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....
