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

[Randy Holmes-Farley]

Randy,

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

[emoji106]. Happy Reefing [emoji3]

 

[Orcus Varuna]

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]

The disassociation constant [emoji848]

 

[Randy Holmes-Farley]

The disassociation constant [emoji848]

[emoji106]

 

[jdmee]

After reading all the answers and explanations I have come up with either:
F. All of the above or
G. None of the above.

So I will go with

H. I have no idea.

 

[Randy Holmes-Farley]

I will give much longer and more explained/simplified answers for the vast majority of reefers reading this thread, but here's a quick answer for those of you who understand the Henderson–Hasselbalch equation.

The answer is (A) and the pH rises substantially (about 0.3 pH units in this question). The reason is that while the Henderson–Hasselbalch equation suggests no rise for a well buffered solution as you dilute it, the pKa of bicarbonate changes a lot between fresh water (10.3) and seawater (35 ppt; 8.9). In 12 ppt seawater it is about 0.3 pH units higher than in 35 ppt seawater, so the whole Henderson–Hasselbalch relationship immediately shifts about 0.3 pH units higher when diluting 35 ppt seawater to 12 ppt seawater with pure fresh water.

This article (Figure 3b) shows the pKa as a function of salinity of seawater:

https://www.sciencedirect.com/science/article/pii/0198014989900642

 

[Randy Holmes-Farley]

As we proceed to look at how to answer this question, let's first figure out exactly how much fresh and salt water we are talking about.

Salinity in ppt of mixing of solutions can be calculated just like any other weight based measurement in solution: it is just the weighted average of the two solutions.

In this case, one solution is 0 ppt, the other 35 ppt, and we want to get to 12 ppt.

The solution is especially easy when one of the solutions is 0 ppt and you are just dilution the starting 35 ppt:

CiVi = CfVf
where Ci = initial concentration (35 ppt), Vi is the initial volume (arbitrary, let's say 1 L), Cf is the final concentration (12 ppt) and Vf is the final volume (that will give us our answer).

Plugging in 35 ppt, 12 ppt and 1 L for initial volume, we get:

35 = 12*Vf

Solving for Vf we get

Vf = 2.92 L

If the final volume after mixing is 2.92 L and we started with 1 L, then we must have added 2.92 - 1 = 1.92 Liters of fresh water.

In general, 1 part of 35 ppt water and 1.92 parts of fresh water gets us to 12 ppt.

 

[Js.Aqua.Project]

A.

Because:

A. Above 8.1

pH is the measure of the concentration of H+ ions in a solution. When adding pure water to a solution that already has a reduce concentration of Hydronium versus Hydroxide ions (i.e. a basic solution), the resulting solution will have a larger total volume and relatively the same amount of Hydronium ions, thus increasing the overall pH of the solution.

This is all in a perfect environment of course with no gas exchange occurring between the two solutions. In a real world scenario, carbon dioxide will dissolve in pure water forming carbonic acid which will lead to a reduction in pH of the RODI solution since it has zero buffer capacity. Adding the reduce pH RODI solution to the saltwater would cause a reduction in the overall pH of the resulting solution.

and per:

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

Again, this assumes a process the avoids gas exchange.

 

[sghera64]

I will give much longer and more explained/simplified answers for the vast majority of reefers reading this thread, but here's a quick answer for those of you who understand the Henderson–Hasselbalch equation.

The answer is (A) and the pH rises substantially (about 0.3 pH units in this question). The reason is that while the Henderson–Hasselbalch equation suggests no rise for a well buffered solution as you dilute it, the pKa of bicarbonate changes a lot between fresh water (10.3) and seawater (35 ppt; 8.9). In 12 ppt seawater it is about 0.3 pH units higher than in 35 ppt seawater, so the whole Henderson–Hasselbalch relationship immediately shifts about 0.3 pH units higher when diluting 35 ppt seawater to 12 ppt seawater with pure fresh water.

This article (Figure 3b) shows the pKa as a function of salinity of seawater:

https://www.sciencedirect.com/science/article/pii/0198014989900642

I did not feel like buying the article and I could not fully appreciate Poisson & Goyet's conclusions from the abstract. But, from your remarks above, it seems that the commonly published pKa tables are for pure water systems. Meaning, if there are other species present, or if the system is another strong polar-protic solvent, then we should seek different pKa values. Do I have that right?

