# Silver nitrate



## Drewbie (Jan 20, 2011)

Well, it's been a long time since high school chemistry, but by my calculation, cementing 100g of AgNO3 crystals will produce ~63g of Ag, and require ~56g of copper to complete the reaction.

Now, someone tell me where I went wrong please. 

Or if short of time, tell me the correct amounts.


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## lazersteve (Jan 20, 2011)

How about a link where this has all been done before:

Silver Nitrate + Copper Reaction

It's nice to you are enjoying the science behind refining.

Steve


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## Drewbie (Jan 20, 2011)

OK. So I was out on the ratio of silver to copper. B student, 30 years ago will account for that.

But what about the actual amount of silver produced from x amount of silver nitrate?

I ask because if silver shoots to the moon like many expect it to (current gold:silver ratio etc), a quick and ready source of discount quality silver might be the chemical suppliers who have silver nitrate crystals in stock. This all depends on exactly how much silver comes out of >99% pure silver nitrate.


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## lazersteve (Jan 20, 2011)

AgNO3 ~= 169.9 g/M
Ag ~=107.9 g/M

Assuming zero water or contamination:

107.9 / 169.9 = 0.635

therefore

100g AgNO3 ~= 100 x 0.635 = 63.5g Ag

Steve


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## Drewbie (Jan 21, 2011)

Wow, I was bang-on for the silver content. That makes me happy.

Thanks Steve!


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## Sucho (Jan 21, 2011)

lazersteve said:


> AgNO3 ~= 169.9 g/M
> Ag ~=107.9 g/M
> 
> Assuming zero water or contamination:
> ...





equation.. 2Ag+ + Cu -> 2Ag + Cu2+

Cu- 63,5 g/M
Ag - 107,9 g/M

n=m/M - 63,5 g Ag+ / 107,9 g/M = 0.5585 M
m = n.M - 0,5585. 63,5 Cu = 37,37g /2 (because of oxidizing number of copper / 1 atom of copper could reduce 2 atoms of silver) =18,69 g solid Cu is necesarry to reduce 100 g of AgNO3 (63,5g of Ag+) in solution


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## silveras (Jan 23, 2011)

How much nitric acid do I need to dissolve 100g of pure silver?


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## nickvc (Jan 23, 2011)

silveras said:


> How much nitric acid do I need to dissolve 100g of pure silver?




You can find the answer to this two ways....
1.Put your silver in a beaker add a cup full of water and then using a measuring jug slowly add the nitric until it's dissolved all your silver.
2.Use the search function at the top right of your screen which as a new member you will find extremely useful.


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## silveras (Jan 23, 2011)

Thanks. Will do that.


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## Sucho (Jan 23, 2011)

nickvc said:


> You can find the answer to this two ways....
> 1.Put your silver in a beaker add a cup full of water and then using a measuring jug slowly add the nitric until it's dissolved all your silver.
> 2.Use the search function at the top right of your screen which as a new member you will find extremely useful.



i agree with nickvc but i write you mathematical solution

silver in HNO3 equation : *3* Ag(s) + *4* HNO3 (ag) = *3* AgNO3 (ag) + NO(g) + *2* H20 (l)

stechiometric coeficients are very important to solve this problem correctly

Ag - 107,87 g/M
HNO3 - 63,01 g/M
density(HNO3) = 1.5129 g/cm3
n (Ag) = m/M = 0,927 mol

due to stechiometric coeficients you will need 4/3 mol of HNO3 to make 1 mol of AgNO3 from 1 mol of Ag

m (HNO3-100%) = 4/3 n(Ag) . M(HNO3) = 4/3 . 0.927 . 63,01 = 77.88 g 100% HNO3

HNO3 has higher density than water -V = m/density = 77.88 g /1.5129 g/cm3 = 51,48 cm3

but you can buy only 60-70 % HNO3 ( i will count for 65 % )

in grams - 100/65 = 1.5384 this is the coefficient 1.5384 . 77.88 = 119.82 grams of 65 % HNO3

119.82 -77.88 = 41.94 g of water = 41.94 cm3

now we see that 119.82 g of 65% HNO3 has a quantity od 41.94 + 51.48 = 93.42 cm3

we can find that 65% HNO3 has a density of cca 1.4 g/cm3

119.82 / 1.4 = 85.59 ml

here is a little imperfection , i thing its caused by a weak intramolecular interractions(van der Waals, ionic etc.) in azeotropic solution of HNO3/water

if i am wrong, please correct my math


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## lazersteve (Jan 23, 2011)

Here's the way I would solve the problem using Sucho's equation:

*3 Ag(s) + 4 HNO3 (aq) = 3 AgNO3 (aq) + NO(g) + 2 H20 (l)*

Knowing from literature that 1L of ~70% HNO3 is 15.9 moles of HNO3 we can compute:

(4 moles) / (15.9 moles / Liter) ~= 0.252 L * 1000 mL/L = 252 mL of ~70% HNO3 contains the 4 moles of HNO3 required in the equation.

