A simplified means of determining the relative Carbonate Species percentages of natural water via it's pH

[Silver_Is_Money]

For natural water of pH 4.30:

Water's pH = 6.40 + log(mol fraction ratio of HCO3- to H2CO3)

4.30 pH = 6.40 + log(mol fraction of HCO3- to H2CO3)

-2.10 = log(mol fraction of HCO3- to H2CO3)

10^-2.10 = mol fraction of HCO3- to H2CO3

mol fraction of HCO3- = 0.00794 x 100 = 0.794%
mol fraction H2CO3 = (1-0.0.00794) x 100 = 99.206%


For natural water of pH 5.40:

Water's pH = 6.40 + log(mol fraction of HCO3- to H2CO3)

5.40 pH = 6.40 + log(mol fraction of HCO3- to H2CO3)

-1.00 = log(mol fraction of HCO3- to H2CO3)

10^-1.00 = mol fraction of HCO3- to H2CO3

mol fraction of HCO3- = 0.100 = 10.0%
mol fraction of H2CO3 = 1-0.100 = 90.0%


For natural water of pH 6.40:

Water's pH = 6.40 + log(mol fraction of HCO3- to H2CO3)

6.40 pH = 6.40 + log(mol fraction of HCO3- to H2CO3)

0.00 = log(mol fraction of HCO3- to H2CO3)

10^0.00 = mol fraction of HCO3- to H2CO3

mol fraction of HCO3- to H2CO3 = 1 = 50% HCO3- and 50% H2CO3


For natural water of pH 8.40:

Water's pH = 6.40 + log(mol fraction of HCO3- to H2CO3)

8.40 pH = 6.40 + log(mol fraction of HCO3- to H2CO3)

2.00 = log(mol fraction of HCO3- to H2CO3)

10^2.00 = mol fraction of HCO3- to H2CO3

mol fraction of HCO3- to H2CO3 = 100 = 100% HCO3- and 0% H2CO3

 

[Silver_Is_Money]

Edit: The mole fraction of HCO3- is represented by the 'Y Axis' scale.

 

[Silver_Is_Money]

This is the reason why you must only remove 90% of Alkalinity or Bicarb whereby to bring your source water to pH 5.40. And why removing 100% of Alkalinity or Bicarb would bring your water to somewhat below pH 4.30 (albeit that pH 4.30 is the generally accepted and acknowledged zero point).

 

[Silver_Is_Money]

Beyond pH 8.4 requires the use of a different equation. One which uses pH 10.4 in the place of pH 6.4.

NOTE: The graphical display of the carbonate species as seen 2 posts above seems more midpoint centered about 6.3 and 10.3 than for 6.4 and 10.4, but a different chart which I observed was focused upon 6.4 and 10.4 as the midpoints. Perhaps the truth better lies somewhere inbetween, as for 6.35 and 10.35. ?

 

[Silver_Is_Money]

One means whereby to better pin down the pH constants might be to stir a few teaspoons of baking soda into a Liter or Quart of distilled water and then after it is fully dissolved and it has cooled to room temperature, measure it's pH.

If the baking soda water's pH is 8.35, the best formula constants to choose are 6.35 and 10.35, etc...

 

[Silver_Is_Money]

In respect to the methodology which I've outlined above, I must now ask everyone who has interest in this to reflect upon why NaHCO3 (baking soda), which dissociates into pure HCO3- and Na+, has pka's as follow:

pKa1 = 6.37 (with some nominal small degree of textbook +/- variability to be found within a broad scan of technical literature)
pKa2 = 10.33 (again, with some nominal small degree of textbook +/- variability to be found within a broad scan of technical literature)

Pure coincidence, or some sort of relationship?

 

[Silver_Is_Money]

A hint lies in the very definition of a pKa. At the pH associated with any pKa, by definition, there is 50% ion dissociation.

So at pH 6.37 nominally 50% of H2CO3 (or NaHCO3) has dissociated into free H+ (or Na+) and free HCO3-, and at pH 10.33 nominally half of HCO3- has dissociated into free H+ and free CO3--.

One very useful interpretation of this is to say that when pH=pKa, half the compound is dissociated (proof below), and if, for instance, water is in a solution with an anion of base (assuming the neutral base has a high pKa) the water will be deprotonated to form the more stable, neutral, base.

The proof that pKa=pH means 50% dissociation is here:

pKa=-log([H+][A-]/[HA])=pH=-log[H+] removing the logs

[H+][A-]/[HA]=[H+], therefore [A-]=[HA], showing 50% dissociation.

https://orbisophchemistry.wordpress.com/2015/07/20/ph-ka-and-pka/

 

[Silver_Is_Money]

One means whereby to better pin down the pH constants might be to stir a few teaspoons of baking soda into a Liter or Quart of distilled water and then after it is fully dissolved and it has cooled to room temperature, measure it's pH.

