So this is why it's generally thought that PH controllers do more harm than good?
Great explanation by the way
Yes, that's right. The theory of the drop of pH in 1 unit is mainly bullshit (see below). I am glad you found it useful.
I've wondered where the 3 ppm came from as well. I think the level will be higher than the
<"theoretical ~0.5 ppm CO2">, mainly because the air in a room has more CO2 than the atmospheric level (of 400ppm).
We have a CO2 monitor in the lab. and it usually reads at 600 - 800ppm CO2.
Hi Darrel,
You are right in that sense. CO2 in a room can be higher, but still, 700 ppm are about 1 ppm of CO2 in equilibrium with water, far beyond from the claimed 3 ppm.
I have been doing some work in the last weeks about CO2, and I got where the problem is coming from. Essentially, the 3 ppm story comes from two (wrong) assumptions:
a) Most tap waters have an alkalinity close to 18 dkH.
b) The carbonates equilibrium does not involves exchanges with atmosphere.
The assumption (a) could have been true many years ago in some areas. Nowadays, many people use RO, or mix of RO and tap water, and most people focused in planted tanks seems to go towards soft water strategies rather than hard waters. However, even if this was not true, assuming most waters have 18 dkH is just very simplistic and then, wrong.
The second assumption is also quite worrying. There is a huge difference, when computing the equilibrium, between considering that there is not CO2 exchange with atmosphere and when there is some. The results of 3 ppm just comes from assuming that CO2 concentration is in fix equilibrium with the alkalinity.
For instance (embrace yourselves, some equations are coming), the carbonates equilibrium has mainly four stages:
1. Dissolution of CO2 from air into water, which is determined by Henry's Law and roughly estimated as:
[CO2]d = PCO2(air) * 10^(-1.43).
2. Conversion of dissolved CO2 into carbonic acid by combination with water molecules
CO2(d) + H2O <==> H2CO3.
Only a small amount of dissolved CO2 becomes into carbonic acid. More specifically, [CO2](d) = 650*[H2CO3], so very few molecules. However, conversion to carbonic acid is so fast that any secondary reaction reducing H2CO3 means that more CO2 dissolved in water is taken to compensate the loss. In practical terms, all the dissolved CO2 can then be considered as carbonic acid, common practise in this kind of equilibriums. In this way:
[CO2](d) + [H2CO3] = [H2CO3*],
being [H2CO3*] the term used in the carbonates equilibrium.
3. Part of the dissolved carbonic acid releases protons to water, becoming into bicarbonate:
[H2CO3] <==> [H+] + [HCO3-]
The generated protons reduce the pH what reflects the acidic effect of dissolving CO2 in water. As usual in equilibriums, molecules of carbonic acid are splitting and recombining all the time. However, in balance, there is certain amount of it always split, which is determined by an equilibrium constant, K1:
[H+]*[HCO3-] / [H2CO3*] = K1
As K1 remains fixed, the amount of bicarbonates depends on pH and concentration of dissolved CO2:
[HCO3-] = K1 * [H2CO3*] / [H+]
What means that at higher pHs, the amount of dissolved bicarbonates, for the same concentration of CO2, increases. And the other way around.
4. However, this is not the end of the story. Part of the bicarbonates also disassociates generating protons and carbonates:
[HCO3-] <==> [H+] + [CO3(2-)]
In the same way than before, the process happens in both directions all the time and there is certain amount of carbonates what is expressed through an equilibrium constant, K2:
[H+]*[CO3(2-)] / [HCO3-] = K2
Then, amount of carbonates is determined by the pH and the amount of bicarbonates existing in water:
[CO3(2-)] = K2 * [HCO3-] / [H+]
Again, by increasing the pH we increase the concentration of carbonates and vice versa.
Now, all these expressions relates the carbonates equilibrium. By itself it is impossible to solve, as CO2 dissolving in water changes pH and [H2CO3*] at the same time, so it is not possible to solve the equations. This is where alkalinity plays a role. Alkalinity, TA, is the estimation of the capability of water to absorb protons. In pure water, with all the alkalinity due only to carbonates and bicarbonates, the expression is:
TA = [HCO3-] + 2*[CO3-] + [OH-].
