# Silver recovery as silver chloride



## saadat68 (Jun 17, 2017)

Hi
I want to precipitate my silver as silver chloride from silver nitrate solution. Yes I know cementation is better but I must use chloride method. ( Because I have Mercury in my solution and I want to get silver without it )

In my process I can not estimate how much silver is in silver nitrate solution. So according to below table I will loss significant silver 

http://www.axgig.com/images/14088847220784620735.jpg

Yes I can drop a copper bar in solution after separating silver chloride but I can not do this too ( Because Mercury will cement with silver)  

Is there a way to recovery all of silver as silver chloride with my conditions ? For example use HCl or KCl instead of NaCl
I need something to add 30% excess of stoichiometric to solution and recovery of above 99% of silver.


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## jimdoc (Jun 17, 2017)

Didn't keeping all your posts about silver in one thread sound good to you?


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## Topher_osAUrus (Jun 17, 2017)

Wouldnt mercury chloride ppt with an stoichiometric excess of NaCl?..


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## saadat68 (Jun 18, 2017)

Topher_osAUrus

I don't know :? 
Can someone say how to determinate NaCl/HCl amount to precipitate all of silver ?


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## goldsilverpro (Jun 18, 2017)

Topher said:


> Wouldnt mercury chloride ppt with an stoichiometric excess of NaCl?..


Assuming it is in water and is a simple compound like Hg2(NO3)2. Mercurous (Hg+) essentially would all precipitate stoichemetrically because HgCl (actually, Hg2Cl2) is nearly insoluble. Mercuric (Hg++) wouldn't because HgCl2 is fairly soluble. Unlike mercury, silver has only one valence form - Ag+.



saadat68 said:


> Can someone say how to determinate NaCl/HCl amount to precipitate all of silver ?


One Mole (M) of dry, pure NaCl (58.45g) will precipitate one M of silver (107.88g) from a silver nitrate solution.

HCl is a little more iffy. It depends on the strength and the temperature. 

One Mole of HCl weighs 36.46g. One M of HCl will precipitate one Mole of silver from solution.

A typical strength of commercial HCl is 37%. This contains 12.00 M of HCl. One liter @ 72F will precipitate 12 x 107.88 = 1295g of silver

A typical strength of muriatic acid is 31.45% HCl. This contains 9.96 M of HCl. One liter @ 72F will precipitate 9.96 x 107.88 = 1074g of Ag

This is only a guide and it is worthless if you have no idea how much silver you have or the strength of the HCl. Even if you know, i almost impossible to add the absolute correct amount09I always add the Cl source in increments, adding smaller amounts towards the end point. After each addition, I stir it and let it settle for a few minutes and then add one drop of the salt water or HCl with the eyedropper close to the solution to try and keep the drop at the surface. If you see more milkiness where the drop is, you need more Cl-. As you approach the point where most of the silver has precipitated, the AgCl produced will be finer and will settle much more slowly. For complete settling, it might take overnight.

I just thought of something else. If there is copper dissolved along with the silver and you are using HCl, you will find that, when all the silver has already precipitated, the drop of HCl will turn a clear yellow color.


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## saadat68 (Jun 18, 2017)

> Assuming it is in water and is a simple compound like Hg2(NO3)2. Mercurous (Hg+) essentially would all precipitate stoichemetrically because HgCl (actually, Hg2Cl2) is nearly insoluble. Mercuric (Hg++) wouldn't because HgCl2 is fairly soluble. Unlike mercury, silver has only one valence form - Ag+.



Thank you for your post and tips 
And about mercury I add these from an article about getting silver from silver oxide cells: 

In the light of the results obtained through dissolution experiments, optimum leaching condition is as follows: 5 mL, 1.0 M HNO3, 45 °C, 1 h, 200 rpm, 250 mg cell powder. It was observed that mercury
is entirely converted to Hg+ 2 ions under this condition, and did not
precipitate as mercury (II) chloride. Even though silver could be
dissolved more than 99% at 25 °C using 2.0 M nitric acid solution
mercury was observed to be converted to mercury (I) ions, which coprecipitates
as mercury (I) chloride with the silver chloride during the
precipitation process. This precipitate would present an undesirable
impurity in the final product.
---------------------------
At room temperature, a silver precipitation efficiency of 99.9% was
obtained after 60 min at 200 rpm. It is worth mentioning that mercury
remained in the solution because mercury was in the form of Hg
(NO3)2 (Petrucci and Harwood, 1997), which made it possible to
precipitate silver selectively


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## saadat68 (Jun 19, 2017)

> optimum leaching condition is as follows: 5 mL, 1.0 M HNO3, 45 °C, 1 h, 200 rpm, 250 mg cell powder.