If the baking soda water's pH is 8.35, the best formula constants to choose are 6.35 and 10.35, etc...

I should emphasize that time is of the essence here in catching the precise solution pH reading, because the CO2 from the air around us is continually acidifying the test Liter/Quart of baking soda and water solution, and thus it is slowly yet continually shifting the solutions pH downward with the progression of time.

 

[meaulnes2]

Is there some demonstration of the law you base your article on, namely
Water's pH = 6.40 + log(mol fraction ratio of HCO3- to H2CO3) ?

 

[Silver_Is_Money]

It evolves from Ka's (dissociation constants) and pKa's [whereby pKa = -log(Ka)] and an arrangement (or rearrangement) of the Henderson-Hasselbalch equation which states:
pH = pKa + log [base/acid] (where base and acid are molar concentrations).

You will often see this as pH = pKa + log ([A-]/[HA])

Based upon relevant pKa's It is probably closer to:
pH = 6.37 + log [base/acid] for the relationship between HCO3- and H2CO3, and:
pH = 10.33 + log [base/acid] for the relationship between HCO3- (or the H+ within HCO3-) and CO3--

This might help you: How to Calculate Buffers

The key is to study the evolution of the Henderson-Hasselbalch equation.
https://en.wikipedia.org/wiki/Henderson–Hasselbalch_equationhttp://saintssnrchem.weebly.com/ka-and-pka.htmlhttps://orbisophchemistry.wordpress.com/2015/07/20/ph-ka-and-pka/

 

[meaulnes2]

Thank you Larry I will look at this thoroughly. BTW could you point me to some good book likely to help me consolidate my understanding particularly by improving the foundations. I recall you I am not a chemist but I am an electrical engineer and can read English rather easily (just to indicate what I can or cannot understand ).

 

[Silver_Is_Money]

Thank you Larry I will look at this thoroughly. BTW could you point me to some good book likely to help me consolidate my understanding particularly by improving the foundations. I recall you I am not a chemist but I am an electrical engineer and can read English rather easily (just to indicate what I can or cannot understand ).

I don't have any books to recommend. I'm sure others will be able to help here.

 

[meaulnes2]

Thanks and don't worry. The light arises little by little. Still lacks familiarity with these concepts but it's coming.

 

[Silver_Is_Money]

Thanks and don't worry. The light arises little by little. Still lacks familiarity with these concepts but it's coming.

My university level chemistry textbooks date to the early 70's at best. There are certainly more up to date books out there today.

 

[meaulnes2]

After some readings to improve my knowledge of this domain, I need to come back to you.
I'm still a beginner in the field but I'm a bit confused by the way you phrase things, although your calculations give exact results within a few thousandths.
From what I think I understood (but I still have the excuse of being a neophyte if I'm wrong), the molar fraction of a constituent is the ratio between the quantity of a constituent (expressed in moles) and the total quantity of all the constituents of the mixture (also expressed in moles).
The Henderson-Hasselbalch equation states that
pH = pKa + log ([A⁻]/[AH])
where [A⁻] represents the concentration of the conjugate base and [AH] the concentration of the non dissociated acid.
However, it seems to me that the ratio [A⁻]/[AH] is not the molar fraction of [A⁻]. This molar fraction would rather be for me [A⁻]/([A⁻]+[AH]).

With the examples you have taken, the thing goes relatively unnoticed because the numerical values are such that there is only a minor variation on the results.
But the fact becomes more apparent if we take, for example, water at pH=7.5
It comes
7.5=6.4 +log([HCO₃⁻]/[H₂CO₃*]
10^1.1=12.58
In this case, one could not say that the mole fraction is 1258% but only that [HCO₃⁻] is 12.58 times larger than [H₂CO₃*].

We would thus have, molar fraction of [HCO₃⁻] = 12.58 / (12.58 + 1] = 0.926 or 92.6% which seems to appear on the curves on the distribution of species as a function of pH.

I hope you understand my remark (and excuse me if I'm wrong) because the trouble came to me while reading Palmer and Kaminski's book, when I wanted to understand where they got the values of 0.93 est 0.01 as the load value on page 93. These values, straight out of the blue, seem to me to correspond to what you describe in this article. Am I right?

