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potassium hexacyanoferrate prep?


chemhero

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Hi everyone

 

I'm planning an experiment to make prussian blue, from scratch. The part going from K4Fe(CN)6 to prussian blue is easy peasy, I can find alot of literature on this.

 

But going from iron (II) sulphate to K4Fe(CN)6 is a little harder for me to find information on. Anyway, using my mind (a dangerous thing!) i figured that FeSO4 in water, then add KCN and some acid, with some heat(90 deg.C ?) and then letting it cool, the cyanoferrate i want will fall out of solution.

 

Would this actualy work? Or am I barking up not even a tree, let along the wrong tree?

 

Matt

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What you suggest certainly will make the hexacyanoferrate (II) ion in solution, but you need excess of KCN in solution. Initially, you'll get a precipitate, but on addition of more solution of KCN, this precipitate will dissolve.

 

Adding acid does not seem like a wise idea to me, especially if you are heating as well, this may be extremely dangerous, due to the creation of HCN. The hexacyanoferrate ion is stable, even to moderately concentrated acid, but the free cyanide of the excess KCN definitely will form HCN.

 

If your goal is making prussian blue, then there is no need to isolate the hexacyanoferrate (II). In this case, simply add acid (in a fume hood!) and add an iron (III) salt. The dark blue prussian blue will precipitate from the solution.

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Ah, ok then - leave the acid! I was a little unsure as what little literature I've found has suggested the use of an acid to help along, but this idea didnt sit well with me because of the HCN formation (as you say) and also the possible formation of H4[Fe(CN)6], not the potassium salt.

 

Thanks for your help, I just needed to bounce the idea off someone!

 

Matt

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But now I have another question.

 

Do you do this, just for fun or for educational purposes, or do you really want to have K4Fe(CN)6. For almost all people in the world, obtaining K4Fe(CN)6 is quite easy, it even is mentioned on eBay every now and then. On the other hand, obtaining KCN is VERY hard (and for good reasons). I know of a few people who do their best to make KCN (or NaCN) from K4Fe(CN)6 and hence I find it funny to read that you want to go the other way around.

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Its for an assignment; how to prepare these compounds. I realise getting potassium hexacyanoferrate is easy, thats why I had to figure out its formation myself! I wont be making it, but its still intresting to know how to make one of the basic pigments from scratch

 

Fun stuff!

 

Matt

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  • 1 year later...

As explained previously by woelen, the use of a fume hood or a well- ventilated area, should be sufficient in dealing with any cyanide that is released. An alternative would be use of a closed chamber apparatus, however this then produces the added complication of the disposal of the cyanide. It's much easier, in this case, to simply use the methods prescribed above

You should also use excess potassium cyanide, my guess would be about 5 times as much (we're refering to molar ratios of course), as the iron sulphate, or whatever iron salt you plan to use. Acid, as explained before, can be used to catalyze the reaction. And voila!

 

(Seriously speaking, I don't see how you couldn't simply achieve that through examinations of the previous posts, as all of this was specified.) Happy "chemical- ling!"

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Thank you,do you know how long you heat it for and what temperature? also cal you explain please why excess NaCN is required? and if I need to isolate cyanoferrate or I just add iron and acid to the solution as woelen said.

 

Thanks,


Merged post follows:

Consecutive posts merged

sorry I forgot to ask if FeSO4 in water is in the form of FeSO4.7H2O?

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For the greater benefit of clarity, I shall address all of these issues in turn. Firstly, the excess of KCN, is required, because if you are to make the stoichiometric calculations, the following reactions take place:

 

2KCN + Fe(II) SO4 > K2SO4 + Fe(CN)2

 

Fe(CN)2 + 4KCN > K4Fe(CN)6

 

As you can see, first, two moles of potassium cyanide, react with one mole iron (II) sulphate, to produce one mole of potassium sulphate, and one mole of iron (II) cyanide. The iron cyanide, then reacts with the excess 4 moles of KCN to produce potassium hexacyanoferrate, as shown below. So, knowing the stoichiometry, by molar mass, the ratio is 6 moles of potassium cyanide, per one mole of iron sulphate. For macroscopic benefit, you can simply translate from moles to grams, and then find how many grams of potassium cyanide and iron sulphate are needed for the reaction.

Secondly, technically speaking, the resulting solution of hexacyanoferrate, can easily be used in the making of prussian blue, but if you want to extract it from the solution, simply do what woelen said, adding acid and iron.

For your final merged post, you are right to say that it hydrates, to form iron sulphate heptahydrate, and this can interfere with your calculations, and as such, if you purchased the chemical, simply check the label to say whether the iron sulphate is hydrated. If you're still unsure, simply heat it to a fair bit higher a temperature than 100 degrees centigrate, and this ought to dehydrate your iron (II) sulfate.

In regards to the temperatures required by the reaction, I doubt it would be too high. I'm a little lazy to make any calculations, but I'ld guess that just a light heat, around 80, 90 degrees celsius is all the heat you'll need. Cheers!

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