RE: Question for chemists

From: Eric Neilsen ^lt;[email protected]>
Date: 06/24/05-08:49:34 AM Z
Message-id: <20050624144934.EF6D6B597BB@spamf3.usask.ca>

Loris, When mixing your palladium solutions it is always or nearly always
recommended that you first start by placing your salt Xcl ( where x is Na,
NH4, K, Li) into your volume of water. As Ryuji states, this allows the
PdCl2 to form it's partnership with the salt and become available to you for
platinum/palladium printing Ware/Zia/Traditional. If you are going to assure
that all your palladium chloride gets into solution properly, error on the
side of slightly too much LiCl, or NaCl, or ... And as Ryuji informed you,
the LiCl is hydroscopic and unless well kept, can lead to errors.

What does adding too much XCl do? In my experience it slows the speed of the
coating solution and adds a bit more effect of the particular salt being
used. For instance, Sodium based palladium prints are warmer than Ammonium
based print, so too much NaCl while mixing will lead to a slower print and
one that is slightly warmer than a mixture or proper molar concentrations.
Can you always see it? NO! Way back in my early printing days, I was using a
formula given too me for mixing palladium solution. Like many, I simply
followed the instructions as written. It wasn't until years later that I
saw conflicting formula and started pushing my pencil to get to the bottom
of it. Well, the earlier formula had been incorrect in that it had nearly
twice the amount of NaCl. This clearly explained to me why I was getting
clogs in my burets; the NaCl was coming out of solution and solid crystals
were plugging up the tip. Now when a suggestion is made to use a little
extra XCl, in no way am I suggesting that you double the amount but an extra
.3 to .5 grams may not be too much if 5 g is required to fulfill the
equation. The palladium solutions have if I recall correctly a very high
saturation and don't require warming to keep them in solution, although
there are those that suggest it, I don't believe that it is required.

I have always used a different mix than B&S has listed when mixing Li based
palladium salt. Using Wares published information, and playing with molar
concentrations, I use a .7 M concentration of my FO, AFO mixes. This is
also why when possible I use Ammonium based Platinum salt in place of the
Potassium based salt. It can reach a molar concentration of .7 where the
Potassium based only reaches a concentration of .457M and requires warming
to maintain that in solution.

I mix my Li Solution with these weights:

1.3g LiCl + 2.5 g PdCl2 with water to make 20 ml

Hope this helps

EJ Neilsen
Eric Neilsen Photography
4101 Commerce Street
Suite 9
Dallas, TX 75226
http://e.neilsen.home.att.net
http://ericneilsenphotography.com
 

> -----Original Message-----
> From: Ryuji Suzuki [mailto:rs@silvergrain.org]
> Sent: Friday, June 24, 2005 8:57 AM
> To: alt-photo-process-l@sask.usask.ca; loris_medici@mynet.com
> Subject: Re: Question for chemists
>
> 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.
>
> Remarks:
>
> 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 <loris_medici@mynet.com>
> 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 08:49:58 2005

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