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

[Otrips]

Assuming "before any gas exchange may occur" also includes respiration or decomposition of any organisms that may be in the natural seawater.

You would have to add (about) 1.9 parts (o salinity) to every 1 part 35ppm water to achieve 12ppm salinity.
A neutral PH will not counteract the Base.
A base of 8.1 is 10.1 times higher than 7.

So I'm going with B.

 

[Ted_C]

Learned more about chemistry in this post then I did in all the classes I took in high school hahaha. New at the Reef game too, so I’m just happy to read through the discussion.

You don't learn this in high school chemistry. This is covered in College Chemistry (actually for me in the 90's - it was analytical chemistry 2 where this topic was covered).

 

[Johniejumbo]

I’m going to pick A. Because that’s what my magic 8 ball said. It’s how I make all of my reef tank decisions.

 

[KMSReefer]

I choose option B. My reasoning is below:

Calculate pOH:
pOH = 14 - pH
pOH = 14 - 8.1
pOH = 5.9

Calculate starting Molarity of OH-:
[OH-] = 10^(-pOH)
[OH-] = 10^(-5.9)
[OH-] = 1.2589e-6 mol/L

Start with arbitrary volume, ie: 1L

Calculate total volume after dilution based on the salt concentrations:
C1V1=C2V2
35*1 = 12*V2
V2 = 35/12
V2 = 2.9167 L

Calculate diluted Molarity of OH-:
C1V1 = C2V2
C2 = C1*(V1/V2)
C2 = 1.2589e-6*(1/2.9167)
C2 = 4.3163e-7 mol/L

Calculate new pOH:
pOH = -log(OH-)
pOH = -log( 4.3163e-7)
pOH = 6.3649

Calculate new pH:
pH = 14 - pOH
pH = 14 - 6.3649
pH = 7.6351

Comparing to given options(only considering B & C):

Note that the calculated [OH-] = 4.3163e-7 mol/L

B. About 7.7
pOH = 14 - pH = 14 - 7.7 = 6.3
[OH-] = 10^(-pOH) = 10^(-6.3) = 5.0119e-7 mol/L
Difference = 0.6956e-7 mol/L

C. About 7.5
pOH = 14 - pH = 14 - 7.5 = 6.5
[OH-] = 10^(-pOH) = 10^(-6.5) = 3.1623e-7 mol/L
Difference = 1.154e-7 mol/L

Therefore, the calculated pH is closest to answer B (About 7.7)

 

[ReeferAl]

I'll take a stab at it.
As others have said, DI water has no buffering capacity. This would suggest that the pH would not change. When the 2 water samples are mixed, the total volume of water will be greater but it will no longer be in equilibrium with room air since the DI water was free of any CO2. So the combined water will now be slightly CO2 "deficient" until it equilibrates with room air again. But, the water is in a "closed container", preventing equilibration so the pH will be above 8.1.

Allen

 

[Joe Rice]

H+ ions in 1 liter of original solution = inverse log (8.1) = 125,892,541
H+ ions in 1 liter of RO/DI water = inverse log (7.0) = 10,000,000

liters of RO/DI water needed to reduce salinity from 35 to 12 = (35-12)/12 = 1.917

H+ ions in mixed solution = 125,892,541 + (1.917)*10,000,000 = 145,059,207

pH of mixed solution = log (145,059,207/(1+1.917)) = 7.6966 ≃ 7.7

 

[GoVols]

A lot of great discussion folks!

When I begin to explain the answer later today or tomorrow morning, I'll start by discussing very simple "models" of the problem (say, starting with RO/DI and sodium hydroxide solution at pH 8.1) and then stepwise increase the complexity.

Note there are two big jumps in complexity that get close enough to true seawater.

 

[Gloryofthereef]

B

 

[Joe Rice]

Well, botched that first attempt. Forgot that pH is the log of 1 over the H+ ion concentration.

H+ ions in 1 liter of original solution = 1/inverse log (8.1) = 1/125,892,541 = 0.00000000794 (moles)
H+ ions in 1 liter of RO/DI water = 1/inverse log (7.0) = 1/10,000,000 = 0.0000001

liters of RO/DI water needed to reduce salinity from 35 to 12 = (35-12)/12 = 1.917

H+ ions in mixed solution = (0.00000000794) + ((1.917) * (0.0000001)) = 0.0000001996

concentration of H+ ions in mixed solution = 0.0000001996/(1+1.917) = 0.0000000684

pH of mixed solution = log (1/0.0000000684 ) ≃ 7.165

Which is none of the answers so the real answer must have something to do with dissociation constants and all sorts of other stuff I've forgotten. Perhaps the bicarbonate ions combine with some of the H+ ions to push the pH higher?

Can't wait to read the explanation.

 

[Leishman]

B. Don't recall what i did on my adding machine, but the final reading was 7.726

 

[Randy Holmes-Farley Verified Community Expert]

We are well within the approximations of the Henderson-Hasselbalch equation

Funny you should mention that. Not only is it highly relevant, but I was on a college tour with my younger daughter on Monday (Washington University in St. Louis), and the tour sat in on a couple of minutes of a chemistry class. That equation was on the board.

 

[AllSignsPointToFish]

I'm going with D, even though all of my calculations indicate a pH of around 7.15-7.2, depending on significant figures. The pH is definitely NOT below 7 and has to be lower than 8.1, so choices A and D are eliminated due to caveats concerning no dissolved CO2, etc.

