Reef Chemistry Question of the Day #45 Brain Teaser on Salt Solutions

Randy Holmes-Farley

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Reef Chemistry Question of the Day #45 : Brain Teaser

Today we’ll start the week with a question that I’ll call a brain teaser. It requires deeper thinking than most of the typical questions, and for that reason, if the answer does not come to mind without help, feel free to look up things that you think might be useful. It is not as if it is exceptionally advanced chemistry, and does not require measurement data of any sort, it’s just a question that involves thinking through several different processes to come to the answer.

So, you are a mad scientist, or at least you play one at home. You set up an experiment with some left over reef supplies where you take a salt bucket and place 4 glass jars flat on the bottom. Each jar has an internal volume of 500 mL. To each jar you add:

Jar 1. 100 mL of RO/DI water
Jar 2. 100 mL of RO/DI water plus 0.1 g (1.7 millimoles) of anhydrous sodium chloride
Jar 3. 100 mL of RO/DI water plus 0.1 g (1.1 millimoles) of anhydrous magnesium chloride
Jar 4. 100 mL of RO/DI water plus 0.1 g (0.83 millimoles) of anhydrous magnesium sulfate

After adding the ingredients, you seal the salt bucket and leave it in a constant temperature room. After a few years, you decide the experiment is complete, and you come back, open it up, and look inside.

What do you see?

The possible answers are:

A. There is no apparent change, and the relative volumes in jars 1, 2, 3, and 4 is 1 : 1 : 1 : 1
B. There is no water left in any jar, and just dried salts in Jars 2,3,4.
C. Jar 1 is empty, and the relative volumes in jars 2, 3, and 4 is 1 : 1 : 1
D. Jar 1 is empty, and the relative volumes in jars 2, 3, and 4 is 1 : 0.65 : 0.49
E. Jar 1 is empty, and the relative volumes in jars 2, 3, and 4 is 1: 0.97 : 0.49
F. Jar 1 is empty, and the relative volumes in jars 2, 3, and 4 is 1 : 1.5 : 1
G. Jar 1 is empty, and the relative volumes in jars 2, 3, and 4 is 1 : 2 : 2
H. Jars 1, 2 and 3 are empty (except dried salts in some), and all of the fluid is in jar 4
I. Jars 1, 2 and 4 are empty (except dried salts in some) , and all of the fluid is in jar 3
J. Jars 1, 3 and 4 are empty (except dried salts in some), and all of the fluid is in jar 2
K. Jars 2, 3 and 4 are empty (except dried salts in some) , and all of the fluid is in jar 1
L. Something different (what?)

For purposes of this question, assume the salt solutions are “ideal” and the answer does not involve any measurement data that is not supplied. Assume no gas exchange out of or into the sealed bucket. Assume that the system has reached equilibrium.

Good luck!























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glennf

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No takers yet? :)

I'll leave this one up a bit longer than usual so folks can have time to look things up if they want. :)

If you insist randy, nice one:thumbup:

I. Jars 1, 2 and 4 are empty (except dried salts in some) , and all of the fluid is in jar 3

Reason:
Magnesiumchloride (also calcium chloride) is being used as a moist absorber.
 

Steven@

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H. and I assume I am wrong as usual, but you sure made me think about this one. Great question!
 

leptang

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My guess is L. If you put water into anhydrous its no longer dry, so jar 2, 3, 4, hydrated with no other changes or crystallization might form. Jar 1 didn't change at all.
 
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DFW

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L? I have done no research on this, but have seen water in a glass after it has set for a few days. It starts looking real funky, and some of the H2O evaporates, which won't happen in this sealed bucket. I will say that Each of the jars still has it's water, and the salts in jars 2, 3, and 4 will be visible, like when you leave water sitting in a glass for a few days.
 
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Randy Holmes-Farley

Randy Holmes-Farley

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And the answer is.... E. Jar 1 is empty, and the relative volumes in jars 2, 3, and 4 is 1: 0.97 : 0.49

Here's the rationale...

Any substance that dissolves in water reduces the vapor pressure of the water. And the more of it that dissolves, the more the vapor pressure is lowered. One way to think of it is that you are "diluting" the water, so there is less to go into the vapor phase. A second way to think of it is that the water likes to be around things like dissolved salts more than it likes to be around itself. These are actually slightly different concepts and science is often more complicated than the simple view expressed in freshman chemistry class, but this is what folks are taught:

Vapor pressure lowering is a colligative property, just like freezing point depression and boiling point elevation. Colligative properties are those which are dependent, to a first approximation anyway, only on the number of things present, not their chemistry. So the vapor pressure above water is lowered by an equal amount by any ion dissolved in the water, and the more ions you have, the more it is lowered.

Note that in this problem, while the weight concentrations are the same, the molar concentrations vary. Molar counts the number of ions present, not the weight. SO we have to look to the molar concentration to understand vapor pressure lowering. Further, sodium chloride makes two ions when it dissolves, as does magnesium sulfate, but magnesium chloride (MgCl2) makes three ions, so we have to account for that.

So, for the concentration of ions in the three cups we have:

Jar 1. 0
Jar 2. 1.7 millimoles NaCl = 3.4 millimoles of ions
Jar 3. 1.1 millimoles MgCl2 = 3.3 millimoles of ions
Jar 4. 0.83 millimoles MgSO4 = 1.66 millimoles of ions (Mg++ and SO4--)

So, what does this problem have to do with vapor pressure lowering???

Well, with several containers in the same closed system, vapor will slowly move around entering and leaving cups until an equilibrium is reached,a nd that will only happen when the final vapor pressure is in equilibrium with any liquid water around.

So water will leave the cups with the highest vapor pressure and enter those with the lowest, until they are all the same.

Cup a has no salts, so it is always pure water, and always has a higher equilibrium vapor pressure than the salt cups. So the only way equilibrium is reached is for all the water to leave cup 1, and end up in Cups 2, 3, and 4.

Cups 2, 3, and 4 will be in equilibrium only when the total concentrations of ions in each is exactly the same.

Looking at the number of millimoles of ions in each in cup tells us how much water must be in it.

So suppose that we arbitrarily assign cup 2 a final volume of X mL. Then its concentration is 3.4 millimoles /X mL.

So what about cups 3 and 4?

They must have the same concentration, so call cup 3's volume Y mL. Then we know that

3.3 millimoles/Y mL = 3.4 millimiles/X mL

rearranging we get

Y mL = (3.3/3.4)x X mL or 0.97 x X mL

likewise for cup 4.
It must have the same concentration, so call cup 4's volume Z mL. Then we know that

1.66 millimoles/Z mL = 3.4 millimiles/X mL

rearranging we get

Z mL = (1.66/3.4)x X mL or 0.49 x X mL

We now know the ratio of volumes must be 1 : 0.97 : 0.49.

Anyway, the reason I posted this problem was for the intuitive idea of moisture moving through the air and into salt solutions, and the fact that we can actually know what will happen with fairly simple chemistry. :)





 
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