# Need pure platinum & pure palladium for standard solutions



## kadriver (Nov 23, 2011)

Hello:

I am searching for about 1 gram each of the purest samples I can get of platinum and palladium to make some standard solutions for testing purposes.

I would prefer powder or granules.

Can anyone help me with these items?

Thank you - kadriver


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## Lou (Nov 24, 2011)

I have H2PtCl6 and H2PdCl6 available and numerous salts beside.

PM for info.


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## kadriver (Nov 28, 2011)

Lou:

Can you breifly describe the proceedure for making standard solutions with the materials that you have posted here.

I could not find these mentioned in Hoke's book.

She refers to making a standard solution using "a definite weight of metal dissolved in a definite amount of liquid... such as 100 milligrams to 100 cc of solution."

I am not familiar with moles or other scientific units of measure.

Sorry for the lack of knowlege - I am certain that what you have is probably what I need.

I just need a little instruction on how to use these items to prepare my standard solutions.

Thank you - kadriver


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## samuel-a (Nov 29, 2011)

kadriver 

For the samples Lou gave above:

If you dissolve 1 gram H2PtCl6 in 100ml HCl, you will produce a solution that contains 0.476 grams of platinum metal.
That corresponds to molar concentration of: 0.00244 g/mol Pt

If you dissolve 1 gram H2PdCl6 in 100ml HCl, you will produce a solution that contains 0.3309 grams of palladium metal.
That corresponds to molar concentration of: 0.00311 g/mol Pd


edit: more info, red = added


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## kadriver (Nov 29, 2011)

samuel

Thank you.


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## lazersteve (Nov 30, 2011)

samuel-a said:


> kadriver
> 
> For the samples Lou gave above:
> 
> ...


 

A correction to Sam's statement :

mole Pt / mole H2PtCl6 =

195.084 / 409.81= 0.476 

So one gram of H2PtCl6 contains 0.476 grams of Pt, which is :

0.476 g / 195.084 g per mole = 0.00244 moles of Pt.

Dissolving the full gram in 100 ml of water gives you 100.411 ml of solution as the density of H2PtCl6 is 2.431 g/ml meaning 1 gram is 1/ 2.431 = 0.411 ml. For a nice round 100ml you could simply use 99.589 ml of water.
The molarity is expressed in moles per liter (1000ml) so the math is:
0.00244 moles * (1000 / 100) = 0.0244 M = 0.0244 moles of Pt per liter = 4.76g Pt per liter.

Or simply 4.76g / 1000 ml = 0.00476 grams of Pt per ml.

Pd is calculated the same way using the appropriate numbers for Pd and it's compounds.

Steve


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## samuel-a (Nov 30, 2011)

Good 'catch' Steve :| 

Seems that i neglected to include one H when caculated, next time i'll use the copy/paste option rather then relay on my memory.


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## lazersteve (Dec 1, 2011)

The main problem was not an H, it was that your answer of 0.0024 g/mol was off by a factor of 10 ( should read 0.024) and your units of molarity should be moles per liter not grams per mole. 

Steve


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## samuel-a (Dec 1, 2011)

I stand corrected, once again.


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## kadriver (Dec 1, 2011)

Thanks, now I need some of the chemicals refered to in this post.

Lou said he had some, if this is what I need, then i will get them from him.

I will have to study these equations. A mole is a furry little critter with no eyes and a funny looking nose - right?

I will get beefed up on some basic chemistry so I can talk intelligently about these things - sorry for my lack of knowlege.

kadriver


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## butcher (Dec 2, 2011)

Kadriver, 
Moles have eyes they just cannot see good above ground.

Kidding aside, we all need more study, or we would not need this wonderful forum keep up the good work, we all make goof-ups, and then we learn how we need more study and GRF.


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## kadriver (Dec 5, 2011)

Lou:

I sent you a PM about the chemicals you have.

It may not have gotten to you.

If you will, could you please email me with the proceedure to get these from you.

My email is [email protected]

Thank you - kadriver


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## kadriver (Dec 19, 2011)

I bought 2 grains (.129 grams) each of fine platinum and palladium for making my standard test solutions.

I got them from Ebay, inexpensive and arrived quickly.

I'll dissolve each in a few drops of aqua regia and then add 129ml distilled water to get 100mg metal in 100 ml solution.

kadriver


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## tensor9 (May 21, 2012)

lazersteve said:


> A correction to Sam's statement :
> 
> mole Pt / mole H2PtCl6 =
> 
> ...



This is actually incorrect. Volume addition of two different substances is not necessarily linearly additive. The interaction of the solute and the solvent (or between two liquids) must be taken into account. For example, mixing 5.00 mL of water and 5.00 mL of ethanol does NOT yield 10.00 mL of mixture. (There is actually a contraction.) The same goes for dissolving solids. Approaching infinite dilution, the density of the solution and pure solvent will be close, but as concentration increases, so does the density, in a not so predictable manner.

