gold question

[ericrm]

you probably have already seen it yourself, when you add smb to a gold bearing solution , for +- long time gold is not yellow solution and not brown powder. it seem to disapear but what is actually happening could gold just disapear and not precipitate, or disapear and precipitate only half of it.

it is very weird to watch in a pure gold case ,solution is yellow ,than turn clear like water than form a brown precipitate, what is happening in the clear state?

 

[FrugalRefiner]

Eric,

I'm going to take a guess at this based on some things butcher has written recently. I'm going to guess that when the ionic gold in the yellow solution is reduced to a metal it changes to the clear phase. But the tiny clusters of metal are so small you can't see them. As they bounce into each other they agglomerate into larger and larger clumps, they become visible and we see the yoohoo phase.

Like I said, that's just a guess. I'm sure someone will correct me if I'm wrong.

Dave

 

[samuel-a]

Very good question.
Personally, i never really given much thought to that matter.


I can't really accept the suggested explanation given by FrugalRefiner (respectfully). Even at the single digit nano scale, metallic gold dispersion is colorful as far as i know.
My assumption always was that, as the SMB is added there's a range in the redox potential as it shifts at which some intermediate gold compund(s) is produced which form colorless solution. Most likely Au (I) species.

But again, i never really researched that angle. Quite interesting!

 

[FrugalRefiner]

I had to go back and find butcher's post to see if I remembered it correctly.

Gold that was dissolved in solution, and has begun the precipitation process, where gold ions begin to gain electrons and form atoms of gold metal particles which combine in solution to form particles of clusters of these invisible gold atoms, as these small gold atoms move in solution these gold atoms crash into each other, or collide they combine and coagulate into each larger particles, which grow in size, clusters of atoms clash to form larger particles,as these particle's grow towards forming larger particles and then crystals of gold large enough to be visible in solution these larger particles of gold are then pulled to the bottom of the vessel by gravity.

His complete explanation was in this thread: Dust on the bottom after adding SMB.

Dave

 

[butcher]

As soon as we add a reducer to solution the gold is reduced.
(the reduction is basically instant).
We may need to stir the solution for it to react with all of the gold, or it could take a few more seconds for the gold ions and the reductant (SMB) to come into contact.
If we added enough SMB (and did not have free oxidizer, organics or other problems in solution), all of the gold would be reduced to metal in an instant.

If we limited the amount of SMB or reducing agent, because of the limited reactant, we would not reduce all of the gold in solution, but as much gold as the SMB used could react with.

At first the gold Ions reduce by gaining an electron and the atom of gold now has a full shell of electrons, but we cannot see these small gold atoms with our eyes (the gold metal atom is just too small), but if there is no problem where gold would form a colloid, these tiny gold atoms join together, stick together as they collide with each other, and when enough of them form a particle big enough, and then the particles join together into clusters of gold atoms, we will begin to see them in solution,the solution seems to turn brown from the clusters of gold atoms in solution, when these clusters of gold metal atoms get large and heavy enough they sink to the bottom of the vessel as precipitated gold.

 

[solar_plasma]

As far as I understand you guys, the two explanations do not exclude each other by definition since Au⁺ already is a reduced form if we start with Au⁺⁺⁺⁺, isn't it? So, couldn't it be both, colourless Au⁺ ions AND atomic unvisible Au⁰ atoms?

 

[butcher]

Gold ions itself does not color a solution much, we may see a yellow or orange if very concentrated, but the as far as coloring the solution, I do not consider it like other metals like copper that give a strong color to solution.

But with stannous chloride the reflection of light on the (reduced gold atoms joined in colloidal particles), the colloidal gold is very colorful violet solution, but it is these invisible particles (small clusters of gold atoms that reflect the light, it is the lights reflection we see , not the many atoms of gold joined into a tiny cluster of gold atoms).

If we add enough reductant (SMB) to reduce all of the gold in solution, we would convert all of the Au+++ (gold Ions) to gold metal atoms Au instantly.

But if we do not use enough reductant (SMB) then yes we can have both gold Ions still in solution and and the gold atoms combining to begin precipitating in solution.

If you used enough SMB you could test a sample of the the solution with stannous chloride even before the gold atoms came together and precipitated (or before you seen the brown powder), and the stannous chloride would not show positive (because the gold was reduced back to atoms of gold.

but if you did not use enough SMB and still had unreduced gold in solution you could test for those gold Ions with the Stannous chloride test.

Remember the stannous chloride test will not test for elemental gold (even if these atoms of gold have not gotten big enough to see, or if they are still floating around in solution).

The stannous chloride test works by reducing the gold Ions, and in the solution only allows the gold atoms to combine and to grow to certain size clusters of colloidal gold which will reflect the violet light.
Gold colloids of different sizes will reflect different colors of light, or different sizes of colloidal gold solutions look different colors.

The nice thing about the tin chloride solution is it will only make one size of colloidal gold, and consistently gives us the violet color if we had gold ions in the solution we tested.

.

 

[NeMonstr]

Colloidal gold is sometimes red. This depends on the particle size.