The mass of silver in the equation Sucho provided is:

3 mole x 107.9 g/mole = 323.7 g Ag dissolved in the equation above.

To determine the number of grams dissolved per mL we have:

*Eq. 1)* 323.7 g / 252 mL ~= 1.285 g Ag per mL of ~70% HNO3

or inversely stated:

*Eq. 2)* 252 mL / 323.7 g = *0.778 mL ~70% HNO3 per gram of Ag*

Steve

PS: I use a different equation (*Ag + 2 HNO3 → AgNO3 + NO2 + H2O*) than Sucho and therefore get a different result (* Eq 1*= 107.9 g / 126 mL = 0.856 g Ag per mL of ~70% HNO3 and *Eq 2* = 126 mL / 107.9 g = *1.17 mL ~70% HNO3 per gram of Ag*).


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## Sucho (Jan 23, 2011)

Steve, which equation is correct? isnt it an equilibrium of two possible reactions ?

// your approach that 1 liter 70% HNO3 has 15.9 Mol is better...you dont have to deal with interractions between molecules in HNO3/H2O solution
i am only using my brain, mobile phone as a calculator and periodic table and i am only human :roll: 

whats your opinion ?


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## lazersteve (Jan 23, 2011)

Sucho said:


> Steve, which equation is correct? isnt it an equilibrium of two possible reactions ?
> 
> // your approach that 1 liter 70% HNO3 has 15.9 Mol is better...you dont have to deal with interractions between molecules in HNO3/H2O solution
> i am only using my brain, mobile phone as a calculator and periodic table and i am only human :roll:
> ...



I use the equation I posted in the footnote.

The equation is from wiki Silver Nitrate

The actual amount of nitric required varies on temperature, actual nitric acid concentration, reflux, and the amount of excess water in the reaction. I find the reaction equation in wiki is a little higher than what I witness experimentally. I run my reactions at below the boiling point, with a watch glass cover for reflux, and diluted with an equal volume of distilled water.

Steve


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## Sucho (Jan 23, 2011)

yes...these are only theoretical numbers and there is a lot of things that could affect nitric consumption...
everything depends on experience

// to equation...i thing that there is only NO presented in reaction , and when its released from the solution its oxidized by atmospheric O2 to NO2.
you can see it when you are boiling your watch glass covered solution- fumes beyond the solution are brown, but you can see throw...but when you uncover your container , fumes become much darker and with larger quantity.i thing this depends on oxidation of NO( its invisible gas) with atmospheric 02 - NO2 - dark brown gas.


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## lazersteve (Jan 23, 2011)

Honestly, I don't know what the correct equation is, I simply use the one from wiki. 

During my silver dissolution reactions I see brown fumes for the entire reaction inside of the covered beaker. Only a slight bit of NOx escapes and becomes clear as it absorbs into the air around the beaker. In fact I judge the end point of the nitric as when the red vapors clear inside the covered beaker and when all of the small fizzing bubbles stop emitting from the solution, or when all of the base metal and/or silver is digested. As the reaction progresses the brown fumes get darker red-brown until the nitric is all used up. I work outside so perhaps the sunlight is what causes the fumes in the beaker to be red-brown?

Here's a photo of the reaction I'm doing today:







Steve


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## Sucho (Jan 23, 2011)

i was thinking about test...during reaction on your photo, insert a tube from aquarium bubbler through bucket sink in the middle of brown air inside your covered bucket....then turn on your bubbler..if it will get darker - its oxidation with 02


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## ultlordx (Apr 8, 2011)

I guess this would be a good place to place this thought. So the reaction of silver with nitric acid is as follows.

3 Ag + 4 HNO3 → 3 AgNO3 + 2 H2O + NO

NO gas will react with oxygen (present in air) to form NO2 gas

2 NO + O2 → 2 NO2

NO2 gas will react with water to form more nitric acid along with nitrous acid (HNO2)

2 NO2 + H2O → HNO3 + HNO2

Nitrous acid (HNO2) rapidly decomposes through the following reaction

3 HNO2 → HNO3 + 2 NO + H2O

Now everything is back to nitric acid except for the NO which can undergo the cycle once more starting from the second reaction.
Based on these equations, assuming an ample supply of oxygen and water, isn't it theoretically possible to consume ALL of the nitric acid without losses to the atmosphere in the form of NO and NO2? Assuming you have the proper pressure vessel and your system is closed, this might be true. 

