[Tittywahah]

Assuming a 36% or near a 12M starting point:  To make a 1M solution in 1 litre is 86 mLs.  So 5x86 is a 5M solution in 1 Litre.  Since I was working in mLs and not Gaseous equivalents I said 5moles in 1 litre and 5moles in 10 litres contains the same amount of dissolved gas.

[Hunter2]

Ok agreed.

But it doesnt solve the problem copperchloride will rise with every etch treatment. And if you add your 5 M in 10 liter of course the cocentration is 0.5 M but the etching strength also will go down. It takes only longer.

[Tittywahah]

Ok back to beginnings.  I said earlier that I never use diluted HCl that is absolutely pointless.  I always put in the 12M when my calculations and titrations indicate the need.  Copper chloride needs to be at a certain density to work efficiently.  1.2 is my average bench mark, the density increases as the copper 1 and 2 species increases. This dissolved copper 1 and 2 takes quite some time to increase in volume, we are talking a 2 litres in 18 months as I use it every 2 to 3 weeks for etching large pieces around 4 to 8 inches at some considerable depth. Etches can take 3 to 16 hours at a time for me.  If the density (Specific gravity, this is the dissolved copper whether it be in the form of CuCL or CuCl2) rises above 1.3 all I have to do is to add distilled water But this is so minimal as to be negligible, you shgould not need to add water for many many weeks as this only decreass the Specific gravity which has an influence on etch speed.  It takes weeks before I need to add water, I think I have only ever added water about three times in two years. As I mentioned I have 1.5 litres after a year and half but I began by making a 800 mL solution mixing copper oxychloride and HCl, this is how I kick-started my second operation. My first operation was through copper metal, HCl and Air.

[Borek]

if you have a 5 moles in 1 litre at a 5M (molarity) concentration then you have 5x86mLs = 430mLs of HCl and 570mLs water.

No, it is NOT 430 mL of HCl, it is 430 mL of HCl solution.

I am ignoring the fact that it doesn't make sense even if explained this way, as 12 M solution of HCl already contains water so the final solution contains more water than 570 mL that you added.

[Tittywahah]

if you have a 5 moles in 1 litre at a 5M (molarity) concentration then you have 5x86mLs = 430mLs of HCl and 570mLs water.

No, it is NOT 430 mL of HCl, it is 430 mL of HCl solution.

I am ignoring the fact that it doesn't make sense even if explained this way, as 12 M solution of HCl already contains water so the final solution contains more water than 570 mL that you added.

Honestly Borek, a little bit of nit picking.  I think it was obvious what I meant:  To make a 1 litre 5M solution from a 12M concentration HCl you place 430mLs HCl and top up with 570mLs water.  Of course I know that 37% HCl contains 63% water, just as 98% sulphuric acid contains 2% water.  Now what on earth was wrong with what I said previously? really.

[BigJohnny]

ok well all of the debate aside, I still don't understand the strength of the final product when making reagent grade.

If you start with 10M 31.45% muriatic acid in one container, and in the other you have distilled water, after a week you end up with a solution of 5M muriatic acid (at what percentage now??) and 5M reagent grade HCL.

So how does the molarity come into play here? Which of the two end products is stronger? Wouldn't this effectively mean I'm making 2 solutions of 5M, 15.72% muriatic acid??

[Borek]

I think it was obvious what I meant:  To make a 1 litre 5M solution from a 12M concentration HCl you place 430mLs HCl

No, it is not obvious. "430 mL of HCl" doesn't unambiguously mean 430 mL of 12 M HCl. First meaning is "430 mL of gaseous hydrogen chloride", second meaning is "430 mL of some uknown solution of HCl".

Besides, you can't say that the diluted solution contains 570 mL of water. 1L of 5 M solution weighs 1082 g, and contains 182 g of HCl, so it contains 900 g of water. 900 g of water is 900 mL, not 570 mL.

This is not about being nitpicking, this is about being precise without risk of misinforming/confusing others.

