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I couldn't find anything online so I'm asking the question here.

I use distilled water for some species of crabs which should have a carbonate hardness of 0.

Now I want to add plants to the aquarium and therefore CO2.

I know there is a table for KH-CO2-pH but it only goes to 0.1 KH while my water should have KH 0.

How will CO2 injection affect pH? I'm looking for specific numbers.

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    adding co2 to water without any buffer effect will make pure carbonic acid.using destilled water in a fish tank is crazy in my opinion. Mar 2 '18 at 17:34
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It sounds as though you already have a good knowledge of how water chemistry works.

In basic terms, KH (carbonate hardness) affects how much the pH can change within the water. The presence of the carbonic acid in the water stops the pH from fluctuating wildly when it is influenced by CO2 or by other pH influencers.

You'll probably be familiar with the classic advice given when attempting to alter your pH: Chances are, you will need to lower your KH so that the changes to the pH have more impact on the actual pH of the tank.

If your KH is 0 that means your pH can be very easily influenced by any factors that would affect your pH. In reference directly to CO2 check out this picture from practical fish keeping:

CO<sub>2</sub> Vs KH Vs pH

You'll see that the mg of CO2 per litre that you need for the pH to fall to 6 at a KH of 0.5 is only 15mg compared to the 87mg you might need for the same pH at a KH of 3.0.

'Okay fine, but what's the answer to my question?'

Injecting a small amount of CO2 into your water will have a big affect on the pH due to the low KH. This may not actually be a problem if you have very limited sources of pH influencers but there is a risk.

I don't know about crabs, but generally very low KH levels are not advised with fish because the pH can swing wildly as a result of pH influences. These can be substrate, decorations and many others. If your environment is very stable (free from potential pH altering sources) carefully injecting CO2 may be feasible for you but low KH and CO2 injection always make me a little nervous.

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Assuming that we have pure, distilled water at 25 °C:

  • its KH is 0 because it has nothing dissolved in it (more specifically, no carbonate nor bicarbonate ions);
  • its initial pH is 7 (which is neutral pH at at 25 °C).

Having said that, we must consider that water - unless it is stored in a sealed and airtight container, which aquarium certainly isn't - dissolves some amount of the air constituents. We could calculate the pH after it has been exposed to air (and thus atmospheric CO2) and has achieved equilibrium with respect to the amount of CO2 dissolved in it. We ignore the rest of air constituents: most prevalent ones - N2, O2 and Ar - don't dissociate in water to form acids nor bases. We also assume that the air is not polluted with any significant amounts of acidic, industrially emitted toxins - like SO2 and nitrogen oxides.

According to my calculations (presented below), applying CO2 injection in distilled water (KH = 0) would result in a HUGE pH swing: from about 5.61 to 4.735. Because pH scale is logarithmic with a base of ten, this pH swing means about 7.5-fold increase in acidity of the water (10-4.735/10-5.61 ≈ 7.5).

For reference: if live animals inhabit the aquarium, pH of aquarium water should not change more than ± 0.2 units over 24 hours, or else it is too fast and unsafe for the well-being of the animals. Plants are more resistant to side effects of pH changes, but are not immune to them.

Please note that all the resulting values are approximate. I trust my calculations to be correct, but they ultimately could not be more accurate than the specific constants used in them - and these constants had all been measured experimentally, with a limited precision as well.

What is more, distilled water in aquarium isn't going to stay completely pure and retain its parameters. Even if we consider a scenario with an inert substrate, etc. then still, many pH-affecting compounds will appear in the water column as a result of biological activity and metabolism of its inhabitants, like ammonia excreted by aquatic animals and decomposing matter and hydrogen ions produced and released by nitrifying bacteria. Also, in my calculations I completely ignore the carbon dioxide both produced and consumed by aquarium inhabitants; I am sorry but taking it into account would be close to impossible because there would be too many unknown variables.

According to Henry's law:

The amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid. The proportionality factor is called KH (Henry's law constant).

For CO2 this constant has the value of KH = 29.41 L ⋅ atm ⋅ mol-1. It is a bit unfortunate how this constant is marked in this context; thus, please don't confuse KH (carbonate hardness) and KH (Henry's constant) and please remember that they are two separate, unrelated things.

According to Dalton's law of partial pressures:

In a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

and taking the ambient concentration of CO2 in the air in 2021 (around 417 ppm) and the atmospheric pressure (we are using atmospheres as pressure units, so in these units atmospheric pressure is by definition 1 atm), we have the result that the partial pressure of CO2 in the air is pCO2 = 4.17 ⋅ 10-4 atm.