Anyway, the way to do this properly is to do it empirically, using volumetric glassware. Weigh out the desired mass, quantitatively transfer it to a volumetric flask, then dilute to a total desired volume (appropriate to whatever vol. flask was used).


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## lazersteve (May 21, 2012)

Good point in general, but by what degree does the variation take place with water and the compounds mentioned in the post quoted?

With respect to the compounds mentioned is there a contraction, expansion, or neither ? 


Steve


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## Oz (May 22, 2012)

tensor9 said:


> This is actually incorrect. Volume addition of two different substances is not necessarily linearly additive. The interaction of the solute and the solvent (or between two liquids) must be taken into account. For example, mixing 5.00 mL of water and 5.00 mL of ethanol does NOT yield 10.00 mL of mixture. (There is actually a contraction.) The same goes for dissolving solids. Approaching infinite dilution, the density of the solution and pure solvent will be close, but as concentration increases, so does the density, in a not so predictable manner.
> 
> Anyway, the way to do this properly is to do it empirically, using volumetric glassware. Weigh out the desired mass, quantitatively transfer it to a volumetric flask, then dilute to a total desired volume (appropriate to whatever vol. flask was used).


By the above am I to assume you have a formal education in chemistry? We could use further posts from those that are chemists.


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## tensor9 (May 22, 2012)

A few things.

With dilute solutions such as this, the deviation is actually not that large. However, it's hard to know if there is a contraction or expansion. So, the most accurate (and easiest) way to do it is to add the solid, then dilute to the mark in a volumetric. This way you only have to add one volume of water and there is no need for measuring out multiple volumes of water... and you'll get better results.

Another thing. Hexachloroplatinic acid is ridiculously hygroscopic, so accurately weighing it is hopeless. I'm not an expert on noble metal processing, so I don't know how it is ACTUALLY done in industry, but I have an idea of how you could possibly do it. According to Inorganic Syntheses (best journal ever -- if it's in there, it works), you can convert ammonium hexachloroplatinate to hexachloroplatinic acid by bubbling chlorine gas through the solution. Ammonium hexachloroplatinate is not hygroscopic and can be dried in an oven at at 120 C. (The literature prep. actually recommends this.) So it seems to be at least a passable "primary" standard for platinum. Possibly weigh the ammonium salt then bubble chlorine through the solution to actually get it to dissolve. I THINK it might be possible to oxidize the ammonium ion with aqua regia in order to convert to chloroplatinic acid, but I don't have any data to back that up.

Also you could dissolve the metal, which can be accurately weighed, in aqua regia, quantitatively add to a volumetric, then dilute. (Of course, don't forget to invert the flask multiple times!)

Anyway, yes I'm an inorganic chemist in graduate school, hoping to obtain my Ph.D. in the next several months or so.


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## lazersteve (May 22, 2012)

So for all practical purposes all we have to go by are the theoretical numbers when postulating what the true numbers would be in practice. I've found that it always helps to perform the theoretical math up front so as to better understand what to expect in the lab. That's all the math I typed above was meant to convey, not the actual observed results. It's very exciting when you find out that your math requires adjustment due to real world results. This is what brings science alive for me.

As for the accuracy of any theorectical computation or physical experiment there will always be errors of some magnitude in both, as long as the margin of error is acceptable to the person performing the experiment and that the desired end results are achieved, that is what matters in the end. With this in mind, I feel my post above was *correct* as stated in the context of the original question barring slight variations due to human, mathematical, procedural, environmental, and mechanical errors.

It's nice to have new members who pay close attention to detail, welcome to the forum. I have enjoyed the OrgSynth site for years, long before I ever refined a single grain of gold I was using the site as a primary reference for many of my lab experiments. It's nothing short of awesome.

Steve


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## tensor9 (May 22, 2012)

Not everything can be perfect, of course. The point is to minimize error, especially with respect to making a standard solution. My point is that it is easier and more accurate to do it properly.

For the Pt source, use a pure, stable source to weigh. This allows you to know exactly how many moles of solute you have.
Just dissolve the stable source in whatever manner required, quantitatively transfer into a volumetric flask, then dilute to the mark, and mix thoroughly. Easy. This allows you to know the exact volume of total solution.
Exact number of moles (by measurement) in an exact volume (by measurement) leads to an exact concentration and eliminates the need for approximations in principle. This is of course assuming a good balance and a good volumetric.

This generalizes to the preparation of all standard solutions. 

Good to be on the forum. Thanks.


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## racergold (May 28, 2012)

The solution you are discussing with HCL, I have a quart of palladium electro plating solution that I have no paperwork for and the company is no longer in business. Could this solution be what you are discussing? Is there a simple test or possibly some industry standard for this type of solution?


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## Lou (May 28, 2012)

Usually that is Pd seeder solution for electroless nickel. Commonly used for nickel then chroming a plastic substrate.

Hope that helps.


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