Author Topic: hexachloroplatinic acid hexahydrate  (Read 1953 times)

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flipper

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hexachloroplatinic acid hexahydrate
« on: April 21, 2002, 11:12:00 AM »
Can I make this myself from Platinum and how do I do that. I have searched and searched but this IS my last hope to find something.  ;)

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PrimoPyro

  • Guest
Aqua Regia
« Reply #1 on: April 21, 2002, 01:57:00 PM »
I'm pretty sure it's made by the action of aqua regia on platinum. Aqua regia is a very concentrated mixture of nitric acid and hydrogen chloride. Just how concentrated, I don't know.

                                                   PrimoPyro

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Belial

  • Guest
3:1
« Reply #2 on: April 21, 2002, 02:04:00 PM »
Aqua Regia = 3:1 of HNO3 & HCl


flipper

  • Guest
found it
« Reply #3 on: April 21, 2002, 05:32:00 PM »
I don't know if it's the sme but I think the procedure is described in this linkje

Largescale methamphetamine production via catalytic hydrogenation (Zipped Word docs, 1MB)

(https://www.thevespiary.org/rhodium/Rhodium/archive/cookbook.zip)


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terbium

  • Guest
Dissolving platinum
« Reply #4 on: April 21, 2002, 06:49:00 PM »
The Merck Index says that aqua regia is 18 ml (concentrated) nitric acid in 82 ml (concentrated) hydrochloric acid. Exact proportions are not critical. It is important that the acids be the concentrated acids - 70% HNO3 and 35% HCl.

I do not mix all of the nitric with the hydrochloric at once. To dissolve platinum I place the platinum wire, strips etc. in a beaker on a hot plate, cover with hydrochloric acid, heat to near boiling and begin adding the nitric acid in small portions. When each portion of nitric acid is added brown fumes of nitric oxides will begin to be evolved (good ventilation from an exhaust fan is required!). When the brown fumes of nitric oxides stop coming from the beaker of hot acid then another small portion of nitric acid is added. If the platinum metal was in the form of wire or small clippings then it will fairly quickly dissolve under this treatment. If the process takes so long that the volume of liquid in the beaker diminishes before dissolution is completed then the volume is made up by the addition of more hydrochloric acid. When dissolution of the metal has been completed then the orange solution is boiled down to near dryness to drive off excess HCl. The solution can then be made up in volume with water and the platinum precipitated as ammonium chloroplatinate.

Post 55334 (missing)

(terbium: "Platinum is good too.", Chemicals & Equipment)

flipper

  • Guest
What SWIM wanna make with it
« Reply #5 on: April 21, 2002, 09:21:00 PM »
I wanna make that sulfided catalyst from patent 5498585.

A) Preparing the catalyst precursor (Pt+H2O2)/C (without sulphidizing)

To prepare a catalyst with 3% platinum load, 97 g of activated carbon (dry weight) were stirred into distilled water with a stirring speed of 300 rpm. 12 g of a 25% strength aqueous solution of hexachloroplatinic acid hexahydrate (corresponding to 3 g of Pt) and 3 ml of a 30% strength aqueous solution of H2O2, corresponding to 2 moles of H2O2 per mole of platinum, were added to this suspension.

Then the suspension was heated to 80.degree. C. and sodium carbonate was added with continuous stirring, to precipitate barely soluble hydroxides. To reduce the precipitate, 1.8 ml of 37% strength formaldehyde solution was then added. The temperature of the suspension was also kept constant at 80.degree. C. during reduction. After reduction, the catalyst was filtered off on a nutsch filter and washed with DI water.

B) Preparing the sulphidized catalyst according to the invention (Pt+H2O2+S)/C.

To prepare a catalyst with a 3% platinum load according to the invention, a catalyst was first prepared in accordance with part A and then sulphidized with dimethyl sulphoxide in accordance with example 1 in DE-PS 21 50 220. The total sulphur content of the sulphidized catalyst was about 0.4 moles of S per mole of Pt.

Catalyst with the same platinum load but with double or four-fold the sulphur content were prepared by appropriately increasing the amount of sulphidizing agent.

Catalysts with different platinum loads were made up using the same procedure as above with a simultaneous increase in the amounts of hexachloroplatinic acid, caustic soda solution and sulphidizing agent.


DE-PS 21 50 220 is a patent very hard too get. Anybody knows what it says. I'm especially interested in example 1.  


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Organikum

  • Guest
Try this
« Reply #6 on: April 22, 2002, 04:49:00 AM »

http://l2.espacenet.com/espacenet/bnsviewer?CY=gb&LG=en&DB=EPD&PN=GB1397584&ID=GB+++1397584A++I+




thats the GB version of what you are searching.

Is there a working electrochemical way to dissolve Pt and Pd avoiding the fumes and dangers of aqua regia?

Why difficult to get?
ORGY

flipper

  • Guest
another question
« Reply #7 on: April 22, 2002, 04:05:00 PM »


Example 1B

To prepare a catalyst with a 3% platinum load according to the invention, a catalyst was first prepared in accordance with part A and then sulphidized with dimethyl sulphoxide in accordance with example 1 in DE-PS 21 50 220. The total sulphur content of the sulphidized catalyst was about 0.4 moles of S per mole of Pt.




Do I have too calculate how much DMsulphoxide or must I just follow this example.


EXAMPLE 1

100 grams of platinum-activated carbon catalyst (F 103 R of Degussa) having a platinum content of 1% were suspended in 400 ml of water and heated to 60.degree.C. After addition of 1 gram of dimethyl sulfoxide the mixture was stirred further for 15 minutes at 60.degree.C. The catalyst was then filtered with suction and washed. It can be added in the wet condition.




100 grams of Pt+H2O2/C (without sulphidizing)
contains 3% Platinum load. That's 3 gram Platinum. That's 0,015 mole. Every mole Pt must have 0,4 mole of S. That's 0,006 mole of S. That's o,19 grams of S. Molar mass of DMSO is 78.1288 That of S is 32.066 So S is 41,04% S present in DMSO. So if there is 0.19 grams of S in DMSO Then there is 0,46 grams of DMSO needed for this reaction. Am I right??????  ::)


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Organikum

  • Guest
calculation vs. imagination
« Reply #8 on: April 22, 2002, 05:59:00 PM »
Sorry flipper I´m to lazy to recalculate but reading the reference it´s obvious that there is per gram platinum one gram DMSO used. Per four gram palladium are three gram DMSO used.
The inventors didn´t calculate the way you do I bet.

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ORGY