Please note that 417 ppm is the average ambient concentration of CO2 in the air all over the world; inside inhabited buildings it is going to be higher than this value (as Wikipedia says, CO2 concentration in an enclosed office room could increase to 1000 ppm within just 45 minutes). For reference, elevated CO2 concentration in the air results in unpleasant and stuffy feeling in the chest, tiredness and general perception of the air being stale.

In water, the concentration of dissolved CO2 (being dissolved in water is indicated by "(aq)" postfix) is:

        [CO2 (aq)] = pCO2/KH = 4.17 ⋅ 10-4 atm / 29.41 L ⋅ atm ⋅ mol-1

        = 1.418 ⋅ 10-5 mol ⋅ L-1

In aquarium hobby it is much more common to express the concentration of dissolved CO2 in parts per million (which is equivalent to miligrams per liter) rather than moles per liter; the molar mass of CO2 is 44.01 g ⋅ mol-1 = 44010 mg ⋅ mol-1 so the result after conversion is:

        44010 mg ⋅ mol-1 ⋅ 1.418 ⋅ 10-5 mol ⋅ L-1 = 0.6241 mg ⋅ L-1 = 0.6241 ppm

We won't use this value of 0.6241 ppm in further calculations because the formulae need values expressed in moles per liter; I just converted it for those people who are more familiar with CO2 expressed in ppm to immediately see a get a feeling about how much this is.

Part of the dissolved CO2 reacts with water to form carbonic acid:

        CO2 (aq) + H2O ⇌ H2CO3

and the equilibrium constant of this reaction in pure water, as Wikipedia says, is about Khydration = [H2CO3]/[CO2 (aq)] = 1.7 ⋅ 10-3 (no units included because equilibrium constants are dimensionless). As a side note: we don't include the concentration of water [H2O] because we assume that the concentration of water in water (literally that) is constant and equal to 1.

Carbonic acid is a weak acid, which means that it doesn't fully dissociate in solution. It is diprotic (which means it is able to donate two H+ ions, a.k.a. protons, per one acid molecule in aqueous solution), so it dissociates in two stages. The first dissociation H2CO3 ⇌ H+ + HCO3- has a related equilibrium constant of about Ka1 = [H+]⋅[HCO3-]/[H2CO3] = 10-3.6. For reference, sometimes equilibrium constants are expressed in logarithmic scale, analogous to pH scale; in this specific case Ka1 = 10-3.6 could be alternatively expressed as pKa1 = 3.6.

The second dissociation HCO3- ⇌ H+ + CO32- has an equilibrium constant of about Ka2 = [H+]⋅[CO32-]/[HCO3-] = 10-10.3 which means that its contribution to the total H+ levels (at thus to the pH value) will be negligible and we could completely ignore the second dissociation in this case (we start at the pH of 7 which means that the initial concentration of hydrogen ions is [H+] = 10-7 mol ⋅ L-1 and this is many orders of magnitude higher than 10-10.3).

Having said that, we now consider the overall reaction to be:

        CO2 (aq) + H2O ⇌ H2CO3 ⇌ H+ + HCO3-

We need to calculate the concentration of hydrogen ions to get the pH value; to get the combined equilibrium constant of the overall reaction Koverall = [H+]⋅[HCO3-]/[CO2 (aq)] we need to multiply the two constants Khydration and Ka1 related to each step in the overall reaction; thus we get:

        Koverall = Khydration ⋅ Ka1 = 1.7 ⋅ 10-3 ⋅ 10-3.6 = 4.27 ⋅ 10-7

Based on the equation H2CO3 ⇌ H+ + HCO3- and ignoring the additional H+ ions resulting from the self-ionization of water (H2O ⇌ H+ + OH-) we could see that the amount of H+ and HCO3- resulting from this reaction is identical; we could assume that the concentrations of H+ and HCO3- are going to be equal ([H+] = [HCO3-]); this simplifies the equation:

        Koverall = [H+]⋅[HCO3-]/[CO2 (aq)]

to:

        Koverall = [H+]⋅[H+]/[CO2 (aq)] = [H+]2/[CO2 (aq)]

After manipulating the equation to finally get the concentration [H+]:

        [H+] = ( [CO2 (aq)] ⋅ Koverall )1/2 = ( 1.418 ⋅ 10-5 ⋅ 4.27 ⋅ 10-7 )1/2

        = ( 6.055 ⋅ 10-12 )1/2 = 2.461 ⋅ 10-6 mol ⋅ L-1

and because the calculated [H+] of 2.461 ⋅ 10-6 mol ⋅ L-1 is much higher than 10-7 mol ⋅ L-1 resulting from self-ionization of water, we were allowed to ignore the self-ionization of water in this case.

