:10

Yes, but I am going to use my air pump, I use it for copper chloride, not in the mood for complicating stuff right now, too many things not going according to expectations, annoyance levels higher than usual; frustration barometer has just entered orange, and "throw whole lot in bin" emergency release valve has just entered critical.

blogfast25 - 7-6-2014 at 11:14

My only attempt ever to oxidise a ferrous salt (sulphate in this case) to Fe(III) using an air pump failed. Sadly I predict the same will happen here...

CHRIS25 - 7-6-2014 at 11:21

Really? I always use an air pump with HCl and Cu, never had any issues. So what are you suggestions then?

aga - 7-6-2014 at 13:36

Gently heat 3% OTC peroxide for ages to get a slightly stronger H₂O₂ solution.

CHRIS25 - 7-6-2014 at 13:55

Quote: Originally posted by aga

Gently heat 3% OTC peroxide for ages to get a slightly stronger H₂O₂ solution.

Ages ago I watched a chemist (Video) put peroxide in the freezer of a lab, and then dripped out a few of the first drops that melted saying that this was Highly concentrated. That is all I remember.

blogfast25 - 8-6-2014 at 04:22

Quote: Originally posted by CHRIS25

Really? I always use an air pump with HCl and Cu, never had any issues. So what are you suggestions then?

It seems I lost a post here, for some reason.

Don't compare apples and oranges: what works for Cu + HCl doesn't necessarily work for this system.

If you have nitric acid, use that to oxidise the ferrous iron to ferric iron. You need 1/3 (one third) of a mol of nitric acid per mol of ferrous iron.

That reaction n requires some care to carry it out properly and safely though.

Another way of preparing FeCl3, without special oxidisers, is as follows. Start from a known quantity of ferrous sulphate heptahydrate or ferrous chloride. Precipitate with ammonia as Fe(OH)2, filter and wash filter cake with plenty water. The filtrate will contain ammonium sulphate or ammonium chloride, which can be recovered.

Place all of the filter cake in a borosil container and heat gently on a hot plate, always making sure the material remains moist (add small amounts of water to prevent it from drying out). The Fe(OH)2 will oxidise quite quickly to Fe(OH)3 by air oxygen, simplified:

4 Fe(OH)2 + O2 + 2 H2O ===> 4 Fe(OH)3

The material will change to reddish brown. This oxidation reaction is much faster than FeCl2 in acid solution, like you've been trying.

Then dissolve the Fe(OH)3 in excess hot HCl 37 %

[Edited on 8-6-2014 by blogfast25]

CHRIS25 - 8-6-2014 at 10:34

After some thought I decided that I need to test this out. I will (given the weather) pump oxygen into the solution for as long as is possible, noting times and amounts. Then re-do titration and ferricyanide tests, let's see how much oxidizes, since we now know how much is already oxidized. Curious.

blogfast25 - 8-6-2014 at 12:15

I have a few ml of the OS 'ferric chloride' solution left and will determine TOTAL Fe molarity again, next week end, using a different type of titration.

[Edited on 8-6-2014 by blogfast25]