Also if you are using the copper to cement out your silver, you end up with a solution of copper nitrate. When heated, copper nitrate will decompose into an oxide of copper and NO2 gas, which could could be converted back into acid.

So I guess the question is, is it not theoretically possible to recover all nitric acid and reuse it in a perfect world?

I plan to run some tests eventually using a distillation flask as my vessel (I don't want to run the reaction in a closed vessel at this point), and bubble the resulting gasses through water. The water should react to form acid. I also believe I may have to bubble air through the water at the same time in order to present the oxygen (aquarium pump would work). The problem lies in producing small enough bubbles and allowing them to remain in the solution for as long a time as possible so they completely react, such that only air depleted of oxygen is leaving the system. I guess you would know if it is working properly if the gas leaving is clear and does not have any hints of the orange or brown color characteristic of NO2.

Any thoughts, suggestions, or corrections here would be appreciated.


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## suhailkhan8547 (Apr 10, 2011)

Hello everybody, I have 150ml of silver nitrate. 
how to recover silver from this solution. 
It has approximate 60 grams of pure silver.
what's the way for it?


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## Harold_V (Apr 11, 2011)

suhailkhan8547 said:


> Hello everybody, I have 150ml of silver nitrate.
> how to recover silver from this solution.
> It has approximate 60 grams of pure silver.
> what's the way for it?



Assuming you're using subsilver2 as your board display option, top right hand side of the page shows this:

FAQ *Search* Members User Control Panel 

This is a question that should not be asked. If you haven't read enough to know the answer, you should not have prepared silver nitrate. 

Harold


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## talalstuvs (Jun 2, 2011)

I want to make 10% of silver nitric i have silver and Nitric acid .Now what should i do to get 10 % silver Nitric . forget to tell you i use 10% silver nitric for silver electrolysis process. 

Example 
X ml Nitric
X g silver 
water or eles ???
waiting for ur reply .


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## Oz (Jun 2, 2011)

You have this 10% thing all over the forum in many different ways and venues. Normally I would call that spamming the forum, but by reading your posts I do not think that is your intent.

So we are at an impasse, help us understand what your ultimate goal is with this 10% thing. Is it 10% nitric, 10% silver nitrate, 10% silver? 

Give us the details of what you wish to accomplish so we can help you in an efficient and safe manner.


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## Fournines (Jun 2, 2011)

talalstuvs said:


> waiting for ur reply .



I looked for "ur" in the memberlist, but the name is not registered. You might be waiting a while for that reply from them.


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## talalstuvs (Jun 2, 2011)

Oz said:


> You have this 10% thing all over the forum in many different ways and venues. Normally I would call that spamming the forum, but by reading your posts I do not think that is your intent.
> 
> So we are at an impasse, help us understand what your ultimate goal is with this 10% thing. Is it 10% nitric, 10% silver nitrate, 10% silver?
> 
> Give us the details of what you wish to accomplish so we can help you in an efficient and safe manner.




Salam my question is that I want to make silver nitrate solution because I can't find silver nitrate here . Now what should I do to get silver nitrate solution ? And its should be 10 % not hight I use it for electrolysis silver .


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## goldsilverpro (Jun 2, 2011)

A 10% silver nitrate solution is 100 grams per liter. Silver nitrate is 63.5% silver. Therefore, a 10% silver nitrate solution contains 63.5 grams of silver per liter. It takes about 1.22 ml of 70% nitric acid to dissolve a gram of silver. To prevent silver nitrate crystals from forming upon cooling, dilute the required amount of nitric with an equal amount of distilled water before dissolving the silver. Heat the solution towards the end but don't boil it. When all the nitric has reacted, add enough distilled water to bring it to the proper volume. Filter before use if there are any solids (dirt and/or undissolved silver) present.

Notes: If you can't tolerate free nitric in the solution, use the calculated amount of nitric with a little heat at the end of the dissolving process, but use a little extra silver. That way, most all the nitric will be consumed. Remove the undissolved silver.

This is not an exacting process, due to slight variations in the nitric strength and your dissolving setup. If you require an exact 10% solution, you'll have to analyze the silver in solution and adjust the solution volume accordingly, either by evaporation (if the silver content is too low) or the addition of distilled water (if the silver content is too high).

If this is to be used as a standard electrolytic Thum or Moebius silver cell solution, you can tolerate a little free nitric and the silver content can vary, say, by 10-15%. A 10% silver nitrate solution contains about 7.7 tr.oz. of silver per gallon.