Based on that, we finally get that pH of distilled water in equilibrium with atmospheric CO2, but without supplementing CO2 via injection, is:

        pH = -log10[H+] = -log10( 2.461 ⋅ 10-6 ) = 5.61

Looking for the pH values of rainwater in my search engine gives more or less similar results, which confirms that the calculations are correct.

Now we are going to calculate pH of water with CO2 injection to see how much will the pH drop. Most aquarist sources I could find agree that the optimal CO2 concentration for planted aquarium we should aim for, that is both beneficial to plants and safe for animals, is about 35 ppm (which is equivalent to 35 mg/L). After converting it to molar concentration, we get that it is equal to:

        [CO2 (aq)] = 35 mg ⋅ L-1 / 44010 mg ⋅ mol-1 = 7.953 ⋅ 10-4 mol ⋅ L-1

After using the same exact formula as before, we get that:

        [H+] = ( [CO2 (aq)] ⋅ Koverall )1/2 = ( 7.953 ⋅ 10-4 ⋅ 4.27 ⋅ 10-7 )1/2

        = 1.843 ⋅ 10-5 mol ⋅ L-1

which means that after installing CO2 injection system in this distilled water aquarium and reaching the CO2 concentration of 35 ppm, we get the pH value of:

        pH = -log10( 1.843 ⋅ 10-5 ) = 4.735

Addressing the subject about the pH/KH/CO2 tables having no data for KH of 0: first off, these tables implicitly assume that CO2 and KH are the only potential sources of acidity and alkalinity (which is not true, but it's still a good enough approximation), thus if the KH was to be 0 then it would be impossible to get any pH values above 7.

In order to calculate the dissolved CO2 concentration required to drop distilled water's pH from 7 to 6, we reuse the previously derived formula:

        Koverall = [H+]2/[CO2 (aq)]

and we rearrange the terms to calculate the unknown [CO2 (aq)]:

        [CO2 (aq)] = [H+]2/Koverall

We know the [H+] in advance from the definition of pH scale because we a priori assume the final pH to be 6:

        pH = -log10[H+] = 6

        [H+] = 10-6 mol ⋅ L-1

Plugging both relevant values into the formula we get the resulting molar concentration of:

        [CO2 (aq)] = ( 10-6 )2 / 4.27 ⋅ 10-7 = 2.342 ⋅ 10-6 mol ⋅ L-1

which we convert to mg/L using CO2's molar mass of 44100 mg ⋅ mol-1:

        2.342 ⋅ 10-6 mol ⋅ L-1 ⋅ 44100 mg ⋅ mol-1 = 0.1033 mg ⋅ L-1

and this means that with KH of 0 it would require only about 0.1 mg/L of dissolved CO2 to drop pH to 6 - and just merely being exposed to the atmosphere itself is going to put even more CO2 into the water than that - about 0.6 mg/L, dropping pH to around 5.61 as calculated in earlier paragraphs. It appears that 0.1 mg/L is the lower limit of precision in the table from Henders's answer, so I guess it would not be really useful to include these numbers.

However, if anyone wanted this mysterious riddle solved (I certainly did), the missing numbers for KH = 0 are around 0.1 mg/L for pH 6, decreasing below the precision limit of the table in Henders's answer until exactly 0 mg/L for pH 7, and N/A for above pH of 7:

pH Dissolved CO2 in ppm
6.0 0.1033
6.2 0.04112
6.4 0.01636
6.6 0.006516
6.8 0.002594
7.0 0
>7.0 N/A

What is more, since dissolving CO2 produces bicarbonate and carbonate anions, introducing CO2 into the water would actually raise the KH by a tiny bit and it would not be 0 anymore; it would be around 0.007 KH if I am correct. However, this fact could be ignored for this answer's purpose and I include it rather as food for thought in this context.

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Distilled water and CO2 will kill fish and I expect crabs. I have more or less done it twice by accident. I would never use distilled, but our tap water has fairly low mineral content: I put water in an empty aquarium (no gravel, rocks or plants) for several hours while working on their home tank. A few hours later most were dead; pH tested below 6.2.

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