If you use the pan method of measuring silver, as is typical with both
Hach and Hanna instruments, for measuring silver content, you will get
results for the ionic silver only. This is similar to what you get with
a conductivity test, neither measure the colloidal portion.
To get the complete silver content it is necessary to digest the
silver. Typically adding a small amount of nitric acid will produce
silver nitrate from the particles (as well as the Ag2O), and can thus be
measured with these colorameters.
However nitric acid is nasty to handle, and dangerous. The reaction
also gives off nitric oxide, which is not healthy, so this should
probably be done outside, or in a fume hood.
I am looking for a method of digestion that can use common chemicals
found around the house. There are very few things that will "dissolve"
silver metal. The only one I can find that is not a health hazard is
H2O2, hydrogen peroxide.
When you add H2O2 to CS, you get two reactions. First it will react
with all the Ag2O ionic silver, and produce an Ag2 colloid. Then it
will react with the silver particles, and produce Ag2O, which then
reacts with the H2O2 to produce Ag2. Finally it will react with the Ag2
particles to produce Ag2O. So we end up in the end with a mix of Ag2O
and Ag2 particles, as each are continually changed from one to the
other, and back again by the redux reaction.
This would end up causing only about half of the silver to be measured
by a colorameter, since about half will be Ag2 particles.
So I was thinking, what if we first add a little baking soda to the CS.
This will immediately cause the Ag2O to become silver carbonate, which
is soluble at the concentration of CS. This leaves the colloidal
particles, and the addition of a small amount of H2O2 should cause that
to convert to Ag2O, which will then immediately be scavenged by the
NaHCO2 (baking soda), producing more silver carbonate.
If my thinking on this is correct, this should quickly, easily, and
safely convert all CS to 100% silver carbonate.
Any chemists here who can confirm this?
Thanks,
Marshall
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