Vinegar to kalkwasser ratio

Adam Obenauf · Sep 28, 2016

So I see that the typically recommended amount of vinegar/kalk to use in topoff water is 3 tsp of kalk and 45 ml of vinegar per gallon. I have some questions regarding that.

1. Does that 3 tsp of kalk really fully dissolve with only 45ml of vinegar? It takes roughly 250ml of vinegar to fully neutralize a tsp of kalk to calcium acetate so how is that possible? How was this calculated?

2. How much vinegar would I need to dissolve 4 tsp of kalk per gallon? How would I calculate the vinegar requirements for other levels of kalk dissolution?

I ask because 45ml per gallon of topoff is way less than my tank normally consumes. My tank uses 80ml of vinegar a day and 0.4 gallons of topoff water a day. So I would need 200ml of vinegar per gallon of topoff water to meet my carbon dosing needs just from topoff water. At the same time I am only able to meet about half of my Ca/Alk from fully saturated Kalk. So to use only kalk for dosing I would need at least 4 tsp per gallon. If dissolving more is really as easy as adding more vinegar than I could kill two birds with one stone with way. It's strange that no one ever goes higher considering most tanks need more of both than can be provided with the standard 3 tsp 45 ml formula.

Randy Holmes-Farley · Sep 29, 2016

You do not need to neutralize all of it. Without any neutralization you still get about 2 tsp per gallon to dissolve, so you are looking to only dissolve that extra teaspoon. Full neutralization is NOT desirable as it will rapidly grow bacteria if the pH is not kept very high (excess calcium hydroxide) or very low (excess vinegar).

Craig Bingman first suggested it as far as I know, and does some calculations here, which are not much different than yours:

http://web.archive.org/web/20030418...twork.com/fish2/aqfm/1999/oct/bio/default.asp

He says 45 ml per gallon (selected for other reasons relating to CO2) will boost the amount dissolved by 36%, so crudely, 2 tsp/gal --> 2.72 tsp/gal, which folks jsut round to 3 tsp per gallon.

IMO, I don't prefer to mix them for a variety of reasons, not the least of which it is a lot of organic carbon dosing that you may or may not actually need at exactly that level.

Adam Obenauf · Sep 29, 2016

Thanks for the link. Unfortunately he doesn't show any of his calculations so I have no idea how he reached those numbers.

I have had no adverse reaction to dosing 80ml of vinegar a day to my 50 gallon system with 1600ml of evaporation per day (around 4.5x his recommended dosage). I've seen increased skimmate production from doing it and a lot of other reefers I know dose even more. Should I back off anyways?

I also don't understand his rational for that limit when he says:

As you can see, complete oxidation of about 12 milliliters of vinegar per liter of limewater would provide enough carbon dioxide to balance the inorganic carbon removed from the aquarium when the calcium and alkalinity in a liter of saturated limewater is converted into calcium carbonate.

But what about CO2 being driven off from surface contact?

Randy Holmes-Farley · Sep 29, 2016

The concern is not the vinegar per se (I dosed more than 1 mL per gallon), but how low nitrate might get and whether corals may begin to starve.

His initial calculation is based on an overall pH neutral effect on the tank, taking the calcium hydroxide and making it into calcium carbonate with the CO2 from the vinegar metabolism. That equivalence is not important, IMO.

Adam Obenauf · Sep 29, 2016

So to go back to my original question, is there a way for me to calculate the vinegar requirements to dissolve a given amount of kalk per gallon of water?

Randy Holmes-Farley · Sep 29, 2016

Yes, that can be done. It will be close to what we've already discussed.

The solubility of calcium hydroxide is determined by the multiplication product

[Ca++]*[OH-]*[OH-] = 5.5×10–6 at 25 deg C.

In the absence of vinegar, the moles of calcium hydroxide dissolved per liter is, say, Y. So [Ca} = Y and [OH-] = 2Y

y*2y*2y = 5.5×10–6

solving for y we get

4Y^3 = 5.5×10–6

y = 0.018 M

Since mw of calcium hydroxide is 74 g/mole

solubility is 0.018 x 74 = 1.3 g/l.

