w0lvez
Well-known member
How much palladium metal do I need in order to make a gram of PdCl2? What is the best way to make PdCl2? I tried dissolving Pd using AR and HCL but I only get almost 1/2 of the weight of the metal. Is that normal?
How much Palladium metal needed to make 1g PdCl2?
- Thread starter w0lvez
- Start date
Help Support Gold Refining Forum:
From Georg Brauer's Handbook of Preparative Inorganic Chemisty:
"Palladium (II) chloride
The anhydrous salt is prepared by heating loose palladium
sponge (contained in a porcelain boat set in a glass tube) to a dull
red heat in a stream of Cl2 . According to Krustinsons, the decomposition
pressure of PdCl3 reaches 1 atm. at 738 °C.
By dissolving finely divided Pd in cone. HCl through which Cl2
is bubbled, one obtains a solution in which both H2PdCl4 and
H2PdCl6 can be detected. Concentrating the solution also yields a
residue of PdCl2. "
There are other methods if you have borohydride handy and use it stoichiometrically.
If you carefully heat the chloropalladous acid in an atmosphere of HCl, it will leave you with fairly good palladium (II) chloride suitable for organic chemistry operations...
Bear in mind that you should probably not attempt any of the above procedure if you can't calculate how much palladium you need to make a gram of palladium chloride:
Palladium weighs 106.42 grams per mol. Chlorine is 35.453 g/mol. There are twice as many chlorine atoms as there are palladium atoms. So, add them all together as follows: 106.42 g/mol + 2(35.453 g/mol) = 177.34 g/mol for the anhydrous material. Palladium constitutes by mass, 106.42 g/mol Pd / 177.34 g/mol PdCl2 *100 = 60% by mass. Can you now see how many milligrams of palladium you need to make one gram of the chloride?
"Palladium (II) chloride
The anhydrous salt is prepared by heating loose palladium
sponge (contained in a porcelain boat set in a glass tube) to a dull
red heat in a stream of Cl2 . According to Krustinsons, the decomposition
pressure of PdCl3 reaches 1 atm. at 738 °C.
By dissolving finely divided Pd in cone. HCl through which Cl2
is bubbled, one obtains a solution in which both H2PdCl4 and
H2PdCl6 can be detected. Concentrating the solution also yields a
residue of PdCl2. "
There are other methods if you have borohydride handy and use it stoichiometrically.
If you carefully heat the chloropalladous acid in an atmosphere of HCl, it will leave you with fairly good palladium (II) chloride suitable for organic chemistry operations...
Bear in mind that you should probably not attempt any of the above procedure if you can't calculate how much palladium you need to make a gram of palladium chloride:
Palladium weighs 106.42 grams per mol. Chlorine is 35.453 g/mol. There are twice as many chlorine atoms as there are palladium atoms. So, add them all together as follows: 106.42 g/mol + 2(35.453 g/mol) = 177.34 g/mol for the anhydrous material. Palladium constitutes by mass, 106.42 g/mol Pd / 177.34 g/mol PdCl2 *100 = 60% by mass. Can you now see how many milligrams of palladium you need to make one gram of the chloride?
w0lvez
Well-known member
OIC, Thanks! I taught the method I used is the problem.
I may have some lab grade PdCl2 on hand if you only need a little.
What are you needing it for?
Steve
What are you needing it for?
Steve
w0lvez
Well-known member
I will use it as activator. I only need 1 gram. How much do it cost? :?:
PM sent.
Steve
Steve
It is used to activate ABS plastics, for its subsequent bath of copper, nickel and finally decorative chrome, it is prepared with tin chloride to form a palladium colloid and activate the polymer.
Truth the preparation is not known here for sure.
Truth the preparation is not known here for sure.