I seem to recall having to use a different pKa table for work I did in DMF (now that I'm thinking about this). It seems reasonable that if the system contains other ionic constituents (boron, sulfur, chlorine, etc.) that have affinities for protons or hydroxides, then there is no way the pKa table for pure water could apply. Is that true?

 

[Randy Holmes-Farley]

Many of you tried to calculate the answer based on similar simple mixing of H+ concentrations. Even with corrections for the recombination of H+ and OH- (see below), this doesn't give a correct answer due to buffering of the seawater by many buffers, especially bicarbonate/carbonate, but lets look at how to do that sort of calculation.

Remember, pH = -log [H+]

where [H+] means the molar concentration of H+. So to determine the pH we need to determine the concentration of H+.

Suppose the question was fresh water and sodium hydroxide solution at pH 8.1. So there's no buffering by things like bicarbonate.

What we then have is:

RO/DI water
[H+] = 10^-7
[OH-] = 10^-7

pH 8.1 Sodium Hydroxide Solution
[H+] = 7.9 x 10^-9 (this comes from pH = -log H+ )
[OH-] = 1.3 x 10^-6 (this comes from the fact that [H+] x [OH-] must equal 10^-14; see below)


BUT, we cannot simply look at combining the H+ from these solutions (like we would for salinity for sodium or any other normal ion) to get an answer.

The reason is that there is always an instantaneous equilibrium between the following:

H+ + OH- <---> H2O

This is governed by an equilibrium constant such that in pure water we always have:

[H+] x [OH-] = constant = 10-14

That is why pure fresh water has some H+ and OH- and why they are exactly equal and happen to be 10-7 moles/L

IF we tried to simply combine these two solutions using 1 L of the sodium hydroxide solution and 1.92 liters of the pure fresh water, we might incorrectly assume we have:

[H+] =?? [1 x [H+ in pH 8.1 water] + 1.92 x [H+ in the pure fresh water]] /2.92

which equals

[7.9 x 10^-9 + 1.92 x 10-7] /2.92 = 2 x 10^-7 / 2.92
[H+] = 6.84 x 10-8 moles/L

So pH = -log[H+] = 7.16, right? Wrong!

This solution also has a lot of OH- in it:

[OH-] =?? [1 x [OH- in pH 8.1 water] + 1.92 x [OH- in the pure fresh water]] /2.92

which equals

[1.3 x 10^-6 + 1.92 x 10-7] /2.92 = 2 x 10^-7 / 2.92
[OH-] =5.1 x 10^7 moles/L

And in this hypothesized mixture, we now have too much H+ and OH- because

[H+ x [OH-] = 6.84 x 10-8 x 5.1 x 10^7 = 3.5 x 10^-14 (it should be exactly equal to 10^14).

Our mixture is unstable, having too much H+ and OH-.

Some of them will recombine:

H+ + OH- ---> H2O
so the numbers of them will be reduced instantly on mixing.

The way we figure out how many is through the equation
[H+] x [OH-] must equal 10^-14
[H+] x [OH-] = 10^-14

And let's assing the random variable X as the amount of H+ and OH- that recombine this way:

[H+ - X ] x [OH- - X] = 10^-14

plugging in our preliminary values from mixing:

[6.84 x 10-8 - X ] x [5.1 x 10^7 - X] = 10^-14

solving for X we get:

X2 - 5.78 x 10^-7X + 3.54 x 10^14

Solving this quadratic, we get

X =4.8 x 10^-8 moles/L

So

[H+] = [6.84 x 10-8 - X ] = [6.84 x 10-8 - 4.8 x 10^-8 ] = 2.04 x 10^-8 moles/L

This final answer gives a pH of 7.69, or about 7.7 (note this is not the correct answer for the original question, but is the correct answer for this simplified mixing problem)

As a check, we can determine that the [OH-] is:

[5.1 x 10^7 - X] = [5.1 x 10^7 - 4.8 x 10^-8] = 4.6 x 10^7 moles/L

and verify that

[H+] x [OH-] = 10^-14
is satisfied:

2.04 x 10^-8 x 4.6 x 10^7 = 9.4 x 10^-15 (which is very close to our 10^-14 target value)

 

[Randy Holmes-Farley]

I did not feel like buying the article and I could not fully appreciate Poisson & Goyet's conclusions from the abstract. But, from your remarks above, it seems that the commonly published pKa tables are for pure water systems. Meaning, if there are other species present, or if the system is another strong polar-protic solvent, then we should seek different pKa values. Do I have that right?

