Calcium formate - a recipe...

[Danchik]

Since it is difficult to buy high-quality calcium formate in Ukraine, I would like to mix it myself...
So I need help in making it... what reagents, etc.
And I would like to invite @Randy Holmes-Farley to the discussion)

 

[Randy Holmes-Farley Verified Community Expert]

It is hard to buy anywhere.

Calcium hydroxide and formic acid would be the ideal diy. Can you get those?

 

[Danchik]

Yes, Randy, it is possible to get it:
- Calcium hydroxide, with classification (pure)
- Formic acid, with classification (pure). Liquid with 85% concentration.

Unfortunately, the highest purity classification is not available(

 

[Danchik]

Added quality certificates with English translation:

 

[Randy Holmes-Farley Verified Community Expert]

To make a formate solution, here’s what I would do. One can do all sorts of calculations, but simplest would be this:

1. Make a solution of however much formate you want in water.

2. Slowly add calcium hydroxide with mixing (will get hot) until no more dissolves. A bit of excess calcium hydroxide solids is fine. Just our off the clear liquid after settling.

100 g of formic acid (molecular weight 46 g/mole) in 1 L will produce a stock solution with an alk of 2,170 meq/L or 5400 dKH.

 

[Doctorgori]

Not to butt in but is this close to a AFR thing or if not, what’s the main difference? (besides the trace elements)

 

[Randy Holmes-Farley Verified Community Expert]

Not to butt in but is this close to a AFR thing or if not, what’s the main difference? (besides the trace elements)

Yes. The difference is the trace elements. And magnesium. I can’t recall if there is any in AFR.

 

[Danchik]

100 g of formic acid (molecular weight 46 g/mole) in 1 L will produce a stock solution with an alk of 2,170 meq/L or 5400 dKH.

Do I understand correctly... 100g of formic acid to bring to 1 liter of osmosis?
But here I do not understand ... what is the concentration of calcium itself in that solution of 1 liter ...?

For example, I am currently mixing a solution of “calcium chloride anhydrous CaCl2” with a concentration of 23.08 g per 0.1 liter.
How do I convert this to calcium formate so that the calcium concentration is the same?

 

[Randy Holmes-Farley Verified Community Expert]

It will be about half the formate concentration, or about half of 2.17 M, which is 1.09 M = 43,400 mg/L calcium.

 

[Danchik]

It will be about half the formate concentration, or about half of 2.17 M, which is 1.09 M = 43,400 mg/L calcium.

I can't figure out how you calculate it..
Is it possible for you to write it down in detail?

And to make the same concentration with CaCl2, do I need to add 200g of formic acid instead of 100g and dilute it in 1 liter of osmosis, or what?

 

[DanyL]

To make a formate solution, here’s what I would do. One can do all sorts of calculations, but simplest would be this:

1. Make a solution of however much formate you want in water.

2. Slowly add calcium hydroxide with mixing (will get hot) until no more dissolves. A bit of excess calcium hydroxide solids is fine. Just our off the clear liquid after settling.

100 g of formic acid (molecular weight 46 g/mole) in 1 L will produce a stock solution with an alk of 2,170 meq/L or 5400 dKH.

Isn’t Calcium Formate solubility far exceeds the solubility of Calcium Hydroxide?

Does it mean that if one would want to take advantage of it, the resulting Calcium Formate solution would need to be evaporated, or somehow distilled back to powder and mixed again at a higher concentration?

I don’t intend to make or use Calcium Formate anytime soon, but it sparked my curiosity.

 

[Randy Holmes-Farley Verified Community Expert]

Isn’t Calcium Formate solubility far exceeds the solubility of Calcium Hydroxide?

Does it mean that if one would want to take advantage of it, the resulting Calcium Formate solution would need to be evaporated, or somehow distilled back to powder and mixed again at a higher concentration?

I don’t intend to make or use Calcium Formate anytime soon, but it sparked my curiosity.

Adding calcium hydroxide solid to formic acid in water immediately makes calcium formate. I do not suggest making kalkwasser first.

 

[Randy Holmes-Farley Verified Community Expert]

I can't figure out how you calculate it..
Is it possible for you to write it down in detail?

And to make the same concentration with CaCl2, do I need to add 200g of formic acid instead of 100g and dilute it in 1 liter of osmosis, or what?

The calcium hydroxide will dissolve until the formic acid is used up. A tiny, tiny bit more will dissolve as the hydroxide that is too small to think about unless you are a chem nerd. lol

Thus, the reaction is:

Ca(OH)2 + CHO2H (formic acid) —> Ca++ CHCO2- (calcium formate) + H2O

Thus, the moles of formate is exactly twice the moles of formic acid.

Formic acid has a molecular weight of 46 g/mole, so 100 g is 2.17 moles.

