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Author: Subject: Calculation solubility - g/L - the proper method of calculation
RogueRose
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[*] posted on 3-4-2018 at 14:53
Calculation solubility - g/L - the proper method of calculation


When Wiki states that something has a solubility of say 50g/100ml in water (of the anhydrous salt), does that mean that if you have 100ml of water, then 50g will dissolve into the solution, or does it mean that in a saturated solution of 100ml, there is 50g of the salt dissolved? The latter would mean that more of the salt will dissolve in 100ml of water than a 50g per 100ml of a saturated solution.

This is especially important to know when calculating needed solvent/water for recrystalizations if you want to use a minimum, which is ideal for greatest purity.
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[*] posted on 3-4-2018 at 14:58


The latter.
The figures cited are per 100mL of solution not per 100mL of solvent.

Thete are I think good reasons for this convention but from the point of view of mixing up solutions it can be awkward.




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RogueRose
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[*] posted on 3-4-2018 at 15:21


Quote: Originally posted by j_sum1  
The latter.
The figures cited are per 100mL of solution not per 100mL of solvent.

Thete are I think good reasons for this convention but from the point of view of mixing up solutions it can be awkward.


Well the wiki just states "Solubility in water" of "solubility in XXXX" and states g/100ml or g/L

Now I understand molar mass and it seems that if it is the latter, then the solubility is calculated in the same way.

On Wiki's page about solubility it states the following:
Quote:

Solubility is commonly expressed as a concentration; for example, as g of solute per kg of solvent, g per dL (100mL) of solvent, molarity, molality, mole fraction, etc. The maximum equilibrium amount of solute that can dissolve per amount of solvent is the solubility of that solute in that solvent under the specified conditions.


So it is saying Grams of solute (say, salt), per kg (or L/ml) or solvent. So the way I read this, if it states, 50g/100ml, then it is 50g of solute (salt) and 100ml solvent. This is not how molar mass is calculated as in this example, you will have an amount greater than 100ml due to the addition of 50g of salt.

Now the reason I'm asking is I've found some discrepancies in solubilities on Wiki. I just did a test of MgCl2 of 100ml of a saturated solution at room temp. This weighed 127g. Upon drying to anhydrous I had 54g of salt so that means 73g of water evaporated. Now the solubility of anhydrous MgCl2 is listed as 54.3g/100ml which is what I got IF is it calculated as the amount of solute dissolved in a 100ml saturated solution. But from the page that defines solubility, it seems that this isn't their definition.

Now I have used the numbers before that are grams soluble in 100ml or 1L of water, and it was accurate listed as such. So in other examples, using MgCl2, then adding 54.3g to 100ml water, would make a saturated solution (again this was for different salts, I think KCl was one which I found was like this) and thus a volume greater than 100ml of a saturated solution.

IDK if there are some mistakes on Wiki when listing solubility where some people don't use the same standard or if they are just miscalculations.

Are there some more detailed sources (texts) that list solubilities for compounds, especially in solvents in addition to water?
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DavidJR
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[*] posted on 3-4-2018 at 15:28


Funny, I was wondering about this today...
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RogueRose
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[*] posted on 3-4-2018 at 15:34


Quote: Originally posted by DavidJR  
Funny, I was wondering about this today...


Well I've learned the hard way that some things on wiki might not be correct and I'm still trying to figure out what the proper way is for this, with regards to wiki and I'm going to test some other salts and see what the numbers are.
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RogueRose
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[*] posted on 3-4-2018 at 15:43


So back to the MgCl2, this is where it gets interesting. It states that the densites are as follows:
2.32 g/cm3 (anhydrous)
1.569 g/cm3 (hexahydrate)

Solubilities in water are as follows:

anhydrous
52.9 g/100 mL (0 °C)
54.3 g/100 mL (20 °C)
72.6 g/100 mL (100 °C)

hexahydrate
167 g/100 mL (20 °C)

If the hexahydrate has a density of 1.569g/cc then 167g would be 106.7cc's of volume, which I think makes the solubility number impossible for it being the amount in a saturated solution of 100ml. I think it has to mean 167g will dissolve into 100ml of water, otherwise it is an impossibliity due to the numbers.

now I have just calculated that my saturated solution of 100ml came out to 54g of anhydrous salt, so that looks like it is the opposite way of calculating the solubility.

