Master Reef Chemist Certification Question!!! (yes, you want this title)

[Randy Holmes-Farley]

Let's have some fun with a question that wraps up several different aspects of reef chemistry into a single, very simple (perhaps deceptively simple) question. It's just the pH of a hyposalinity solution. How much easier can it get? [emoji1]

Get it right for the right reason and you too can be certified as a Master Reef Chemist!!!

Suppose you have normal natural seawater of about 35 ppt salinity (specific gravity of about 1.0264), and you want to make hyposalinity water with it to treat a sick fish. You decide you want the salinity to be about 12 ppt (specific gravity about 1.009)

The pH of the normal natural seawater at 35 ppt is 8.1 in perfect equilibrium with your home air.

Assume you have totally perfect RO/DI water with nothing at all in it besides water. No CO2, nothing. You collected it straight from the DI with no gas exchange with the room air. You know that the pH of this water is ~7.0.

You mix the 35 ppt salinity seawater at pH 8.1 with sufficient RO/DI water at pH 7 to get to the 12 ppt salinity target. All mixing and solutions and measurements are at 25 deg C and are done in a closed container.

The pH of the final hyposaline solution before any gas exchange may occur (e.g., no CO2 entering, no CO2 leaving, no evaporation, etc.) is:

A. Above 8.1 [note this says above 8.1, it implies a rise from the level in the salt water]
B. About 7.7
C. About 7.5
D. About 7.3
E. Below 7.0 [note this says below 7, so it implies a drop from the level in the RO/DI water]

Please explain your choice (if you want full credit)

Searching online for help is encouraged (if needed).

Good luck!












































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[JCM]

I'm not even an amateur reef chemist, but to get the water down to 12ppt you'll use roughly 2/3 ro/di and 1/3 original water. So I'd assume the ph will drop to around 7.3. (8.1 x .33) + (7 x .66) = 7.29 ish. I pick D.

I'm sure a quick Google search would tell me how wrong I am haha

 

[Greenstreet.1]

I know I’m way way off but I pick A. 8.1 because the gas exchange did not happen yet ‍

 

[CNDReef]

C
Not a chemist but hypo runs around that when first setup

 

[themcfreak]

From the words of @Randy Holmes-Farley himself: (your hint gave it away, google is your friend)

3. The pH of the combination of two solutions does not necessarily reflect the average (not even a weighted average) of their two pH values. The final pH of a mixture may actually not even be between the pH’s of the two solutions when combined. Consequently, adding pH 7 pure water to pH 8.2 seawater may not even result in a pH below 8.2, but rather might be higher than 8.2 (for complex reasons relating to the acidity of bicarbonate in seawater vs. freshwater).

The answer is A) higher than 8.1. Salinity has no bearing on the question as far as I can read.

Source: http://reefkeeping.com/issues/2005-05/rhf/index.php#8

 

[Chris86]

The answer is A. The alkalinity will drop, not the ph

 

[Chris86]

Also, this is only before gas exchange occurs. If gas exchange occurs, the CO2 levels in the water will come to equilibrium with the atmosphere. Then the pH would drop.

 

[Frogger]

I will go with 7.3. You need to add 2.9l of fresh water to 1.0l of saltwater to get a specific gravity of 1.009. Assuming that the H+ and OH- concentrations are in the same ratio when diluted I get something close to 7.3pH.

 

[alanbetiger]

Guessing without searching B
The “PERFECT” RODI pH7 water should of had most of the buffer capabilities removed so there for diluting the salts/buffers in the natural seawater. The buffering actions still contained in the natural sewater will still be counteracting the drop so it would be proportional to the ratio of the mix.

 

[DLHDesign]

Simple, huh? Okay; keep it simple.

First order of business; find out how much water we'll be adding, which can be calculation using the general dilution formula:
C1*V1=C2*V2
C1 = initial concentration = 35ppt
V1 = initial volume = 1 (1 gal; 1L, whatever)
C2 = final concentration = 12ppt
V2 = final volume
Re-writing the equation, then:
V2 = (C1*V1)/C2
Plugging it in:
V2 = (35*1)/12
V2 = 2.9166
But since V2 is final volume and I want the added volume, we subtract what we started with to get the addition:
Added water = 1.916666...

