# Confusing acid dilution calculations



## goldsilverpro (Jun 16, 2011)

I just got an email from a guy that bought 35% nitric and wanted to know how much it would take to dissolve a gram of silver. When I answered him, like most everyone else, I just assumed that 35% would be the same as diluting 70% with an equal amount of water. In thinking about this further, I find this to be in error. Actually, when diluting 70% nitric, 50/50, the result is close to 41%, not 35%.

http://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry

Using this chart, here is the math:

What happens when you dilute 70%, by weight, nitric 50/50? The specific gravity (g/ml or g/cc), at 20C, of 70% is 1.4134 (see chart). Therefore, a liter of 70% would weigh 1413.4 grams. Therefore, a liter of 70% would contain 1413.4 x .70 = 989.4 grams of nitric. By adding a liter of water, which weighs about 1000 grams, to a liter of 70% nitric (thereby diluting it 50/50), the total weight would be 1413.4 + 1000 = 2413.4 grams. Therefore, the final acid strength, by weight, of 50/50 70% nitric, would be 989.4/2413.4 = .41 or 41%.

Here again, by the chart, a liter of 41% nitric would contain 1252.7 x .41 = 513.6 grams of nitric. On the other hand, by the chart, a liter of 35% nitric would only contain 1214 x .35 = 424.9 grams of nitric. Therefore, the man I advised would require 513.6/424.9 = 1.209 times more 35% nitric than I told him. I told him to use 2.44 ml of 35%/gram of silver. It would actually take 2.44 x 1.209 = 2.95 ml of 35% nitric to dissolve one gram of silver.

As you can see, the math on this is not intuitive. You have to use a percentage/specific gravity chart like the one above in order to get the correct answer. There are other ways, but this is the easiest.

This works with all acids or any other solution that you can find a specific gravity chart on.

I did this math hurriedly. If anyone finds an error in my logic or my math, please let me know.

I might note that all this can vary slightly, depending on the actual temperatures involved. I used the specific gravities at 20C (68F) for all calculations.

Chris


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## samuel-a (Jun 16, 2011)

Hi Chris 

Thank you for crunching the numbers.
Though, somthing doesn't add up here. When I'll get home later I'll try and figure out what.


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## goldenchild (Jun 16, 2011)

I'm a bit confused too. 

1000 ml of 70% nitric has 700ml of pure nitric

1000ml of water added to the 1000ml of 70% nitric is a total of 2000ml of solution

700ml / 2000m = .35

right? :|


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## Sucho (Jun 16, 2011)

what about doing neutralising titration ? :lol:


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## goldsilverpro (Jun 16, 2011)

> 1000 ml of 70% nitric has 700ml of pure nitric



That's true, BY VOLUME, but it's 70% BY WEIGHT and 1000 ml of 70% weighs 1413.4 grams. Apples and oranges! The WEIGHT of nitric in a liter of 70% = 1413.4 x .70 = 989.4 grams.
My own Edit: That's NOT true, BY VOLUME. A liter of 70%, by weight, nitric contains 1413.4 - 989.4 = 424 grams or ml of water (it's 42.4%, by volume, water). Therefore, a 70%, by weight, nitric solution would be the same as a 100 - 42.4 = 57.6%, by volume, nitric solution.



> 1000ml of water added to the 1000ml of 70% nitric is a total of 2000ml of solution



That's true, but the 2000 ml weighs 1413.4 + 1000 = 2413.4 grams. BY WEIGHT, therefore, the nitric is 989.4/2413.4 = 41%

The problem is that you're combining 2 different materials, nitric and water, which have great differences in their specific gravities.


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## Sucho (Jun 16, 2011)

an international agreement for stating a concentration of acids is in percent w/w or percent v/v?because both calculations are correct.in only depends on apples and oranges...


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## goldenchild (Jun 16, 2011)

Think I get it now. Adding a liter worth of say corn syrup to a liter of water is heavier than a liter of water to a liter of water. Same in volume different in weight. 

