Re: Question for chemists

From: Ryuji Suzuki ^lt;>
Date: 06/24/05-07:57:01 AM Z
Message-id: <>

Loris, what you are looking at is lithium salt of tetrachloropalladic
(II) acid, Li2[PdCl4]. You see there is excess lithium chloride used,
and it's a common way to secure the quantitative yield of this
compound. That is, excess chloride is needed to minimize unreacted
palladium(II) chloride in the solution. (I think PdCl2 would be very
difficult to dissolve in water but it dissolves rather easily in
presence of excess HCl or alkali metal chlorides.)

The reaction you want is 2LiCl + PdCl2 -> Li2[PdCl4] with some
other excess Li+ and Cl-, wherein the presense of the latter will
drive the reaction to the right hand side.

Anyway, when you calculate the quantity of lithium
tetrachloropalladate, you should use the amount of palladium present
in the solution. Using your figures, it would be 0.013 mol or 0.52M.
However, I don't know if practitioners of your process follow the
right way - they may have more conventional way to determine the
amount of sensitizer to add.


Lithium is an alkali metal and I suppose your "LiCl2" is an error in
the literature you consulted. It sould be LiCl, MW 42.4. This compound
is extremely hygroscopic. (That is, it's easy to weigh out an
insufficient amount of this compound if this is not the excess agent
in the reaction.)

Commercial supply of tetrachloropalladic acid comes with excess HCl
for the same reason as above. If HCl isn't in excess, some palladium
will be in PdCl2 form.

Tetrachloropalladic acid and its salts can exist in solution but not
in dry form. If this acid is dried, it will decompose and go back to
PdCl2 form again.

Metals of platinum group is used in very small quantities in silver
gelatin emulsions as a way to chemically sensitize emulsions
(especially platinum) as well as to boost contrast (especially with
rhodium) and to improve reciprocity law (especially iridium,
ruthenium, osmium).

Ryuji Suzuki
"Well, believing is all right, just don't let the wrong people know
what it's all about." (Bob Dylan, Need a Woman, 1982)
From: Loris Medici <>
Subject: Question for chemists
Date: Fri, 24 Jun 2005 15:28:25 +0300
> In Bostick-Sullivan formulary pages, it is stated that Lithium Palladium
> Chloride (Li2PdCl4) solution is made following the formula below:
> Lithium Chloride 1.7gr (LiCl2, molecular weight = 77.8464)
> +
> Palladium chloride 2.3gr (PdCl2, molecular weight = 177.3254)
> +
> Water to make 25ml
> What is the exact chemical reaction equation for making Li2PdCl4?
> Is it:
> 2(LiCl2) + PdCl2 -> Li2PdCl4 + Cl2
> or something else? Does water enter in the equation? If yes, how?
> The mol value for 1.7gr LiCl2 is -> 1.7 / 77.8464 = 0.0218378... When
> you mutiply this value with molecular weight of PdCl2 then you find that
> you have to add around 3.9 gr of PdCl2 - Which obviously is not the
> case. So, I assume my equation above is incorrect. Can you please
> provide me the correct equation formula? I need this in order to mix
> fresh Ammonium Iron(III) Oxalate sensitizer of right strength (I don't
> want to waste Pd by preparing a weaker AFO sensitizer than what is
> needed). Or maybe I shouldn't bother with all this and prepare a 60% AFO
> solution (I guess this is almost a saturated solution)... What will be
> the effects of preparing a sensitizer of extra strength?
> Thanks in advance,
> Loris.
Received on Fri Jun 24 07:56:47 2005

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