They used 65% HNO3 in their research 
Can someone say how much nitric acid in gram or ml I must use for 250 mg powder ? 
I know 1 M Nitric acid 65% is 96.9384 ml + 903.06 ml water


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## goldsilverpro (Jun 19, 2017)

saadat68 said:


> > optimum leaching condition is as follows: 5 mL, 1.0 M HNO3, 45 °C, 1 h, 200 rpm, 250 mg cell powder.
> 
> 
> They used 65% HNO3 in their research
> ...



I think your numbers are quite wrong.

According to the calculator at the top of this page:
http://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry
The weight of 65% HNO3, by weight, @ 22.2C (about 72F) is 1388.13 g/l
The formula weight of HNO3 is 63.02g. This is the weight of 1 Mole (M) of HNO3
In one liter of 65% HNO3, there are 1388.13 x .65 = 902.3 grams of HNO3
*∴*, there are 902.3/63.02 = 14.32 M of HNO3 in one liter of 65% HNO3
*∴*, one M is 1000/14.32 = 69.84 ml of 65% HNO3
And, *∴*, the water in one liter of 65% HNO3 is 1388.13 - 902.3 = 485.83g (or, ml)

In an open container, it takes about 1.22 ml of 70% HNO3, diluted 50/50 with distilled water, to dissolve 1g of silver.
Using the same math as above, one liter of 70% HNO3 contains 987g of HNO3
*∴*, 70% HNO3 is 987/902 = 1.094 times stronger that 65% HNO3
*∴*, it would take about 1.094 x 1.22 = 1.33ml of 65% HNO3 to dissolve a gram of silver.
*∴*, it would take about 1.33 x .250 = .33ml of 65% HNO3 to dissolve 250mg of silver. That's about 6.6 drops from a 20 drop/ml eyedropper. I always determine the drops/ml with water at room temp. when I get a new eydropper. They can vary from about 15 to 25 drops/ml. Handy to know. The glass ones from a pharmacy usually run about 20-25 drops/ml.

*∴* = Therefore

Note: This is about the most it will take. It can take less HNO3/gram by using more dilute HNO3, by covering the beaker, and by using a lower temperature (I think).

I didn't take the time to check my math. If I made an error, let me know. Please note that the math is not linear. You must use the specific gravities of the various concentrations to get an accurate answer. 

Anyone know the number key for "therefore", 3 dots in the form of a triangle?


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## Topher_osAUrus (Jun 19, 2017)

GSP, you are a god with a calculator!

Your sign, dear sir
:arrow: https://en.m.wikipedia.org/wiki/Therefore_sign


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## goldsilverpro (Jun 19, 2017)

Thanks, Topher! 

Simple and straightforward but, at the same time, very confusing math.


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## saadat68 (Jun 19, 2017)

goldsilverpro said:


> Thanks, Topher!
> 
> Simple and straightforward but, at the same time, very confusing math.


Thank you GSP
But I want to know how much nitric acid acid they used in their research. They said 5 ml;1 mole HNO3 for 250 mg of powder.


> optimum leaching condition is as follows: 5 mL, 1.0 M HNO3, 45 °C, 1 h, 200 rpm, 250 mg cell powder. It was observed that mercuryis entirely converted to Hg+2ions under this condition,


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## saadat68 (Jun 19, 2017)

So they used 5 ml of 69.84 ml of 65% HNO3 + 444.33 ml water 
69.84+485.83= 555.67
555.67 - 1000 = 444.33 ml

Right ?

But how much* 65% HNO3* is in 5 ml of that solution ? Can I divide 69.84 to 200 and say in 5 ml of that solution there is 0.349 ml of 65% HNO3? (1000ml/5ml = 200)


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## goldsilverpro (Jun 19, 2017)

Is this what you want?

To make 1 M nitric from 65% v/v nitric. 
1 liter of 65% HNO3 contains 14.32 M of HNO3
There is 1 M of HNO3 in 69.84 ml of 65% HNO3
*∴*, 1 liter of 1 M Nitric is made up from 69.84ml of 65% HNO3 plus 1000 - 69.84 = 930.16 ml of pure water.

There are 63.02g of HNO3 in 1 liter of 1 M HNO3
*∴*, 5 ml of 1M HNO3 contains 63.02 x .005 = .315g of HNO3


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## saadat68 (Jun 20, 2017)

Yes 
Thanks


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