So when we are cutting our 70% nitric we should use less nitric and more water... I think. I wanna say to use 6% less nitric EX. 94ml of nitric to every 100ml of water. But like GSP said, this math isn't very intuitive.


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## goldsilverpro (Jun 16, 2011)

The purpose was to determine how much 35% nitric it would take to dissolve a gram of silver.

I can't see how this would have anything to to do with an international agreement. The concentrated acids we buy here in the US are all based on weight percentage, as far as I know. Hydrochloric is about 35-37%, Sulfuric is about 95-96%, and nitric is about 67-70%. All these percentages are by weight. I haven't carefully read an acid bottle label recently but, if I remember right, it just gives the percent and doesn't say whether it's by volume or by weight. It is assumed to be by weight. If it is different in other countries, I don't know. I would doubt it, though, unless it is designated on the label as being by volume.



> So when we are cutting our 70% nitric we should use less nitric and more water... I think. I wanna say to use 6% less nitric EX. 94ml of nitric to every 100ml of water.



Not true because when we use the term 50/50, a by volume dilution, we are assuming you use 100 ml of 67-70% nitric and 100 ml of water.


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## samuel-a (Jun 16, 2011)

samuel-a said:


> Though, somthing doesn't add up here. When I'll get home later I'll try and figure out what.



Oranges and Appels :mrgreen: 

My bad.


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## goldsilverpro (Jun 16, 2011)

You guys were starting to make me doubt. However, I just couldn't reconcile in my mind that a 70%, by weight, nitric solution was equal to a 70%, by volume, nitric solution (it isn't). I now know I was right on target all along. See my Edit above.

I told you this stuff was confusing.


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## samuel-a (Jun 16, 2011)

goldsilverpro said:


> You guys were starting to make me doubt. However, I just couldn't reconcile in my mind that a 70%, by weight, nitric solution was equal to a 70%, by volume, nitric solution (it isn't). I now know I was right on target all along. See my Edit above.
> 
> I told you this stuff was confusing.



Chris, you are my favorite math freak 8)


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## Oz (Jun 16, 2011)

Interesting point GSP, I had always treated my acid concentrations as V/V.

Since this discussion was centered around nitric I read the label on mine. I use Mallinckrodt 68-70% ACS nitric. I will be darned if I can find anything that indicates if the concentration is V/V or W/W. Even where it is giving the trace contaminates in PPB it says “68-70% HNO3. The fomula weight on the container is no help as it is obviously just for the nitric not the mix as it is FW: 63.01 Even going to the manufacturers website I cannot find the answer.

At the end of the day without knowing if your concentration percentage starting point is V/V or W/W the math becomes meaningless. 

This will bug me, thanks GSP!


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## Sucho (Jun 16, 2011)

i have asked my friends from inorganic department. it is w/w for all aquaeous solutions of acids


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## Fournines (Jun 16, 2011)

Our suppliers sell us all of our chemicals by weight, not by volume.


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## Fournines (Jun 16, 2011)

This calculator also shows 41% if you enter

1413.4 and 70 in the first section, and 1000 and 0 in the second one.

http://www.handymath.com/cgi-bin/chemcalcc.cgi?submit=Entry


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## goldsilverpro (Jun 16, 2011)

Fournines said:


> This calculator also shows 41% if you enter
> 
> 1413.4 and 70 in the first section, and 1000 and 0 in the second one.
> 
> http://www.handymath.com/cgi-bin/chemcalcc.cgi?submit=Entry



Great find, Matt! That's a calculator everyone should bookmark and use. Sure saves doing a lot of confusing math. Personally, though, I enjoy the math because it allows me to visualize what is really going on.


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## goldsilverpro (Jun 17, 2011)

In general, if you combine 2 materials with different specific gravities (SG), the percentage, by volume, of the one with the higher SG, will be lower than it's percentage, by weight.

As a hypothetical thought experiment, to make this clearer, let's say you combined 1000 g of Au powder with 1000 g of Al powder. Then, let's say you could press all the air out, distort all the particles together, and make a solid cake. In this cake, the weight percentage of the gold is 50% but the volume percentage of the gold is only 14%. That is very easy to see how that would occur. In general, it's a similar situation when mixing 2 liquids of different SGs.