I seem to recall having to use a different pKa table for work I did in DMF (now that I'm thinking about this). It seems reasonable that if the system contains other ionic constituents (boron, sulfur, chlorine, etc.) that have affinities for protons or hydroxides, then there is no way the pKa table for pure water could apply. Is that true?

The pKa of every acid changes in salt solutions (say,simple NaCl) because these ions stabilize the higher charge form (e.g., carbonate) more than the lower charge form (bicarbonate). That shifts acid pKa's to higher values. There are specific effects too, such as carbonate being strongly ion paired to magnesium, calcium, and sodium. That stabilizes carbonate more than bicarbonate (which does not ion pair that extensively), and hence serves to shift the pKa even higher. I'll discuss these more later.

 

[nautical_nathaniel]

Many of you tried to calculate the answer based on similar simple mixing of H+ concentrations. Even with corrections for the recombination of H+ and OH- (see below), this doesn't give a correct answer due to buffering of the seawater by many buffers, especially bicarbonate/carbonate, but lets look at how to do that sort of calculation.

Remember, pH = -log [H+]

where [H+] means the molar concentration of H+. So to determine the pH we need to determine the concentration of H+.

Suppose the question was fresh water and sodium hydroxide solution at pH 8.1. So there's no buffering by things like bicarbonate.

What we then have is:

RO/DI water
[H+] = 10^-7
[OH-] = 10^-7

pH 8.1 Sodium Hydroxide Solution
[H+] = 7.9 x 10^-9 (this comes from pH = -log H+ )
[OH-] = 1.3 x 10^-6 (this comes from the fact that [H+] x [OH-] must equal 10^-14; see below)


BUT, we cannot simply look at combining the H+ from these solutions (like we would for salinity for sodium or any other normal ion) to get an answer.

The reason is that there is always an instantaneous equilibrium between the following:

H+ + OH- <---> H2O

This is governed by an equilibrium constant such that in pure water we always have:

[H+] x [OH-] = constant = 10-14

That is why pure fresh water has some H+ and OH- and why they are exactly equal and happen to be 10-7 moles/L

IF we tried to simply combine these two solutions using 1 L of the sodium hydroxide solution and 1.92 liters of the pure fresh water, we might incorrectly assume we have:

[H+] =?? [1 x [H+ in pH 8.1 water] + 1.92 x [H+ in the pure fresh water]] /2.92

which equals

[7.9 x 10^-9 + 1.92 x 10-7] /2.92 = 2 x 10^-7 / 2.92
[H+] = 6.84 x 10-8 moles/L

So pH = -log[H+] = 7.16, right? Wrong!

This solution also has a lot of OH- in it:

[OH-] =?? [1 x [OH- in pH 8.1 water] + 1.92 x [OH- in the pure fresh water]] /2.92

which equals

[1.3 x 10^-6 + 1.92 x 10-7] /2.92 = 2 x 10^-7 / 2.92
[OH-] =5.1 x 10^7 moles/L

And in this hypothesized mixture, we now have too much H+ and OH- because

[H+ x [OH-] = 6.84 x 10-8 x 5.1 x 10^7 = 3.5 x 10^-14 (it should be exactly equal to 10^14).

Our mixture is unstable, having too much H+ and OH-.

Some of them will recombine:

H+ + OH- ---> H2O
so the numbers of them will be reduced instantly on mixing.

The way we figure out how many is through the equation
[H+] x [OH-] must equal 10^-14
[H+] x [OH-] = 10^-14

And let's assing the random variable X as the amount of H+ and OH- that recombine this way:

[H+ - X ] x [OH- - X] = 10^-14

plugging in our preliminary values from mixing:

[6.84 x 10-8 - X ] x [5.1 x 10^7 - X] = 10^-14

solving for X we get:

X2 - 5.78 x 10^-7X + 3.54 x 10^14

Solving this quadratic, we get

X =4.8 x 10^-8 moles/L

So

[H+] = [6.84 x 10-8 - X ] = [6.84 x 10-8 - 4.8 x 10^-8 ] = 2.04 x 10^-8 moles/L

This final answer gives a pH of 7.69, or about 7.7 (note this is not the correct answer for the original question, but is the correct answer for this simplified mixing problem)