[Tittywahah]

Borek this is stupid.  We are NOT doing analytical chemical analysis of empirically derived products in a sanitized bio-chemical lab!  For the purposes of BigJohnny's needs there is absolutely Nothing wrong at all with my directions - if there were then every experiment and mixture I have performed over the last 2 years would never have worked and I have made at least 50 chemicals from crystals to new acids to plain inorganic dry chemicals.  What you are showing does not benefit bigjohnny.  What I have shown is correct and is simply practical. If you want to get analytical then yes do this, but this analytical is irrelevant and has absolutely no practical use whatsoever in this thread so far.  Sorry Borek, I respect your wealth of knowledge and expertise, I do, so I am not disagreeing with your perspective, simply the application of your perspective is not relevant in this context..

BigJohnny:  If you start with 10M  and in the other you have distilled water?  What do you mean by two containers - sorry do not get this.  Anyway, if you have a 10M HCl solution in one litre and this is 31.45% a 5M will be 15.7% in one litre.  When you are working with your HCl work in either Percentage Or Molarity.  They mean both the same thing. For example a 95% sulphuric acid is approx 17.6M in one litre  10% would be 1.76M.  This molarity and percentage is always expressed as in One litre, not 500 mLs or gallons.  Molarity and Percentage is an expression of concentration.  If the consumer would read 2.58M on the package they would not have a clue what it meant.  Hence percentage since we all relate to percentages.  5M and 10M is Not the same thing.  Please do not hesitate to ask questions, we all have to start somewhere, and no one was born with a PHD in chemistry, so no question is too stupid.

[Borek]

If you start with 10M 31.45% muriatic acid in one container, and in the other you have distilled water, after a week you end up with a solution of 5M muriatic acid (at what percentage now??) and 5M reagent grade HCL.

Right. 5 M hydrochloric acid is about 16.9%

So how does the molarity come into play here? Which of the two end products is stronger? Wouldn't this effectively mean I'm making 2 solutions of 5M, 15.72% muriatic acid??

Once the system is in equilibrium concentration of acid is identical in both beakers, so they have both exactly the same strength. Whether their concentration is exactly half of the initial concentration, depends on the amount of distilled water put in the second beaker.

Actually this is kind of calculations where being precise about "amount of water" is paramount.

Let's assume we start with 1.000 L of 10 M HCl. It contains 10 moles (364.6 g) of HCl. Solution density is 1.1569 g/mL, so the mass of the solution is 1157 g, of this 1157-364.6=792.4 g is water.

After the procedure we want to end with 2.000 L of 5 M acid (in separate beakers, but we can treat them combined here). Density of 5 M acid solution is 1.0819 g/mL, so the total mass of the solution needs to be 2164 g. We already have 1157 g of the solution and all HCl we need is there, so we need to add 2164-1157=1007 g of water.

Things get tricky when we try to calculate volumes of both acids - the pure one (reagent grade) and the leftover (muriatic, technical, stock, whatever you want to call it). New beaker contained 1007 g of water, now it contains 5 M HCl solution with a density of 1.0819 g/mL. How much HCl does it contain? Let's call HCl mass m_HCl - then the mass of the solution is

[tex]1007+m_{HCl}[/tex]

volume of the solution is mass over density

[tex]\frac {1007+m_{HCl}} {1000 \times 1.0819}[/tex]

(1000 is a conversion factor between mL and L) and number of moles of HCl is

[tex]\frac {m_{HCl}}{36.46}[/tex]

(36.46 g/mol being molar mass of HCl).

By definition molar concentration is

[tex]C = \frac n V[/tex]

We know concentration to be 5 M, so we can write

[tex]5 = \frac {\frac {m_{HCl}}{36.46}}{\frac{1007+m_{HCl}}{1000\times 1.0819}}[/tex]

solving for m_HCl yields 204 g, so mass of the reagent grade solution is 1007+204=1211 g (or 1211/1.0819=1119 mL), the other beaker contains now 1157-204=953 g (or 953/1.0819=880 mL) of the technical, 5M solution.

[Borek]

Borek this is stupid.  We are NOT doing analytical chemical analysis of empirically derived products in a sanitized bio-chemical lab!  For the purposes of BigJohnny's needs there is absolutely Nothing wrong at all with my directions

It is not recipe that I am disagreeing with, it is your explanation of what is going on that is wrong and confusing. Sorry to say that, but you don't understand what you are doing and that means you are a danger to yourself and others.

For example a 95% sulphuric acid is approx 17.6M in one litre  10% would be 1.76M

No, 1.76 M solution of sulfuric acid is 15.6 %, 10% acid is 1.09 M.