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## WiredOnSite (Sep 29, 2011)

I just purchased 50% Nitric acid v/v & 50% Hydrochloric acid v/v both are in 500ml amber glass bottles. I am going to be refining gold but a very small amount as a test as I do not want to work with these chemicals in the first place. So there for I have some jewelry about 28 grams of 14k. What amounts of each acid do I use, did I purchase the correct chemicals? I also will be getting Nava granular pH Decreaser Granular Sodium Bi-sulfate to precipitate the gold from the solution which I believe is ok. Any help would be appreciated!


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## philddreamer (Sep 29, 2011)

WiredOnSite, not bisulFATE, bisulFITE is what you'll need. You can find it @ beer & wine supplies stores or @ hardware stores as Stump Out by Bonide. 
Do some more reading before starting your batch. You should inquart your 14k before running thru acids & many more steps before adding your SMB.

Take care!

Phil


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## bswartzwelder (Nov 6, 2011)

I am really confused by these numbers. I would like to make real Aqua Regia (not HCl an Sodium Nitrate) and have seen a video which states "use 20 ml of Nitric acid and 60 ml of Hydrochloric acid". Everywhere that I have seen the formulation of Aqua Regia, it has been 1 part Nitric to 3 parts Hydrochloric. How is this affected if I use Nitric acid with a strength of 70%and Hydrochloric acid with a strength of 31% or 32%? I don't indulge in alcoholic beverages, but I'm almost ready to try. Do I use wine with 17.5% alcohol or whiskey at 50% alcohol? :roll:


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## Geo (Nov 6, 2011)

bswartzwelder said:


> I am really confused by these numbers. I would like to make real Aqua Regia (not HCl an Sodium Nitrate) and have seen a video which states "use 20 ml of Nitric acid and 60 ml of Hydrochloric acid". Everywhere that I have seen the formulation of Aqua Regia, it has been 1 part Nitric to 3 parts Hydrochloric. How is this affected if I use Nitric acid with a strength of 70%and Hydrochloric acid with a strength of 31% or 32%? I don't indulge in alcoholic beverages, but I'm almost ready to try. Do I use wine with 17.5% alcohol or whiskey at 50% alcohol? :roll:



if i may,i would like to help save you a few headaches.toss the ratio formulations out the window.when working with hcl acid and nitric acid to make AR,start with the respective amount of hcl only and add nitric acid in small increments at a time. add a few ml's of nitric and wait for it to NOx off before adding more,be sure to look and see if there is still metal to be dissolved before adding more nitric and at the last addition stop with the slightest amount of metal remaining undissolved.just add the undissolved metal to your next batch.this will help cut down on free nitric in your solution and its always better to have free hcl in solution rather than free nitric.


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## goldsilverpro (Nov 6, 2011)

bswartzwelder said:


> I am really confused by these numbers. I would like to make real Aqua Regia (not HCl an Sodium Nitrate) and have seen a video which states "use 20 ml of Nitric acid and 60 ml of Hydrochloric acid". Everywhere that I have seen the formulation of Aqua Regia, it has been 1 part Nitric to 3 parts Hydrochloric. How is this affected if I use Nitric acid with a strength of 70%and Hydrochloric acid with a strength of 31% or 32%? I don't indulge in alcoholic beverages, but I'm almost ready to try. Do I use wine with 17.5% alcohol or whiskey at 50% alcohol? :roll:



The best way is to use the 2 acids separately, as Geo suggested. However, if you want to premix the aqua regia and you have 70% nitric and 31% muriatic (hydrochloric), I would use 4.5 to 5 volumes of the muriatic and 1 volume of the nitric.


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## bswartzwelder (Nov 8, 2011)

This is my second attempt at posting this message. I am having trouble with my phone and DSL lines and Verizon seems to be taking their good old time at fixing it.

