October 15, 2019, 07:26:32 AM
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Topic: pH adjustment of Spirulina culture via dilution  (Read 238 times)

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Offline Elliander

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pH adjustment of Spirulina culture via dilution
« on: September 16, 2019, 07:51:56 PM »
I have been developing a cold tolerant strain of Spirulina, and succeeded, but then I made a miscalculation in adding the baking soda for a new larger system. Spirulina is capable of surviving from mildly acidic to a pH of 12, but grows best at a pH of 9. Keeping it between 8.5 and 9.5 ensures an ideal growth rate. Unfortunately, the pH of around 1500 gallons of culture is at 12. This doesn't kill the culture, but it has kept it from growing much since the last harvest, which is very bad.

Since I am planning to expand into a much larger system, I was hoping that simple dilution would solve the problem, since it would save money on baking soda.

Given that I will do my mixing in a 275 gallon IBC tote, how many gallons of pH 12 culture would I need to mix with pH 7 water to achieve a pH 9 culture within that 275 gallon IBC tote?

Phototoxicity is a potential problem if the dilution is too great, but I can potentially settle Spirulina out and draw water out from the top prior to mixing to mitigate this problem. If the dilution is too great though I may have to combine dilution with food grade acidifier.

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If I wanted to correct by adding an acid, I would use the following equations:

Molarity of culture = 10^-[pH] = 10^-[12] = 1e-12

(M_1)(V_1) + (M_2)(V_2) = (M_3)((V_1) + (V_2))

Where “M_1” is the molarity of the acid, “V_1” is the volume of the acidic solution, “M_2” is the molarity of the water and “V_2” is the volume of the water, and M_3 is the expected molarity at the target pH of the total solution.

But I can't seem to find the right equation via dilution in my notes. Extrapolating from the above, I would assume to use the following, but I want to be sure before I do anything:

Molarity of culture = 10^-[pH] = 10^-[12] = 1e-12
Molarity of Water = 10^-[pH] = 10^-[7] = 1e-7
Molarity of Goal = 10^-[pH] = 10^-[9] = 1e-9

(1e-7)(V_1) + (1e-12)(V_2) = (1e-9)((V_1) + (V_2))

Where  1 of the water (in this case, since the water is reducing the pH) and 2 of the culture.

since I am looking for a total of 275 gallons, I need to rearrange to solve for one volume and plug in another until I get closer to the 275 gallons I am looking for?  Or, I guess, 1040.99 Liters.

V_1 = ((M_3)(V_2) – (M_2)(V_2))/((M_1) – (M_3))
V_1 = ((1e-9)(V_2) – (1e-12)(V_2))/((1e-7) – (1e-9))

and then if I plug in, say, 50 gallons of culture - or 189.271 liters, it says I would need the following liters of water to sufficiently dilute to a pH of 9:

V_1 = ((1e-9)(189.271) – (1e-12)(189.271))/((1e-7) – (1e-9))

V_1 = 1.90991645455 liters?

um... That doesn't seem right. What am I doing wrong here? Can I not use the equation in this way? Or is there another equation I could use?


Offline hypervalent_iodine

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Re: pH adjustment of Spirulina culture via dilution
« Reply #1 on: September 17, 2019, 12:42:41 AM »
Baking soda is not a strong base, so it is not as simple as C1V1=C2V2. Moreover, you will have other components in there that will alter the pH. The best way would be to add HCl and monitor the pH as you go.

Offline Elliander

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Re: pH adjustment of Spirulina culture via dilution
« Reply #2 on: September 17, 2019, 08:11:37 AM »
You have a good point that baking soda is not a strong base. It has a pH of 9, so it should not have - by itself - managed to bring the pH above 9. There are nutrients in the culture as well, but none that should have been any more alkalizing. So then what could have caused this? I could see how the pH would gradually reduce since the Spirulina could consume the carbon, but what would cause it to increase like this? I've never seen this before, so assumed it a miscalculation on my part. Now I am not so sure.

In any case, that mystery aside, pH is currently at 12 and I need to correct that.

I realize that I can add a strong acid, but I would rather use dilution, and even if I was going to use a strong acid to correct I would still want to know exactly how much to use rather than randomly try. The strain can survive a high pH, but it won't survive a low pH and I figure that through dilution the absolute worst I'd get is a pH of 7.

 C1V1=C2V2 wouldn't seem appropriate here anyway. That's used to calculate unknown quantities where two solutions/mixtures are proportional. That would be useful if I was trying to figure out how much baking soda is in a cup, given how much I put in a tank, but I don't really need to know that.

Isn't there a formula for the amount of dilution with water required to lower the pH by 1? If there isn't I could take varying sample sizes and dilute until I find a good dilution ratio for what I have and hope it scales up properly.

EDIT: Is it possible that carbon dioxide from the air could be causing this via too much aeration? I just read an interesting article that asserts that in nature the carbonic acid is broken down by algae (maybe in this case cyanobacteria?) to raise the pH.

https://principia-scientific.org/carbon-dioxide-makes-alkaline-water-experiment/ 

However, what's the highest such an effect could go?
« Last Edit: September 17, 2019, 08:24:44 AM by Elliander »

Offline Borek

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Re: pH adjustment of Spirulina culture via dilution
« Reply #3 on: September 17, 2019, 08:39:43 AM »
Isn't there a formula for the amount of dilution with water required to lower the pH by 1?

For a simple solution of a strong (preferably monoprotic) acid or base - yes, it is possible to derive such a formula. Won't be trivial to use, and changing pH by 3 units (12 :rarrow: 9) will require diluting the solution 1000 times (doesn't have to be the same number for other starting pH values).

But you don't have a simple solution, with a well defined composition. Your solution contains buffers, and pH of a buffer solution doesn't change during dilution (it is actually more complicated, but will do as a first approximation here).

Just like hypervalent_iodine I don't see other reasonable approach than neutralization with a strong acid.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Babcock_Hall

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Re: pH adjustment of Spirulina culture via dilution
« Reply #4 on: September 17, 2019, 10:39:24 AM »
 I don't know whether the identity of the counter-ion is or is not important in terms of growth of the cells, but you could also use sulfuric acid.  If you are worried that the local pH around a strong acid will be too low before mixing is complete, you could use a somewhat weaker acid such as KHSO4 or acetic acid.  There is also cost to consider.

Offline hollytara

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Re: pH adjustment of Spirulina culture via dilution
« Reply #5 on: September 17, 2019, 10:56:46 AM »
I often use citric acid as a mild acid for acidifying sensitive solutions.  The spirulina might also be able to use it as a food source. 


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