Now that I had that number (phew!) I can just plug things into this calculator.

But why is that the case - how did that calculation work, I wondered...
Took a look at the code of the page. Here's the relevant calculation:

dd=aa-lod((eval(bb)+eval(cc))/bb);

Where "lod" is getting the log of a base-10 number and "aa", "bb", "cc", and "dd" are references to the values in the form above (I left the final value - dd - blank before hitting submit).
So the calculation is:
Resulting Ph = Orig Ph - log ( Total Volume / Starting Volume )
Where "log ( Total Volume / Starting Volume )" represents the change in Ph resulting from the volume change (pH being a logarithmic scale). (Note that I would have been fine with the "V2" final volume number had I understood the math first...)

So - why?!?
Did some more reading and found that water is a pretty unstable little beast. It's constantly trying to balance itself, while also (literally) tearing itself apart. Sheesh! But the general point is that the solution will reach pH equilibrium based on the number of "free floating" hydrogen ions that have resulted from the dissociation process...
At this point, I feel like I've gone about as far into "simple" as I can...

Answer: "B - About 7.7" (because my result of 7.635 is closer to "about 7.7" than it is to "about 7.5"...)

 

[Frogger]

Sorry I change my answer because the OH- concentration of saltwater is over 10 times that water. Diluting 1l of saltwater with 2.9l of saltwater would only lower the ph to 7.7

 

[jduong916]

So I'm assuming a mixing equation twice.

First we need to know the volume of RO/DI water needed to go from 1.0624 to 1.009 specific gravity. Assume a volume of 1 was used for the 1.0264 specific gravity solution.

RODI = 1.00 specific gravity

1(1.0264)+VolRODI(1.00) = (1+VolRODI)(1.009)

VolRODI = 1.93

now with the two volumes and given PH values another mixing equation

1(8.1)+1.93(7)=PH(1+1.93)

PH = 7.38 = answer D

Hope this is right

 

[NanoCrazed]

How about "None of the above"? The pH should be about 7.64 so between B and C. But closer to B.

It's a two part problem... the first is you want to dilute 35ppt to 12ppt. So, if you have say 1L of 35ppt solution, you would need to add about 1.917L of water to get 2.917L at 12ppt. (C1V1 = C2V2) C=concentration , V = Vol

Knowing that we're adding 1.917L of water at pH of 7, you would then figure out the dilution factor of the pH based on the addition of this water. Being that 8.1 is a basic solution, you need to figure the amount of [OH-] ions per vol and then calculate the concentration based on volume change.

pH = 14 - pOH. so in this case, pOH = 14-8.1=5.9 (since pH of the 35ppt solution is 8.1)

C(new) = 10^-5.9 * (1L/2.917L) ... take the -Log of this give 6.36 then pH(new) = 14-6.36 = ~7.64

In basic English, you're adding lower concentration (i.e., water) to a higher concentration solution and increasing the volume of liquid, the effect is lower salinity and lower pH due to less base ions [OH-] per volume of solution.

 

[Randy Holmes-Farley]

Folks electing choice A or choice E should note that it says above 8.1 or below 7. I edited the question to make that more clear.

 

[Chris86]

It’s still A. Diluting the solution will increase the carbonic acid which will be buffered increasing the pH. After gas exchange, the CO2 will then come into equilibrium with the atmosphere, which will decrease the pH

 

[Greenstreet.1]

Sticking with A. No running now [emoji23]

 

[themcfreak]

I, also, will be sticking with A. The article I linked in my original answer clearly says that the pH of mixing 2 solutions may not be an average of the two. And that the acidity of bicarbonate in saltwater can contribute to a higher pH. And since 7.0 water is not a buffer, that makes sense, as it shouldn't 'buffer' the solution to a lower pH.

 

[alanbetiger]

Can’t edit my post. Still probably wrong haha Meant the drop will NOT be proportional to the mixing ratio....

 

[reef_ranch]

D. Using a pure pure dilution ratio analysis. The questions seems to be framed to remove all that CO2 in the air/degassing issues.

 

[mcarroll]

I think A because the the ratio of H atoms has gone up relative to everything else that matters.