Thanks to Geo and GSP for your answers. I will be using the information when I start using Nitric acid instead of the HCl/Sodium Nitrate solution.

I have started reading Hoke and see where she refers to using copperas. Also, she hasn't mentioned Urea yet. Are these differences because of new developments in the chemistry field since the book was written? I graduated High School in 1968 and have degrees in Electrical and Electronic Technology and Nuclear Engineering Technology so my lack of college chemistry courses is a little strange, but I have always loved any science classes. Since all the most knowledgeable people on this forum highly recommend Hoke, I will continue reading her book, but I am wondering how to sort out the outdated stuff from that which is in use today? I don't want to spend a lot of timetrying to fully comprehend a poiunt she is trying to make only to find it is not still considered relevant.


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## jimdoc (Nov 8, 2011)

bswartzwelder said:


> I have started reading Hoke and see where she refers to using copperas. Also, she hasn't mentioned Urea yet. Are these differences because of new developments in the chemistry field since the book was written? I graduated High School in 1968 and have degrees in Electrical and Electronic Technology and Nuclear Engineering Technology so my lack of college chemistry courses is a little strange, but I have always loved any science classes. Since all the most knowledgeable people on this forum highly recommend Hoke, I will continue reading her book, but I am wondering how to sort out the outdated stuff from that which is in use today? I don't want to spend a lot of timetrying to fully comprehend a poiunt she is trying to make only to find it is not still considered relevant.



Although Hoke's book is from the 1940's, I would say that it is all still important to learn the basics. She won't mention urea because it is not needed when you use her methods. You can learn some newer work around for electronic scrap here on the forum, computers were not around for Ms. Hoke to be refining in her time. And she had no problem finding nitric acid and any other supplies. Most of the newer methods have been to get results with chemicals you can easily find. Just don't use gasoline for incineration like she says to do, that one will be a mistake. All her other advice will get you on the right track.

Jim


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## philddreamer (Nov 8, 2011)

Bswartzwelder, when making AR it's not so critical to use more of HCL. Some use a 4:1 HCl/Nitric; I use more @ times. 
If a make Poorman's AR I use 5:1 & a bit of peroxide, heat, & then I add the sodium nitrate in small increments until I get a nice reaction. 

I find that it's more important to calculate the amount of nitric needed in relation to the material to be digested, that way you won't have too much nitric left over & now have to deal with it, properly, like evaporating it, which is time consuming. 

Take care!

Phil


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## kurt (Feb 15, 2012)

OK - this one goes completely over my head - sooo - I am using 67% nitric - how much water do I add (to lets say 30 ml of 67% nitric) to get a true 50/50 dilution.

Kurt


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## Geo (Feb 15, 2012)

to get a true ratio of acid to water 50/50, you have to do it by weight and not by volume.

nitric acid at 68% weighs 1.42 g/ml

water weighs 1 g/ml

im not good with math but its a simple equation to figure out how much of each to add for X amount of volume.

30 ml x 1.42 = 42.6 of nitric acid + 42.6 ml of water = 72.6ml of true 50% nitric acid


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## lazersteve (Feb 15, 2012)

50/50 Nitric acid is actually in the neighborhood of 35% concentration. The term 50/50 comes from the fact that you mix the 70% nitric acid with an equal volume of water.

I find it easier to work with acids in their molar form. 70% HNO3 is approximately 16 M (M = moles per liter).

So 50/50 nitric acid (a slang term we use here) is 16 / 2 = 8 M

61% HNO3 is :

61/70 = 0.87 

the concentration of 70% HNO3.

0.87 x 16 M = 13.92 M

To get to the desired 8 M we need 

8 / 13.92 = 0.575 L x 1000 mL/L = 575 mL of 61% acid

and

1000mL - 575mL = 425 mL of water.

So to get 35% acid from your 61% acid add 425mL of water to 575mL of acid.