As a check, we can determine that the [OH-] is:

[5.1 x 10^7 - X] = [5.1 x 10^7 - 4.8 x 10^-8] = 4.6 x 10^7 moles/L

and verify that

[H+] x [OH-] = 10^-14
is satisfied:

2.04 x 10^-8 x 4.6 x 10^7 = 9.4 x 10^-15 (which is very close to our 10^-14 target value)

I was off a little bit

 

[Randy Holmes-Farley]

So in the previous post i showed what happens if you try to look at simple mixing of a pH 8.1 solution with a pH 7 solution, neither of which have buffers in them.

I mentioned that there were two big complexities in seawater. The first is that seawater does have buffers in it. A lot of them, in fact.

They serve to try to "hold' the pH at a specific level, even if something is trying to lower the pH.

For example, if I take NSW and add a little vinegar or muriatic acid to it, the pH hardly drops, while adding the same acid to the ph 8.1 sodium hydroxide solution might drop the pH like a rock.

let's look at a specific example that I tired experimentally a few years ago.

Take a salt mix at 35 ppt, which I measured to have an alkalinity of 6.3 dKH.

Stir to equilibrate with normal air. The pH was 8.1.

I added 0.5 meq/L of acid (H+). The pH dropped to 6.91. That'as a substantial drop, but not nearly as low as if I had added that same amount of acid to the pH 8.1 sodium hydroxide solution of the previous example. In that case, it is easy to predict that the pH will drop to about pH 3.3 (doing it just the way I did above for RO/DI being added to this same water). 3.3!

So the buffers in the seawater halted the drop in pH, and the buffers far overroad the effects of simple H+ and OH- combination.

What exactly happened?

All of these ions in the water took up H+ to slow the drop in pH when the acid was added:

CO3-- + H+ ---> HCO3- (carbonate to bicarbonate)
HCO3- + H+ --> H2CO3 (bicarbonate to carbonic acid)
B(OH)4- + H+ ---> B(OH)3 + H2O (borate to boric acid and water)
MgOH+ + H+ ---> Mg++ + H2O (magneisum monohydroxide to magneisum ion plus water)
Si(OH)3O- + H+ ---> Si(OH)4 + H2O (silicate to silicic acid plus water)
PO4--- + H+ ---> HPO4-- (phosphate to monohydrogen phosphate)
HPO4-- + H+ ---> H2PO4- (monohydrogen phospahte to dihydrogen phosphate)
H2PO4- + H+ ---> H3PO4 (dihydrogen phospahe to phosphotic acid)
there are organic compounds buffering too, but they will be variable and typically small quantities unless you dose vinegar, etc.

It would be an incredible mess to have to worry about all of these at once. It can certainly be done with a computer and a shreadsheet, but its too much work for the purposes of this question and answer.

In fact, we will simplify the question by taking the biggest contributing buffer system. In the original question, involving seawater at pH 8.1 combined with RO/DI water, the first buffer above is the biggest impactor on pH of these two solutions as they mix:

CO3-- + H+ ---> HCO3- (carbonate to bicarbonate)

That fact is true partly because bicarbonate and carbonate are present at very high concentrations relative to the others, and due to the fact at pH 8.1 we are closer to the pH at which this chemistry mostly happens (say, pH 8 to 10) compared to the second buffer system:

HCO3- + H+ --> H2CO3 (bicarbonate to carbonic acid)

which mostly happens in the range from 5 to 7.

So in an upcoming post, I will show how we can calculate how the pH change on mixing of RO/DI water into seawater is controlled by the carbonate/bicarbonate buffer system.

 

[five.five-six]

I’ll stab at B 7.7

Not wanting to spend too much time, strike that ANY TIME calculating

But pH (potential of hydrogen) is a log 10 scale and we did 4X the H20 consentration or there about and 8.1-.4= 7.7


How’s that for the wrongest answer in the whole thread?

 

[Randy Holmes-Farley]

I’ll stab at B 7.7

Not wanting to spend too much time, strike that ANY TIME calculating

But pH (potential of hydrogen) is a log 10 scale and we did 4X the H20 consentration or there about and 8.1-.4= 7.7


How’s that for the wrongest answer in the whole thread?