Do others see this and understand what I am saying?

[Edited on 4-3-2018 by RogueRose]
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[*] posted on 3-4-2018 at 15:59


Well, to be fair, both approaches can be used.
https://sciencing.com/units-solubility-measured-8420688.html
IME, expressing concentration and solubility in terms of solution is far more common than expressing in terms of solvent. And the reason is simple. It greatly simplifies any calcuations related to the use of that solution. For example the dilution formula, c1v1=c2v2 -- this simply does not work if c is expressed in terms of solvent.

Quote:
Solubility Expressed with Units of Solution
When expressing solubility with units of solution -- that is, after the solute has already been added to the solvent -- it is important to note that the weight of the solution will change as solute is added. Solubility units that incorporate units of solution include grams of solute per 100 grams of solution or grams of solute per liter of solution. Another way to express solubility is in moles of solute per liter of solution; this ratio is called "molarity."




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[*] posted on 3-4-2018 at 16:35


Interesting, I wondered this for years. And I still don't know. That's why am I theory hater. The best way to truth is through practice. Worst of all there is much wrong data in wikipedia and various pdfs. Theoretical verification shows that data is wrong. In crc handbook of chemistry and physics 97th they say that mass of hydrogen is 100784 (without dot).
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[*] posted on 4-4-2018 at 01:52


Quote: Originally posted by RogueRose  
So back to the MgCl2, this is where it gets interesting. It states that the densites are as follows:
2.32 g/cm3 (anhydrous)
1.569 g/cm3 (hexahydrate)

Solubilities in water are as follows:

anhydrous
52.9 g/100 mL (0 °C)
54.3 g/100 mL (20 °C)
72.6 g/100 mL (100 °C)

hexahydrate
167 g/100 mL (20 °C)

If the hexahydrate has a density of 1.569g/cc then 167g would be 106.7cc's of volume, which I think makes the solubility number impossible for it being the amount in a saturated solution of 100ml. I think it has to mean 167g will dissolve into 100ml of water, otherwise it is an impossibliity due to the numbers.

now I have just calculated that my saturated solution of 100ml came out to 54g of anhydrous salt, so that looks like it is the opposite way of calculating the solubility.

Do others see this and understand what I am saying?

[Edited on 4-3-2018 by RogueRose]


What is the density of saturated MgCl2? It could be the density is higher than 1.569g/cc, making the whole 167 g/100ml possible. If you have 167 grams of the hexahydrate, and add water untill everything dissolves, but never more than what would give a total volume of 100 ml you are fine with these numbers.

If a saturated solution has a density of e.g. 175 grams/100 ml, you would have to add 8 grams of water, which could possibly dissolve the whole thing, and in the mean time give you a total volume of 100ml. Could be.

Normal convention is that e.g. 200 per liter solubility is 200 grams of a compound dissolved to a volume of in total one liter. The density of the solution and the amount of solvent is not specified and unknown in this case.

There are a lot of strange examples when going about this; 500 ml water and 500 ml ethanol doesn't give a liter. Also, (your MgCl2 example included), 500 ml of compound dissolved in 500 ml of solvent doesn't have to give a liter of solution, it rarely happens that 500 ml of A and 500 ml of B gives one liter of C.

So; more than a kilo of compound with a volume of more than a liter can be dissolved in a solvent weighing less than a kilo, giving a solution with a volume of less than a liter.

Edit: ah I see now I'm sort of repeating j_sum1

[Edited on 4-4-2018 by Tsjerk]
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XeonTheMGPony
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[*] posted on 4-4-2018 at 03:22


Just a reminder any ol idiot can edit wiki, few schools here would give you an immediate fail if you cited wiki for ANYTHING for a reason. Best to dig through the library and find a good book on material solubility, been ages and can't recall the one I found but I did find one.
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[*] posted on 4-4-2018 at 09:50


I think the main reason we come across these two differing methods is: one is better for determining the solubility, and the other is better for replicating a solution. As a chemist it is a good thing you noticed this discrepancy before being told about it.

This will affect how you interpret solubility data for the rest of your life. It has, for myself.
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