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?