It's the right answer to the wrong question: lol

https://www.reef2reef.com/threads/m...ou-want-this-title.361380/page-5#post-4481813

"This final answer gives a pH of 7.69, or about 7.7 (note this is not the correct answer for the original question, but is the correct answer for this simplified mixing problem)

 

[Frogger]

I am not a Chemist so I am not going to argue with your logic but:
So Randy you are saying that a salt water mix with a specific gravity of 1.009 has a higher pH than a saltwater mix of 1.026?
What about a salt mix with a specific gravity of 1.004 would it be higher again?
What's the cut off point to start lowering the pH because we know that the RO water has a pH of 7.

 

[Randy Holmes-Farley]

I am not a Chemist so I am not going to argue with your logic but:
So Randy you are saying that a salt water mix with a specific gravity of 1.009 has a higher pH than a saltwater mix of 1.026?
What about a salt mix with a specific gravity of 1.004 would it be higher again?
What's the cut off point to start lowering the pH because we know that the RO water has a pH of 7.

At 1.004 it would likely be high higher than at 1.009 (and especially, higher than at 35 ppt), yes, but it is not easy to calculate how little salt it takes to give a higher pH than the starting 35 ppt seawater. It may not take much at all. I'm going to go through more calculations to get to the correct answer, but think of it this way so it doesn't seem so unexpected:

Bicarbonate is a far stronger acid when there are salts around then when there are not. It is about 25 times as strong in seawater than in fresh. As you add fresh water, you weaken that acid, and in effect, allow the pH to rise. You can essentially think of it as removing acid from the seawater by adding freshwater.

The fact that this result is so unexpected by most reefers, despite the fact that I've been saying it for many years, is part of the reason for posting this question and going through all of the calculations needed.

 

[Frogger]

Thanks for taking the time to clarify this.

 

[Randy Holmes-Farley]

Soooooo.

Now that we are at the point where we know that buffers are impacting the pH of our mixed solution of 35 ppt seawater and RO/DI water, how can we determine the pH?

Luckily, mathematical chemists have done the heavy lifting for us already, resulting in the Henderson–Hasselbalch equation:

https://en.wikipedia.org/wiki/Henderson–Hasselbalch_equation

pH = pKa + log ([A-]/[HA])

We know the pKa (for now we will assume it is the 35 ppt standard seawater value of 8.915 at 25 deg C).

The [A-] is the base form of whatever buffer system we are considering. In this case, that is carbonate.

The [HA] is the acid form of the buffer system, in this case, bicarbonate.

So we have:

pH = pKa + log ([carbonate]/[bicarbonate])

pH = pKa + log ([CO3--]/[HCO3-])

If we wanted to, we can calculate the ratio of [CO3--]/[HCO3-] in our starting seawater sample:

8.1 = 8.915 + log ([CO3--]/[HCO3-])

-0.815 = log ([CO3--]/[HCO3-])
0.153 = [CO3--]/[HCO3-]
[HCO3-] = 6.5 [CO3--]

Which is as we know for seawater at this pH, there is a lot more bicarbonate than carbonate.

But, back to the pH issue.

Now we look to dilute the sample with RO/DI water. Both carbonate and bicarbonate drop in concentration by a factor of 2.92 (see post #86)

This the equation after dilution becomes:

pH = pKa + log ([CO3--]/[HCO3-])
pH(diluted) = pKa + log ([carbonate initial/2.9]/[bicarbonate initial/2.92])

Importantly, the concentration change (2.92) is exactly the same on the top and the bottom of the fraction of this equation, and they exactly cancel out:

pH(diluted) = pKa + log ([carbonate initial/2.9]/[bicarbonate initial/2.92]) = pKa + log ([carbonate initial]/[bicarbonate initial])

WHICH IS EXACTLY THE SAME AS BEFORE THE DILUTION!!

This is a critical concept: the pH of a buffer system is not changed by simple dilution with pure water

The only limitations to this conclusion are:

1. The pKa is assumed to be unchanged (we will come back to this later)

2. The buffer has to be strong enough (concentrated enough) that it swamps out the effects of H+ and OH- as well as some more very minor effects. This limits the theory so that you cannot keep on endlessly diluting with buffer with no change in ph. Eventually, the buffer is unable to overide the natural amount of H+ and OH- present in pure water. But that is actually far below where we are in seawater. Bicarbonate is about 0.002 molar and carbonate is about 0.0003 M in seawater, while the H+ in RO/DI water is 0.0000001 M. Thus, as folks have mentioned earlier in this thread, the buffer easily outweighs the natural levels of H+ and OH- in pH 7 water, and probably would do so for much more diluted seawater, maybe by a factor of at least 100 down from 35 ppt (that is down from 0.002 molar bicarbonate and 0.0003 M carbonate).