Steve


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## lazersteve (Feb 15, 2012)

Sorry I thought you were using 61%, for 67% it's:

67/70 =0.957

0.957 x 16M = 15.31 M

8 M / 15.31 M = 0.523 x 1000 mL/L = 523 mL

1000mL - 523 mL = 477mL H2O

477 mL water to 523 mL 68% HNO3.

Steve


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## Geo (Feb 15, 2012)

i understand that there is water in the acid to start with, at 68% that leaves 32% water and its only mols of nitrogen dioxide that constitute the concentration of the acid. i never did well in chemistry, but i was wondering why my formula is wrong. i trust you because you know worlds more than i do about it but here is my problem. i go to wiki and it tells me that acid has a specific weight and can be measured and you can calculate density by weight, so if you use weight to calculate density then cant you use weight to calculate amount of acid in a solution?


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## lazersteve (Feb 16, 2012)

Geo,

I've always used molarity to solve these kinds of problems. The shortest answer I can provide you is that the density of acids vary with concentration. As an example check the sulfuric acid concentration chart I posted in the data section. I'm not saying you are wrong; in fact I did not check your answer. 

I was simply sharing the way I do this sort of calculation.

Steve


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## Meh (Feb 16, 2012)

Yes, this is quite confusing to a noob who is still trying to figure some of these basic things out.

I was able to find a table of specific gravity for HNO3 at various temperatures here: http://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry

I plotted the 15*C column data points, this really shows how non-linear the change in SG is with concentration. Interestingly though, it's reasonably linear in the 30 to 70% range.



Once I finally caught on that % concentration referred to the weight of the pure acid content to the weight of the solution and not the volume of the pure acid to the total volume of the solution, I was able to make sense of the dilution calculations.

For what it's worth, I think Geo's post from yesterday morning is probably as good a method as anything. (My days of calculating Molar concentrations in acids ended with my undergrad chemistry courses 33 years ago.)

Al


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## goldsilverpro (Feb 16, 2012)

Meh said:


> For what it's worth, I think Geo's post from yesterday morning is probably as good a method as anything. (My days of calculating Molar concentrations in acids ended with my undergrad chemistry courses 33 years ago.)





Geo said:


> to get a true ratio of acid to water 50/50, you have to do it by weight and not by volume.
> 
> nitric acid at 68% weighs 1.42 g/ml
> 
> ...



Sorry, it won't work. 

The 30 ml is only 68% nitric. Therefore, 30 ml of 68% nitric contains 42.6 x .68 = 28.97 g of HNO3 and 42.6 x .32 = 13.63 g of water. To make a 50%, by weight, solution from 30 ml of 68% HNO3, you would add 28.97 - 13.63 = 15.34 ml of water. You end up with 30 + 15.34 = 45.34 ml of solution containing 28.97 g of HNO3 and 28.97 g of H2O - a true 50%, by weight, HNO3 solution.


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## NoIdea (Feb 16, 2012)

Hello All - when working with reactions, it always pays to work with molar concentrations along with a picture of your reaction. For working out dilutions try this link, very handy in deedy :lol: 

http://www.trimen.pl/witek/calculators/stezenia.html

Cheers 

Deano


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## Meh (Feb 16, 2012)

goldsilverpro said:


> Meh said:
> 
> 
> > For what it's worth, I think Geo's post from yesterday morning is probably as good a method as anything. (My days of calculating Molar concentrations in acids ended with my undergrad chemistry courses 33 years ago.)
> ...



Might be terminology - I interpreted Geo's calculation to be diluting 70% (and close enough to 68% acid) acid BY 50% to yield 35% HNO3. The calculation appears to work well in that context. But you're right, it would not yield 50% HNO3 by weight.


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## goldsilverpro (Feb 16, 2012)

Very good on the graph! I've wanted to do that for a long time but was too lazy. Now, the formula can be easily derived.

Assuming it's linear (it's pretty close) between 0% and 70%, I derived the formula from the graph:

Specific Gravity = (.006 x % nitric, by weight) + 1

I've never see that formula written anywhere. It's good to know and easily memorized. Now I won't have to go to handymath.com every time I need the SG of a nitric solution. Now, with that formula, I can calculate any nitric dilution.

