Preciptiating Gold by Volume - How to determine

[rusty]

for those using sulfur dioxide gas this would be a no brainer, there is no way you could over dope your gold chloride with precipitant.

I unfortunately do not have the luxury of having sulfur dioxide my choices at the moment are sodium metabisulfate, oxalic acid, laboratory produced sulfur dioxide, hydrogen sulfide or a metal exchange.

My goal is that have a nice clean gold drop without the hitchhikers for simplicity for my first precipitation I'll use sodium metabisulfate, here is where I'm at a loss in determining just how saturated my gold solution is to know how much precipitant to use.

Is there an easy way to determine the gold content of a solution, at this point all my gold chlorides from each batch processed have been added as one lot.

I let everything settle out for a week, this paid off in spades, I was surprised at how much silver chloride settled out, then after the silver chloride was filtered out the addition of sulfuric acid which produced a wee bit of lead precipitate.

If there is no easy way to determine the gold content of my solution, I'll take a measured sample then proceed to precipitate the gold from that then weight it up using those numbers to calculate the amount of sodium metabisulfate needed for my larger gold drop.

Also I'm thinking that once I have evaporated this 10,000 ml of gold chloride to a syrup, I would now know how many concentrated milliliters there are this would be a good starting point --- gold content using this data gleaned from wikipedia gold chloride solubility in water 68 g/100 ml - cold.

Still have to research the percentage of gold metal content of gold chloride.

Getting ready to evaporate the gold chloride.

 

[butcher]

In Aucl3 we have one mole of Au (gold) and 3 moles of Cl (chlorine)
molecular weights
AuCl3 = 303.32 g/mol
Au = 196.96 g/mol
Cl = 35.5 g/mol
Cl3 =(3 x 35.5)=106g/mol
(note we have 3 chlorine in AuCl3)


% composition of Au in AuCl3 =
197g/303.5g X 100 = 64.9%

% composition of Cl in AuCl3 =
106.5/303.5 X 100 = 35%

note this is not considering water or acid in solution

 

[g_axelsson]

I think I have read this in Hooke's "Testing of precious metals". My memory of the title could be wrong as I'm writing from a hotel room. Anyhow, the principle is easy and is used in many areas, not just precious metal refining.

Make a reference solution of known strength. Take 1 part of this solution and dilute with 9 parts water. Now you have a solution of 1/10 of the first solution. Do it one or two times more (whatever you need is) and create a 1/100 strength from the 1/10 and maybe a 1/1000 from the 1/100 too.

With these reference solutions add your unknown solution to a spot plate and one or more of your reference solutions in another spots. Add equal amounts of stannous to each sample and reference, compare the color between them.
If your unknown solution is too concentrated, dilute it too and compare. You could easily do a number of different strength of your unknown solution until one gives a close match to one of your references. Then it is just a matter of simple math to find the concentration of your unknown solution.

You can save the reference solutions and use later, but the concentration will rise if there is any evaporation so always keep them in closed bottles and remake them if you think some of the water could have evaporated.

Göran

 

[butcher]

I am just thinking out loud.

From Hokes we know 4 fl oz. of 32% HCl and 1 fl oz. of HNO3 will dissolve a troy ounce of gold.
This equates to 118.29ml HCl and 29.57ml HNO3 per toy ounce.

Dividing these figures down with 31.103 grams per troy ounce we get:
3.8ml HCl and 0.95ml HNO3 per gram of gold,
or 4.75ml of aqua regia per gram of gold,

This may be able to be used to get a starting point, this gives us an idea of the amount of gold in solution, If we used up all of the nitric acid dissolving gold, or did not add more than needed, but the solution can hold much more gold in solution than it takes to get it dissolved, we can see this when we concentrate the solution driving off liquids as gases and concentrating it down to an red orange solution in very little liquid, but maybe we could use this as a starting point (guess) of how much gold is in solution, before we concentrate (being sure no more gold will go into solution).

another thought is to take enough of a sample, precipitate the gold, and determine how much gold was in the sample.

maybe specific gravity if we knew the specific gravity of the acid with no gold and compared to specific gravity with gold.

 

[4metals]

for those using sulfur dioxide gas this would be a no brainer, there is no way you could over dope your gold chloride with precipitant.

Actually you can add too much SO2 and end up dropping copper as well.

 

[rusty]

for those using sulfur dioxide gas this would be a no brainer, there is no way you could over dope your gold chloride with precipitant.

Actually you can add too much SO2 and end up dropping copper as well.

So I'm guessing over doping with sodium metabisulfate will drag down copper as well.

Thank you 4metals for setting me straight on the SO2.

 

[4metals]

What actually forms from over adding SO2 either as a gas or as metabisulfite, is copper chloride. It can be seen as small white crystals in the precipitated gold. It is soluble and can be rinsed out with hot concentrated HCl but it is a pain to get it all, best not to form any. The mechanics of what happens is the sulfur dioxide reduces some copper at the very end of the reaction and the copper metal reacts with the high chloride concentration to drop copper (II) chloride.

 

[rusty]

What actually forms from over adding SO2 either as a gas or as metabisulfite, is copper chloride. It can be seen as small white crystals in the precipitated gold. It is soluble and can be rinsed out with hot concentrated HCl but it is a pain to get it all, best not to form any. The mechanics of what happens is the sulfur dioxide reduces some copper at the very end of the reaction and the copper metal reacts with the high chloride concentration to drop copper (II) chloride.

So it would be prudent to spend that extra time removing base metals avoiding this problem.

This is what I have tried to achieve, my chlorides combined look very good with a hint of platinum group ions in the mix.

 

[4metals]

That's why some refIners looking to get highest purity drop their gold twice. The second drop is usually out of a bright red acid which is largely unincumbered with base metals so it is less of an issue.