You first have to calculate the H3O+ concentration in the saltwater, then the moles of H3O+ in the saltwater.

You have to calculate a dilution volume of around 2.9L (assuming 1L of seawater as your basis, 35ppt/12ppt x 1L), then subtract of the volume of saltwater from the total diluted volume to calculate the amount of DI water needed to reach a total volume of 2.9L.

From the amount of DI water, you can calculate the H3O+ concentration by taking the volume of water added (2.9L diluted volume - 1L saltwater volume gives ~1.9L DI water needed) multiplied by 1x10^-7 moles H3O+/L since pH = -log([H3O+]), and pure water has a pH of around 7.

Adding 10^-8.1 (H3O+ concentration of saltwater x 1L) + 1.9x10^-7 moles H3O+ in 1.9L DI water gives you a total of 1.98x10^-7 moles H30+. Dividing by the total solution volume of around 2.9L yields a H3O+ concentration of around 6.82x10^-8 mol/L. pH = -log(6.82x10^-8) ~7.16.

 

[ReefTeacher]

Funny you should mention that. Not only is it highly relevant, but I was on a college tour with my younger daughter on Monday (Washington University in St. Louis), and the tour sat in on a couple of minutes of a chemistry class. That equation was on the board.

Randy,

wish I knew! I live at St Louis and teach at the rival!

 

[DeepBlueSomething]

A) Above the initial 8.1 - the CO2 is the odd actor here. The NSW is at pCO2 of X while the ultra pure, non-gassed RODI has pCO2 of zero. Therefore, the addition of the 'freshwater' raises pH because there the equilibrium is unbalanced. (cite several internet reviews and the 'extent' of my understanding'

 

[kecked]

Ok so I see you state as a condition the RodI is empty of co2 and hence any buffering of any kind. Perfect water that doesn’t exist is in an equilibrium [H+][OH-]=HOH the pka being such that pH is 7. Salvation of the salt doesn’t make any difference and remains neutral. So I have to go with A above or equal 8.1. With a drop in pH over time as the water absorbs CO2 and achieves equilibrium.

Ever use RodI to rinse your ph probe between calibration fluids? Watch what happens next time.

 

[sghera64]

Ever use RodI to rinse your ph probe between calibration fluids? Watch what happens next time.

I was going to contribute that too. So, I'll answer your question. My probe sitting in tank water might read 8.2. When I remove it and dip it into fresh RO/DI water, the pH jumps to like 9.6 and then slowly comes back down. I always thought this was the "dilution effect" taking place on the probe's surface as the pure water "dilutes" the [H+] on the probe's surface.

 

[sghera64]

Ok so I see you state as a condition the RodI is empty of co2 and hence any buffering of any kind. Perfect water that doesn’t exist is in an equilibrium [H+][OH-]=HOH the pka being such that pH is 7. Salvation of the salt doesn’t make any difference and remains neutral. So I have to go with A above or equal 8.1. With a drop in pH over time as the water absorbs CO2 and achieves equilibrium.

Ever use RodI to rinse your ph probe between calibration fluids? Watch what happens next time.

I don't think that I saw anyone mention the use of direct observation in any of the posts. So, I just went over to my tank and grabbed a small sample of water and measured the pH. It was 8.44. Then I added some RO/DI water to it to nearly triple the resulting volume. The pH went to 8.76. I kept stirring and waiting and the pH slowly dropped from 8.76 to 8.69 and just sits there now.

I even soaked the probe in RO/DI water (pH ~7.2) and then put it back into the diluted sample: pH 8.68. At first I thought the pH swing was just a surface electrical phenom. But those usually dissipate pretty quick. It's looking like the answer is (A).

One way to look at it is in the equations I wrote out. As I look at this, adding more water to the system causes the equilibrium to move to the right (Le Chatelier's principle). So, more OH- is present (or less H+) and the pH increases.

But, there is also the other way of looking at it (Ka). In this case, the opposite effect would be indicated (pH would decrease because adding water causes the equation below to shift to the right , or more H+). In the original sample and the "diluted" sample, the concentration of water is very high (~55 M) and changes only slightly from initial to diluted state. I'm just not sure how to determine which equation (Ka or Kb) drives here. Or, maybe this is irrelevant in this case.

 

[Orcus Varuna]

The answer is A as reefteacher stated it’s well within the approximations of the Henderson Hasselbach equation.

In essence by knowing the total and carbonate alkalinity which we can imply as nsw and the chlorinity of the solution (which is about 6.6) you can solve for hydrogen activity and therefore it’s inverse log (pH) at any point in the brackish curve. I don’t have time for the mess of math but the answer is A.

 

[Randy Holmes-Farley Verified Community Expert]

The answer is A as reefteacher stated it’s well within the approximations of the Henderson Hasselbach equation.

One aspect of that equation is not entirely satisfied. I’ll hold off giving the full explanation till tomorrow, but as a hint, what might change from seawater to hypo that you didn’t account for as you plug numbers into that equation? It makes a substantial difference , [emoji3]

 

[Randy Holmes-Farley Verified Community Expert]

I don't think that I saw anyone mention the use of direct observation in any of the posts. So, I just went over to my tank and grabbed a small sample of water and measured the pH. It was 8.44. Then I added some RO/DI water to it to nearly triple the resulting volume. The pH went to 8.76.

As the A-team used to say, it’s nice when a plan comes together. [emoji3]

FWIW, I’m surprised the result is nearly exactly my calculated amount of rise. [emoji23]