 

[Silver_Is_Money]

After some readings to improve my knowledge of this domain, I need to come back to you.
I'm still a beginner in the field but I'm a bit confused by the way you phrase things, although your calculations give exact results within a few thousandths.
From what I think I understood (but I still have the excuse of being a neophyte if I'm wrong), the molar fraction of a constituent is the ratio between the quantity of a constituent (expressed in moles) and the total quantity of all the constituents of the mixture (also expressed in moles).
The Henderson-Hasselbalch equation states that
pH = pKa + log ([A⁻]/[AH])
where [A⁻] represents the concentration of the conjugate base and [AH] the concentration of the non dissociated acid.
However, it seems to me that the ratio [A⁻]/[AH] is not the molar fraction of [A⁻]. This molar fraction would rather be for me [A⁻]/([A⁻]+[AH]).

With the examples you have taken, the thing goes relatively unnoticed because the numerical values are such that there is only a minor variation on the results.
But the fact becomes more apparent if we take, for example, water at pH=7.5
It comes
7.5=6.4 +log([HCO₃⁻]/[H₂CO₃*]
10^1.1=12.58
In this case, one could not say that the mole fraction is 1258% but only that [HCO₃⁻] is 12.58 times larger than [H₂CO₃*].

We would thus have, molar fraction of [HCO₃⁻] = 12.58 / (12.58 + 1] = 0.926 or 92.6% which seems to appear on the curves on the distribution of species as a function of pH.

I hope you understand my remark (and excuse me if I'm wrong) because the trouble came to me while reading Palmer and Kaminski's book, when I wanted to understand where they got the values of 0.93 est 0.01 as the load value on page 93. These values, straight out of the blue, seem to me to correspond to what you describe in this article. Am I right?

You may well be correct, and I applaud your determination and effort here.

I'm certainly no expert on Henderson-Hasselbalch, and at this juncture I've forgotten most of what I formally learned of it. Clearly someone better versed and more up to date in the practical application of HH needs to chime in and guide/assist you here. I'm roughly 47 years post any learning I would have received in this area, and in those 47 years I had never once been called upon to apply HH within the working world, albeit that my entire working life (I'm retired now) was spent within chemical manufacturing and synthesis facilities (variously in such capacities as a lab tech, analyst, dept. supervisor, production scheduler, effluent and air pollution monitoring and control, area supervisor, area manager, materials planner, and lastly buyer/planner, or more technically correct, planner/buyer [since materials planning must precede purchasing]).

AJ deLange appears to have been the genius behind much of the science presented within the book you reference. I'm not worthy to tie his shoe laces. I wish he would chime in here.

Edit: Martin Brungard also contributed to the Palmer and Kaminski book, and he is still active here on this forum. Perhaps he might shed light upon a more proper and correct means whereby to apply HH to the ion fraction curves as witnessed within a carbonate species chart that equates them to water pH.

 

[meaulnes2]

OK, thanks, I understand.
I would be delighted if one of the two could answer the question of what the values 0.93 and 0.01 represent on page 93 of Palmer and Kaminski's book and tell me if these values can actually be accessed by the calculation you propose here instead than by reading a curve.

 

[Silver_Is_Money]

It may be that Martin Brungard's contribution to the book was more in the capacity of editor, but he would need to clarify as to whether or not he additionally contributed in the direct capacity of providing math/science input.

 

[Silver_Is_Money]

Factually the equation expresses merely a mole fraction ratio, as stated directly within, so with the presumption of 6.4 as the constant (whereby I now prefer 6.37 as more correct) the equation (which does not originate from me) is correctly yielding a ratio of 12.59 times more molar HCO3- than H2CO3 for pH 7.5, so I'm not going to fault the output as derived from equation. It may however not be fully complete as expressed, or perhaps more likely, as interpreted. But it still seems to have its place as to usefulness.

 

[Silver_Is_Money]

10^1.1 = 12.59 (rounded)

12.59 : 1 for HCO3- to H2CO3 infers 1 : 12.59 for H2CO3 to HCO3-, or 1/12.59 for H2CO3.

1/12.59 = 0.0794

And when one goes up the 'Y Axis' to meet the value of 0.0794 and then moves horizontally to the right by which to intersect the H2CO3 line, the intersection corresponds to pH 7.5 on the 'X Axis'.

Again, no conflict from the equation as given.

 

[meaulnes2]

It is a ratio of moles but in my understanding the values on the y axis do not represent the ratio of one specie to the other but rather the ratio of one specie to the total which is the definition of the molar fraction . It is why at the cross point it is not 1/1=1 but rather 1/2=0.5.
Thus in the same spirit using pK1=6.38
when [HCO₃⁻]/[H₂CO₃*] (that is not the molar fraction) =13.18 (i.e. 10^1.12) the value to be read of the y axis (that is the molar fraction) is 13.18/14.18=0.929 i.e. ~93%
Putting the image in Inkscape which has horizontal and vertical guides, I find the the same.