Consequently, we conclude from the Henderson–Hasselbalch equation, that the pH of our sample should not change from the 8.1 that it started at.

In reality, this equation is not perfect and the pH would drop down toward 7 by a very small amount (that is, the buffer does swamp out the natural H+ and OH- effects, but it does not actually eliminate them, they are just a very small contribution to the total). I will not do the heavy lifting to calculate it, but maybe something like 8.1000000 to 8.0999999. Well within our limitations of experimental measurement or reefer interest.

I would also note that another limitation is our assumption that the one buffer system we picked is dominating. It is, but the others contribute some too. Some are too weak (low concentration) to contribute much, but they all hold to the Henderson–Hasselbalch conclusion that pH is unchanged on dilution, or they contribute nothing significant to the problem, just as the natural H+ in the RO/DI water does not contribute much.

Next, on to the assumption that the pKa is unchanged....

 

[Randy Holmes-Farley]

Finally, we have:

pH = pKa + log ([carbonate]/[bicarbonate])

and we know that the carbonate to bicarbonate ratio does not change with dilution, but does the pKa?

yes, in fact it changes a lot. All pKa values change on going from freshw ater to seawater at 35 ppt, with a continuum in between/

The pKa of bicarbonate:

HCO3- --> CO3-- + H+

pKa = -log( [CO3--][H+]/[HCO3-])

changes from 10.3 in pure fresh water to about 8.9 in 35 ppt seawater (and drops even lower at higher salinity).

A higher pKa means a weaker acid, and higher pKa means the equation

HCO3- --> CO3-- + H+

is pushed to the left, using up H+ and lowering [H+] and hence raising pH.

Here's why. There are tow main reasons:

1. All of the positively charged ions in seawater (Na+, K+, Ca++, Mg++, etc.) will tend to crowd around the negatively charged buffer ions, bicarbonate and carbonate. That crowding around of opposite charges tends to stabilize the buffer ions, but it stabilizes the higher charges more (due to higher electrostatic interaction to higher charged species), so carbonate is stabilized more than bicarbonate. Stabilizing carbonate more than bicarbonate tends to make the reaction

HCO3- --> CO3-- + H+

proceed more to the right. Hence it makes bicarbonate a stronger acid, produces H+, and lowers pH, and ends up with a lower pKa.

2. There are specific ion pairing reactions that are a step above simple crowding around the buffer ions. THese ions actually becoem attached to the buffer ions for a fleeting instant, stabilizing them even more. Carbonate is actually more than 80% ion paired at any given instant, with most of it being ion-paired MgCO3 (>50%), CaCO3 (~20%), NaCO3-(~20), with the remainder free. Bicarboante is less extensively ion paired (maybe 75% free, 10% NaHCO3, 10% MgHCO3+, few percent CaHCO3+.
Stabilizing carbonate more than bicarbonate tends to make the reaction

HCO3- --> CO3-- + H+

proceed more to the right. Hence it makes bicarbonate a stronger acid, produces H+, and lowers pH, and ends up with a lower pKa.

So for both of these reasons, the pKa is higher in seawater (8.9) than fresh (10.3), and in 12 ppt seawater it is about 9.2 or so.

This, as we dilute 35 ppt seawater to 12 ppt with RO/DI water, the pKa of bicarbonate shifts up by about 0.3 pH units, and so we have

the starting point:
pH (initial) = pKa (initial) + log ([carbonate initial]/[bicarbonate initial])

the final point:
pH (final) = pKa (final) + log ([carbonate final]/[bicarbonate final])

And substituting what we now know that
pKa (final) = pKa (initial) + 0.3
and that
[carbonate final]/[bicarbonate final] = [carbonate initial]/[bicarbonate initial]

we can conclude that
pH (final) = pKa (initial) + 0.3 + log ([carbonate initial]/[bicarbonate initial])

pH (final) = pH(initial) + 0.3 pH units

Hence, the pH rises and the answer is:

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]

 

[Randy Holmes-Farley]

I hope I did that right...