This molarity and percentage is always expressed as in One litre

No, concentration is an intensive property and doesn't depend on the volume. Molarity is defined PER liter, so it is sometimes easier to think about it this way, but percentage is defined as ratio and doesn't require any particular units.

Amount of substance is an extensive property, so it depends on the volume.

[Tittywahah]

On the two parts where you quoted me I made mistakes due to edit typing.  I know that percentage is Not an expression of how much in a litre, that was a typo error in my second edit when I added something.   1.76 as 10% yes I made a thoughtless mistake in my haste.  I am not a danger to myself just because I mis-calculated.  I do not make mistakes when I am doing stoiochemetry, I take my time.  It's just that I was eager to stress that your perspective and argument with me has no practical purposes for bigjohnny.  With all your instructions his question has not been answered still.  So I give up since my input got lost in an array of in-precise reasoning and mathematical calculation. I will leave the answer to the PHD'ers.

[Borek]

With all your instructions his question has not been answered still.

Which question was not answered? I believe I addressed all questions from his last post (plus some).

[Tittywahah]

Borek, I really do not want to appear an antogonist, neither disrespectful nor in-polite, I hope I have not been.  I just feel there has been a little misunderstanding perhaps on my part about exactly what the OP needs to know, maybe I have completely missed the point, and maybe I have cross wired with hunter2's questions and it became a bit confusing.  The only thing I would like to say is that I was trying to keep things absolutely simple, though a bit anethema to the professional, because I was not pedantically observant in my calculations. Yes I made a mistake with one percentage - honestly I should have spent an extra few seconds with a calculator instead of typing off the top of my head.  My fault. If the OP is happy now with your answers then great, if not I will leave him to you to clarify.

[BigJohnny]

BigJohnny:  If you start with 10M  and in the other you have distilled water?  What do you mean by two containers - sorry do not get this. 

Sorry,my bad, I was referring to this video that I didn't even link to.
https://www.youtube.com/watch?v=jv1Ms6Subg4

Anyway, if you have a 10M HCl solution in one litre and this is 31.45% a 5M will be 15.7% in one litre.  When you are working with your HCl work in either Percentage Or Molarity.  They mean both the same thing. For example a 95% sulphuric acid is approx 17.6M in one litre  10% would be 1.76M.  This molarity and percentage is always expressed as in One litre, not 500 mLs or gallons.  Molarity and Percentage is an expression of concentration.  If the consumer would read 2.58M on the package they would not have a clue what it meant.  Hence percentage since we all relate to percentages.  5M and 10M is Not the same thing.  Please do not hesitate to ask questions, we all have to start somewhere, and no one was born with a PHD in chemistry, so no question is too stupid.

thanks, that pretty much cleared up most of my confusion.

In respect to the above video, what the hell is he talking about then? if I understand this topic correctly he isn't purifying anything, nor is he making any stronger acid....so what exactly is he doing?

[Borek]

Sorry,my bad, I was referring to this video that I didn't even link to.

Funny thing is, I have seen this video in the past and I know what he is doing, so I knew what it is all about and I have not noticed you didn't link to the video.

In respect to the above video, what the hell is he talking about then? if I understand this topic correctly he isn't purifying anything, nor is he making any stronger acid....so what exactly is he doing?

He is purifying the acid by a quite clever method, ending with an acid that is about half concentrated as the one he started with.

Hydrochloric acid is a solution of HCl (hydrogen chloride, which is a gas at room temperature) in water.

Hydrochloric acid - when concentrated - is quite volatile (which means - HCl fumes are present over the solution).

When you put an open jar with a concentrated acid in the room, you will be able to smell HCl everywhere.

He puts two separate, open jars in a closed container. One initially contains 10 M HCl solution, the other contains pure water.

What is happening next is that the HCl from the concentrated solution gets airborne and saturates the container. Then, it dissolves in the pure water that is present in the other jar.

This process continues until concentrations in both containers are identical (which we call an equilibrium).

What the exact final concentrations are depends on the initial amount of acid and initial amount of pure water used. I have explained how to calculate it in my earlier post.

Note, that it is HCl that is volatile - most impurities are not, so they stay in the original solution, and only HCl is transferred to the new solution. That's why the purification works.

After you correctly wrote

Wouldn't this effectively mean I'm making 2 solutions of 5M, 15.72% muriatic acid??

I assumed you got it.