Half of that is 1.08 moles.
Calcium has a molecular weight of 40 g/mole, so 1.08 moles x 40 g/mole = 43.2 grams.

43.2 g in 1 L is 43,200 mg/L.

 

[DanyL]

Adding calcium hydroxide solid to formic acid in water immediately makes calcium formate. I do not suggest making kalkwasser first.

That makes perfect sense, thank you.

 

[Danchik]

Thank you @Randy Holmes-Farley for your answer.

But I'm trying to recalculate the data of my replacement solution, but it's not coming out right...
If there was an option to buy calcium formate in ready-made dry form, then I would not have any problems in the calculation

I am diluting a solution (CaCl2) of 57.7 g per 1.0 liter. And I would like to switch to calcium formate, but I want the solution concentration to remain the same.

So the question is how much to dilute:
- Formic acid to 1 liter of pure water (osmosis). I have it listed with a concentration of 85% - should I take this into account...?
- Calcium hydroxide

To ensure that the concentration of the mixing solution remains the same as that of “CaCl2” 57.7 g per 1.0 liter.


P.S. I hope the translator has conveyed my thoughts correctly...

 

[Randy Holmes-Farley Verified Community Expert]

Thank you @Randy Holmes-Farley for your answer.

But I'm trying to recalculate the data of my replacement solution, but it's not coming out right...
If there was an option to buy calcium formate in ready-made dry form, then I would not have any problems in the calculation

I am diluting a solution (CaCl2) of 57.7 g per 1.0 liter. And I would like to switch to calcium formate, but I want the solution concentration to remain the same.

So the question is how much to dilute:
- Formic acid to 1 liter of pure water (osmosis). I have it listed with a concentration of 85% - should I take this into account...?
- Calcium hydroxide

To ensure that the concentration of the mixing solution remains the same as that of “CaCl2” 57.7 g per 1.0 liter.


P.S. I hope the translator has conveyed my thoughts correctly...

Ok, so to clarify, you are not using calcium chloride to make this material, you just want to match the calcium potency of it?

Assuming so, then we need to know what sort of calcium chloride solid material you were using. It is often the dihydrate, for which the solid is only about 77% calcium chloride by weight.

 

[Danchik]

Ok, so to clarify, you are not using calcium chloride to make this material, you just want to match the calcium potency of it?

Assuming so, then we need to know what sort of calcium chloride solid material you were using. It is often the dihydrate, for which the solid is only about 77% calcium chloride by weight.

Yes, you have understood everything correctly. I would like the concentration to match, that is, to be the same
I use ordinary calcium chloride anhydrous (CaCl2)

 

[Randy Holmes-Farley Verified Community Expert]

Yes, you have understood everything correctly. I would like the concentration to match, that is, to be the same
I use ordinary calcium chloride anhydrous (CaCl2)

ok.

57.7 g anhydrous calcium chloride contains 36% calcium by weight, or 20.8 g of calcium.

Put that in 1 l total volume and the calcium concentration is 20,800 mg/L.

The recipe I gave above produces 43,200 mg/L, so we use less formic acid.

If you use 20.8/43.2 = 0.48 times as much (meaning 0.48 x 100 g = 48 grams formic acid), and that is in 1 l total volume, then you have calcium formate matching your calcium chloride solution.

 

[Danchik]

ok.

57.7 g anhydrous calcium chloride contains 36% calcium by weight, or 20.8 g of calcium.

Put that in 1 l total volume and the calcium concentration is 20,800 mg/L.

The recipe I gave above produces 43,200 mg/L, so we use less formic acid.

If you use 20.8/43.2 = 0.48 times as much (meaning 0.48 x 100 g = 48 grams formic acid), and that is in 1 l total volume, then you have calcium formate matching your calcium chloride solution.

Thank you, it's much clearer now

For example, to increase the concentration of calcium formate by a factor of 4, is it enough to multiply the 48 g*4 of formic acid?

Another question about calcium hydroxide, I understand that it is easier to add it to formic acid by eye...
Is there a calculation ... to immediately weigh the calcium hydroxide and add it to the formic acid, where it will dissolve)? Probably it would be better...

 

[Randy Holmes-Farley Verified Community Expert]

Thank you, it's much clearer now

For example, to increase the concentration of calcium formate by a factor of 4, is it enough to multiply the 48 g*4 of formic acid?

Another question about calcium hydroxide, I understand that it is easier to add it to formic acid by eye...
Is there a calculation ... to immediately weigh the calcium hydroxide and add it to the formic acid, where it will dissolve)? Probably it would be better...

Yes on the 4x question.

Yes, you can add it based on weight if you have an idea how much moisture is in each product (which can be hard to know.

The mass ratio of formic acid (2 x molecular weight) to calcium hydroxide (1 x molecular weight) is 2 x 46 to 1 x 74 = 92:74 = 1:0.80