If you add vinegar, you are adding H+, and that H+ is combining with the OH- to form water. So the [OH-] is reduced

If we think of the moles of vinegar added as X moles/L, then the new solubility is:

(Y)*(2Y-X)*(2Y-X) = 5.5 x 10-6

So if you know how much vinegar you add in moles/L (X), you can solve for the new solubility of calcium hydroxide in moles/l (Y).

Adam Obenauf · Sep 30, 2016

Thanks so much.

Randy Holmes-Farley · Oct 1, 2016

You're welcome.
Happy Reefing!

Atu · Apr 25, 2019

Randy Holmes-Farley said:
Yes, that can be done. It will be close to what we've already discussed.

The solubility of calcium hydroxide is determined by the multiplication product

[Ca++]*[OH-]*[OH-] = 5.5×10–6 at 25 deg C.

In the absence of vinegar, the moles of calcium hydroxide dissolved per liter is, say, Y. So [Ca} = Y and [OH-] = 2Y

y*2y*2y = 5.5×10–6

solving for y we get

4Y^3 = 5.5×10–6

y = 0.018 M

Since mw of calcium hydroxide is 74 g/mole

solubility is 0.018 x 74 = 1.3 g/l.

If you add vinegar, you are adding H+, and that H+ is combining with the OH- to form water. So the [OH-] is reduced

If we think of the moles of vinegar added as X moles/L, then the new solubility is:

(Y)*(2Y-X)*(2Y-X) = 5.5 x 10-6

So if you know how much vinegar you add in moles/L (X), you can solve for the new solubility of calcium hydroxide in moles/l (Y).

I've finally found this after having a rough day with the calculations yesterday and today.
Randy, I've a couple of problems with this. First of all

"4Y^3 = 5.5×10–6

y = 0.018 M"

I think Y = 0.011, am I doing something wrong?

Secondly

Of y = 0.011 M then 0.011 x 74 = 0.8 g/l

And now my problem, Wikipedia list the Kps for calcium hydroxide as you've written it, but it says that it's solubility is 1.73 g/L at 20º C and 1.6 g/L at 30º C. I'm guessing the Kps is given at 25º C.

Why this do not match my calculated value 0f 0.8 g/l? Its driving me nuts! I'm supposed to know this...

Atu · Apr 28, 2019

@Randy Holmes-Farley ? I hope it does not bother you that I tag you.

Randy Holmes-Farley · Apr 28, 2019

It doesn’t bother me since I don’t use them [emoji23]let me check the numbers.

Randy Holmes-Farley · Apr 28, 2019

0.011 is correct. My 0.018 is presumably a typo.

The known saturation of calcium hydroxide at 25 deg C is about 0.0102 M. One complication to a simplistic calculation is that there is a significant amount of soluble CaOH+ in solution, allowing more solid to dissolve than the simple Ksp predicts.

Atu · Apr 28, 2019

Randy Holmes-Farley said:
0.011 is correct. My 0.018 is presumably a typo.

The known saturation of calcium hydroxide at 25 deg C is about 0.0102 M. One complication to a simplistic calculation is that there is a significant amount of soluble CaOH+ in solution, allowing more solid to dissolve than the simple Ksp predicts.

At least I’m happy I considered that to be one of the options.
Do you know how could I calculate the original question? How much more calcium hydroxide could I dissolve when adding vinegar?

Thanks for your help

Randy Holmes-Farley · Apr 28, 2019

Atu said:
At least I’m happy I considered that to be one of the options.
Do you know how could I calculate the original question? How much more calcium hydroxide could I dissolve when adding vinegar?

Thanks for your help

A perfect calculation is not trivial when there are multiple forms as a function of pH. As an estimate, I'd just assume that for each mole of acetic acid added, you can dissolve 1/2 more moles of calcium hydroxide. That ignores that calcium is rising and the real value is less than 1/2 moles more calcium hydroxide per mole of acid added.

Atu · Apr 28, 2019

Ok, thank you.
I think im gonna try experimentally for a range of values that I might use. I'm hopping that in a small enough interval this might behave linearly and be able to build a curve out of it. If not possible, at least I'm gonna get a value useful for my case.
If I do this I'm gonna update this thread for future reference.
Thanks again Randy!