Doing quantitative analysis for silver compounds is really rather easy,
if you have much silver present. The biggest problem is that at 5-20
ppm, the quantities are so small, that getting a good weight on them is
difficult. It can be done though with a good balance, forget a postage
scale. These pages might give you some ideas:
http://dwb4.unl.edu/chemistry/smallscale/SS063c.html
http://dwb4.unl.edu/chemistry/picts/SS063All.gif
Anyway, this can be done.
Add a few drops of hydrochloric acid (muriatic acid should be ok) to the
mix. Verify that it becomes cloudy. Let the cloudy part settle, and
decant off the liquid. Dry the power and weight. Take the weight in
grams and divide by 1.33, that will give you the silver content of the
precipitant. Now divide that by the amount of water you had in liters
and multiply by 1,000,000. Add .6 ppm for the dissolved portion, and you
will now know the amount of ionic silver in ppm in the mixture.
Now take the same amount of EIS and boil it dry. Add about 10 drops of
nitric acid (this must be done in a test tube) and reflux the sides
while boiling dry again over a flame (or use a flask over a hot plate).
You now have 100% silver nitrate. Weight this in grams and divide by
1.575 and that will give you the total silver content. Divide that by
the amount of water in liters and multiply by 1,000,000 to get the ppm
of silver.
Warning, nitric acid is very very toxic, breathing the fumes will
liquefy your lungs so boiling it must be done under a fume hood. Don't
even think of using a venta hood over a stove, it will completely eat
the metal of the hood and the metal piping to the outside up (if it even
goes outside). Fumes are not only toxic, but will destroy any
electronics you have in area.
Another way to get the total silver would be to add the hydrochloric
acid as above, and boil it dry. Then subtract the amount you got in
grams in the first part. What remains will be the silver content of the
colloidal part. Divide this by the amount of water in liters and
multiply 1,000,000 for the colloidal part in ppm.
PS, if you measure everything in micrograms instead of grams, you do not
have to multiply by 1,000,000 to get the ppm.
Marshall
Tony Moody wrote:
Hi Tom,
It would be a real boon to have some sort of standard quantitative
analysis of silver content which could be done in a home lab/kitchen. a
titration method maybe? which would involve simple reagents and
glassware. I think another way would be a colorimetric method or
turbidity meter kind of thing which would involve test slides or a
photometer of some sort.
Do you , or anyone have any thoughts on this?
In the sticks,
Tony
On 22 Feb 2010 at 20:11, poast wrote about : Subject : Re: CS>Which layer
of skin for silv
Hello Marshall,
I only did the reduction to 60 uS once just to see what happened. While
the solution was basically clear, you didn't need a laser to observe the
Tyndall effect. A pen light worked fine. As I recall, I had a few larger
sparkle particles in the solution.
I didn't keep it very long because I was using it in some soap.
Overnight, there was no residue settling on the bottom of the jar, but
that was as long as it sat.
I regularly reduce a 10 uS solution to a 20 uS solution. This also shows
a strong Tyndall effect, is clear, and everything remains in suspension.
The main problem I have is not having the equipment to translate uS to
PPM.
Tom
----- Original Message -----
From: "Marshall Dudley" <mdud...@king-cart.com>
To: <silver-list@eskimo.com>
Sent: Monday, February 22, 2010 8:01 AM
Subject: Re: CS>Which layer of skin for silver deposit?
Thus part I don't understand. The ionic portion of EIS is silver oxide
and silver hydroxide, each with a solubility of only 13 ppm, so combined
they have a solubility of 26 ppm. (Actually since they continually
convert from one to the other, and theoretically the hydroxide should be
much more soluble than the oxide, I believer that what really happens is
that with it continually converting from one to the other and back again
when dissolved, the the less soluble one really sets the limit for
both). So just what IS the compound that is forming the 60 uS part. Is
it from carbon dioxide that is absorbed into the air forming silver
chloride? Or if an analysis I made earlier where the silver particles
get cemented together by silver oxide particles produces an ionic form
which increases conduction? I find it curious that the conductivity
increases.
Marshall
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