Confusing acid dilution calculations

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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.
 
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.
 
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.
 
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
 
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
 
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
 
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
 
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
 
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
 
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?
 
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
 
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.
hno3.jpg

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
 
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

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

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.
 
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.)
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

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

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.

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.
 
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.
 
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.
 
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.
 
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
 
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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