If we assume that the expression of TA is correct, then our value of KH in water is just due to these species, and hence, we can use the value in dkH to determine TA, and then having the answer. This requires some numbers. Let's asusme that we have the wrongly considered standard 18 dkH. Now, dkH are german carbonate hardness degrees, which are expressed as equivalent to 17.848 mg/l of CaCO3 dissolved in water. This means that our water, with 18 dkH is equivalent to:
[CaCO3](d) = 17.848 (mg/l) * 18 = 321.264 mg/l
To translate this to other carbonated species, we need to become this into mols, which is done using the CaCO3 molecular weight, i.e. 100.0869 gr/mol. Doing so:
[CaCO3](d) = 321.264 (mg/l) / (100.0869 (gr / mol) * 1000 ( mg/gr) = 0.00321 mol
We need to be, however, a bit careful about the value, as it relates to the alkalinity generated purely by carbonates, i.e. after converting all bicarbonates into carbonates. Expression of TA then needs to be readjusted:
TA = 2*[CO3(2-)] = 0.00321 * 2 = 0.00642 M
Reason why we can do this is because a mol of CaCO3 is equivalent to a mol of CO3(2-). However, [CO3(2-)] is not going to be such value, so still we need to make some approximations. To solve the issue, we can assume that alkalinity has been mainly originated by dissolution of carbonated rocks by acidic attack of rain, which is a fair assumption in most fresh water environments. Then:
CaCO3 + H2O + CO2 = Ca + 2 * HCO3-
This means that resulting concentration of bicarbonates is equalt to two times the concentration of original carbonates. This matches with the value of TA, and then:
TA = [HCO3-] = 0.00642 M
So, now, with this assumption, we have then the resulting concentration of bicarbonates. This, translated to ppms and using the molecular weight of bicarbonates:
HCO3- = 61.01 gr/mol
[HCO3-] = 0.00642 (mol/l) * 61.01 (gr/mol) * 1000 (mg/gr) = 391.68 ppm
Next step is to find out pH, so we can find the value of dissolved CO2 and the value of carbonates in equilibrium. To do so, we need to use the expression of proton balance in water when dissolving bicarbonates salts:
[H+]total = [H+]water + [H+]carbonates - [H+]carbonic acid
The total of protons is equal to the ones found in pure water plus the ones generated by dissociation of bicarbonates into carbonates, minus those one absorbed by bicarbonate ions to form carbonic acid. We can now put this in terms of existing variables. For instance, in pure water:
[H+] * [OH-] = 10^(-14) = KW
This is known as ionic product of water, which reflects the equilibrium between protons and hydroxide ions. In pure water, with a pH = 7, [H+] is equal to [OH-]:
[H+] = [OH-] = 10^(-7)
This means that [H+]water can be replaced by [OH-] as the value is the same. Additionally, we can work the expression to get the value of [OH-] as function of [H+]:
[OH-] = KW / [H+]
Furthermore, from the equilibrium it is clear that
[H+] carbonates = [CO3(2-)]
[H+]carbonic acid = [H2CO3*]
This is true because, in both cases, converting bicarbonate into carbonate or carbonic acid just involves the transfer of a single proton with water. Now, replacing terms in the proton balance equation:
[H+] = KW / [H+] + K2 * [HCO3-] / [H+] - [H+] * [HCO3-] / K1
We have now one equation and a single unknown, the [H+] concentration. Working in the algebra of the expression, we can put in one side all the references to protons, getting the expression:
[H+] =( (KW + K2 * [HCO3-]) / (1 + [HCO3-] / K1) )^0.5
K1 and K2 are equilibrium constants. Their value is rather imprecise and in fact is difficult to tell which ones must be used. However, traditionally in the hobby, the following values apply:
KW = 10^(-14)
K1 = 10^(-6.3)
K2 = 10^(-10.33)
And with [HCO3-] = 0.00642, we can solve the equation and get [H+]:
[H+] = 4.9215 * 10^(-9) M
that translated to pH as
pH = -log10([H+]) = 8.31
Having [H+], we can now determine the amount of dissolved CO2, as
[H2CO3*] = [H+] * [HCO3-] / K1 = 6.304 * 10^(-5) M
As molecular weight of CO2 is roughly 44.01 gr/mol and ppms is passed to mg / l:
[H2CO3*] = 6.304 * 10^(-5) (mol/l) * 44.01 (gr/mol) * 1000 (mg/gr) = 2.77 ppm = ~ 3 ppm.
So, it is from these computations where the value of 3 ppm comes, by assuming that most water has 18 dkH.
The associated concentration of carbonates (to complete the equilibrium), will be then:
[CO3(2-)] = K2 * [HCO3-] / [H+] = 6.102 * 10^(-5) M
And this translated to ppm:
CO3(2-) = 60 gr/mol
[CO3(2-)] = 6.102 * 10^(-5) (mol/l) * 60 (gr/mol) * 1000 (mg/gr) = 3.66 ppm
The interesting part, and source of the error, is considering that the obtained [H2CO3*] value will not change and then, CO2 will remain in water, which allows to get about 3 ppm of CO2 when KH = 18 dkH. This is totally false. By Henry's law, as pointed out, maximum concentration
in equilibrium would be:
[H2CO3*] = 10^(-3.5) * 10^(-1.43) = 1.1749 * 10^(-5) M, which is equivalent to
[CO2](d) = 1.1749 * 10^(-5) (mol/l) * 44.01 (gr/mol) * 1000 (mg/gr) = 0.52 ppm = ~0.5 ppm
The key of this part is the words "in equilibrium", which means that enough time has passed since the addition of the bicarbonates to generate a solution of 18 dkH. The full chain is as follows:
a. Dissolution of the carbonated rocks by acid water from rain produces the bicarbonates, which dissolve in water very fast.
b. The resulting bicarbonates look for equilibrium, fixing pH of water and generating certain amount of carbonate ions and certain amount of dissolved CO2, in a relatively fast process.
c. However, dissolved CO2 are higher than the maximum values associated to the equilibrium with the atmosphere. This means water starts to lose CO2 to the air.
d. The degasification ends when CO2 dissolved in water equals the value in atmosphere. This process does not affect to alkalinity because the loss of CO2 produces more carbonates as bicarbonates disappear due to the change of pH. However, pH increases in the process, so degassed water will have higher pH than solution before the equilibrium is reached:
Note that, the loss of CO2 implies consuming protons and bicarbonates:
HCO3- + H+ <==> CO2 + H2O
As protons are consumed, pH increases and system displaces towards carbonates generation.
It is possible to get the final output for this situation and getting the species in the new equilibrium and pH after degasification, but that is a bit long for now. However, I wanted to show that the assumption of KH = 18 dkH and 3 ppm in water is rather arbitrary and doe snot obeys at all to real science associated to the problem, as most natural waters are indeed in equilibrium with air, i.e. changes in alkalinity are affecting to pH but not much to the CO2 concentrations. There are exception, of course, as waters coming from hydro-thermal processes ending in rivers/lakes before degasification is completed, or local processes like high concentration of organic matter being degraded, or water coming out from karst systems underground. But as a general rule, if no CO2 injection is produced by other means, natural water has 0.5 ppm of CO2 (considering a PCO2 of 350 micro-mol, obviously).
Cheers,
Manuel