The other calculation methods posed on this thread were in error because it was assumed, mathematically, that the specific gravities of the 2 strengths (e.g., 61% and 70%) in question were the same. When you mix something heavier (nitric acid) with something lighter (water), each mixture will be of a different specific gravity. This difference will affect the proper mix for a desired dilution.


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## Geo (Feb 16, 2012)

thanks GSP, i needed to know why the formula i had wouldn't work. i of coarse knew the acid contained water but i didn't subtract the water in the acid from the total amount. i may not ever need to have this degree of accuracy but its nice to know how to do it correctly.


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## tensor9 (May 22, 2012)

Skimming through this thread makes my blood boil a little. Concentration in percentages is just so silly and ambiguous. Is it by mass? By volume? How does the density change? It's not universal by any means. Molarity is a much more useful, unambiguous concentration.

It is the way it is though, so I guess we just have to deal with it.


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## lazersteve (May 22, 2012)

I agree with you tensor, Molarity is the best way to determine concentration. Unfortunately not everyone on the board has a firm understanding of the terms mole or molarity as they apply to chemistry.

Steve


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## tensor9 (May 22, 2012)

And even if they did, acids and such are still reported in such a manner. There's just no escaping it.


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## goldsilverpro (May 22, 2012)

Unfortunately, when you buy acid, it's usually sold as a weight (or, mass) percentage(weight of acid/weight of acid plus water). I never liked it either but that's just the way it is and you have to learn to deal with it. Working in moles would be much, much easier.

Using percentages. How many here can calculate how much water would be added to convert 500ml of 60%(w/w) nitric to 30%(w/w) nitric? It's not 500ml. It's 664ml. Must first use S.G. charts, graphs, or a formula like the one I gave above. Not intuitive.

On the other hand, how much water is needed to convert 500ml of 16 Molar nitric to 8 Molar nitric. It's 500ml. Lots easier. Very intuitive.


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## j2000 (Nov 30, 2015)

What percent "standard" of nitric acid when we wanna make good AR ?

I have 65% NA, and i want to dilute to be 10% NA for 1000mL.
I use calcutor and calcultor said: 116.52mL nitric acid + H20 883.38mL.


This is correct?

Regards

Joel


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## solar_plasma (Nov 30, 2015)

j2000 said:


> What percent "standard" of nitric acid when we wanna make good AR ?



"Good AR" is when you use enough HCl to cover the material to be dissolved and add only as much nitric as needed, though use at least 3-4 parts HCl per 1 part nitric. If you use SMB for the precipitation, it will consume some HCl and you need even more HCl. The concentration of the nitric you are adding is not important.



j2000 said:


> I have 65% NA, and i want to dilute to be 10% NA for 1000mL.
> I use calcutor and calcultor said: 116.52mL nitric acid + H20 883.38mL.



Better and easier to think and calculate in molarity and weight than in % and ml. Why do you want to dillute your nitric. I can't see the necessity. Just add about 1,7ml of your 60-70% nitric per each gramm of gold. Add it in small increments, give it good time to react, even more since your material is in shape of fine powder and stop adding, when the reaction stops (it has stopped when there is no detectable production of NOx gasses anymore).

I have no clue about ores and concentrates, so wait if other opinions or corrections will pop up.


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## goldsilverpro (Dec 2, 2015)

j2000 said:


> What percent "standard" of nitric acid when we wanna make good AR ?
> 
> I have 65% NA, and i want to dilute to be 10% NA for 1000mL.
> I use calcutor and calcultor said: 116.52mL nitric acid + H20 883.38mL.
> ...


I get exactly the same results that you did - 116ml of 65% diluted to 1 liter = 10%


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## rickbb (Dec 2, 2015)

As Solar posted, (and I have learned the hard way), if you mix your AR based a ratio or % type of formula you will be wasting nitric.

Use only the amount needed to dissolve the gold. You will find you need somewhat less than the general rule of 1 to 3 or 1 to 4 ratios.


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