SEARCH AND STUDY THE FORUM!


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## talalstuvs (Jun 3, 2011)

goldsilverpro said:


> A 10% silver nitrate solution is 100 grams per liter. Silver nitrate is 63.5% silver. Therefore, a 10% silver nitrate solution contains 63.5 grams of silver per liter. It takes about 1.22 ml of 70% nitric acid to dissolve a gram of silver. To prevent silver nitrate crystals from forming upon cooling, dilute the required amount of nitric with an equal amount of distilled water before dissolving the silver. Heat the solution towards the end but don't boil it. When all the nitric has reacted, add enough distilled water to bring it to the proper volume. Filter before use if there are any solids (dirt and/or undissolved silver) present.
> 
> Notes: If you can't tolerate free nitric in the solution, use the calculated amount of nitric with a little heat at the end of the dissolving process, but use a little extra silver. That way, most all the nitric will be consumed. Remove the undissolved silver.
> 
> ...



Thank you very much for ur wealthy help sir I undersood all procedure except on pharagraph its is "When all the nitric has reacted, add enough distilled water to bring it to the proper volume" I don't know what does it mean. If ur mean to dilute enough water to get 10% silver nitrate if yes then how what is ratio ?and what is this silver nitrate acid. Which I make ?


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## goldsilverpro (Jun 3, 2011)

To avoid repetition, please explain in detail what you need this solution for and how much of this solution you need. If it's to make a solution for a silver purification cell, whether Thum (horizontal) or Moebius (vertical), the exact makeup is not that critical. If it's for something else, explain what it is and we'll go from there.


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## talalstuvs (Jun 4, 2011)

goldsilverpro said:


> To avoid repetition, please explain in detail what you need this solution for and how much of this solution you need. If it's to make a solution for a silver purification cell, whether Thum (horizontal) or Moebius (vertical), the exact makeup is not that critical. If it's for something else, explain what it is and we'll go from there.


 I want to make 60 ml of 10% silver nitrate for purifining silver my cell is vertical and i am using 3 volt for it now how much silver and how huch nitric acid I should use . Waiting for ur reply .


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## goldsilverpro (Jun 4, 2011)

> I want to make 60 ml of 10% silver nitrate for purifining silver my cell is vertical and i am using 3 volt for it now how much silver and how huch nitric acid I should use . Waiting for ur reply .



Are you sure that "60 ml" isn't a typo error? That must be the smallest silver cell on the planet. The biggest problem with something that small is the anode/cathode spacing. They will be so close to each other that the crystals formed on the cathode can rapidly bridge to the anode, maybe in a matter of minutes. If the electrodes short, you'll most probably blow the rectifier fuse and burn a hole in the anode bag. Then, you'll have to start over.

However, for 60 ml of a 10% AgNO3 solution, here's how I would do it. To make sure you don't end up with a bunch of excess nitric in the solution, I would use the calculated amount of nitric and about 10% more silver than calculated. For 60 ml of 10% AgNO3 solution the calculated amount of silver is 63.5 x .06 = 3.81 grams. To dissolve this much silver, it takes about 3.81 x 1.22 = 4.65 ml of 70% nitric. A 10% excess of silver is 3.81 x 1.1 = 4.19 grams. So:

(1) Weigh out 4.19 g of silver and place in beaker.
(2) To this, add a mixture of 4.65 ml of nitric plus 4.65 ml of distilled water.
(3) Allow to dissolve. This will generate heat but, when the reaction has slowed down, heat the solution to about 80C (don't boil) for about 5-10 minutes minutes to make sure most all of the nitric has been consumed. At this point, there should be a little undissolved silver.
(4) Remove from heat, allow to cool, and add about 30-40 ml of distilled water. Filter (to remove dirt and excess silver) in a very small filter and rinse with a total of not over 10-15 ml of distilled water.
(5) Add enough distilled water to bring the volume of the solution to 60 ml.

Like I said earlier, for several reasons I could name, you won't end up with an exact 10% solution (unless you're very lucky), but it could be close if you're careful.


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## talalstuvs (Jun 5, 2011)

thank you very much sir for your precious help.


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## goldsilverpro (Jun 5, 2011)

Also:
In a cell that small, your anode/cathode areas are probably 2 cm2, at the most. That means you are limited to about 0.08 amps. This would only produce about 0.3 grams of silver crystal per hour.

This is driving me nuts, so I have to ask this question. Why on earth do you want such a tiny silver cell?


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## gold4mike (Jun 10, 2011)

Maybe he wants a "thumbnail" sized Thum cell :?:


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