 

[Silver_Is_Money]

As it turns out, the H-H equation is in essence a rather simplified kludge of sorts which can be off by as much as 50% in its pH predictions in some specific cases, as can be seen here:
The Henderson-Hasselbalch Equation: Its History and Limitations - PDF Free Download

Per this 5 page peer reviewed document, in reality (as for a more exacting solution) 4 independent equations must be satisfied, as opposed to merely 1 equation. With those 4 being presented as:

[H+][OH-] = 10^-14
[H+] = Ka[HA]/[A-]
[HA] + [A-] = (nA + nB)/V
[H+] + nB/V = [A-] + [OH-]

This is all way beyond my (former) pay scale, but you may find something within it.

Edit: My effort to copy and paste brought with it some font corruptions for the equations, which I have quickly attempted to resolve, but to assure equation accuracy one must refer only to the linked dissertation as to the presented formulas.

 

[meaulnes2]

Thank you. Interesting but I am starting to remember why I had a certain aversion for chemistry at school.

 

[Silver_Is_Money]

Thank you. Interesting but I am starting to remember why I had a certain aversion for chemistry at school.

Well, I did cautiously preface this as a "simplified means".

Here are a few juicy tidbits from the paper:

As the scope of its application expanded,
however, the approximate nature of the Henderson–
Hasselbalch equation faded out of curriculum. None of the
textbooks of general or biochemistry chemistry we recently
examined offered a quantitative discussion of the reliability
of pH calculations from eq 1 or its non-logarithmic form.
As we will show, the discrepancy between the exact and
approximate calculations, even at moderate concentrations
and pH values not far from the pKa , can be as much as 50%
(when Ka = 10^-3 and the acid and base are 0.01 M), and many
buffer problems solved through Henderson–Hasselbalch
equation—with the usual interpretation of [HA] and [A-] as
the initial molarities—do not warrant an answer with more
than a single significant digit. Buffer problems that carry two
or more significant figures are often fictional exercises in
general chemistry.

The Henderson–Hasselbalch equation is eminently suited to
calculate the pH of buffers made with acids whose pKa lies
in the range of about 5 to 9, so long as the composition of
the buffer is not highly skewed in favor of one or the other
component. For an acid with a pKa of 7, the approximate
and exact [H+ ] remain within a percentage point of each other
over almost the entire titration range—from 3 mL of NaOH
until 95 mL for the titration of 100 mL of 0.01 M acid.
The Henderson–Hasselbalch equation, however, becomes
unreliable for calculating [H+ ] when the dissociation constant
of the acid departs from 10^-7 by more than two orders of
magnitude. When Ka is 10^-3 , for example, even in buffers
made with equal number of moles of acid and base (i.e., at
the midpoint of a titration), the approximate [H + ] differs
from the exact value by as much as 365% in dilute solutions.
Thus for acids such as HNO2, HF, HCOOH, BrCH2COOH,
ClCH 2 COOH, and lactic acid, whose Ka’s are around or
above 10^-4 , the Henderson–Hasselbalch equation is not
appropriate to derive the titration curves.

 

[Silver_Is_Money]

The moral of this story is that if the hallowed foundational math model presumptions of the very science of Chemistry itself are highly flawed, then how is one to presume that via using such models one can design an accurate software tool for Mash pH prediction?

If the bedrock math model of buffered system pH "prediction" itself is only "relatively" accurate some of the time, and with those accuracy times being both highly idealistic and highly cherry picked, then nothing more than this must or factually can be expected of Mash pH assistant software. But since the dissociation predicted H+ error of the bedrock model is up to 3.65X (365%), and any 1 full pH point drift requires a full 10X change in the quantity of extant H+, then this establishes a pH prediction error limit of:

For example:

For pH to be 5.4, the molar concentration of dissociated H+ must be:
10^-5.4 = 0.000003981 M in free (dissociated) H+

Now, lets increase this by 365% and see what the resultant pH is:
0.000003981 x 3.65 = 0.000014531 M in free H+

And now lets reverse the process and solve for pH:
pH = -log(0.000014531) = 4.84

5.4 - 4.84 = 0.56 pH points of potential error for the H-H math model prediction.

Or roughly the 50% error in pH prediction for H-H as mentioned within the paper.

Whereby: 0.56 x 100 = 56%

If you casually scanned the referenced paper and observed that at one juncture it discussed H-H error as being on the order of 50% and at another juncture it discussed it as being on the order of 365%, and this confused you, this is why. One error is in reference to pH, and the other is in reference to H+.