Another calcium question using Randy's 2 part recipe.

[Sand.man]

Hi all,

I've done a lot of research, but I can't get the answer to my situation.

I bought bulk Calcium Chloride from a fellow reefer in my area, and I forgot to ask if it is anhydrous, or dihydrate... in fact, I totally forgot that two different forms even existed until I mixed the solution.

When I was mixing the calcium component for Randy's recipe, I used recipe 1, and treated the calcium solution as dihydrate. :crazy1: Therefore, I mixed 500 grams of the calcium solution to 1 gallon of water in a plastic milk jug.

As I was mixing it, the solution turned pretty hot, almost too hot to hold the milk jug anymore, and that's when I remembered that there is actually two different types of calcium chloride, and realized my mistake.

Now my questions are, is it safe to assume since the water turned hot while I was mixing the calcium, that it is indeed anhydrous, and not dihydrate? I did find Randy mentioning this before, but I would just like to confirm anyway.

Since the solution turned so hot, do you think it was safe to mix it inside the plastic milk jug? I assume it should have been fine, but maybe it got too hot for the plastic and it might have leeched something?

Finally, since I used the ratio for dihydrate calcium instead of anhydrous, I'm guessing that my solution is more concentrated. Is it safe to assume that since the recipe calls for a 20% deduction in volume when substituting dihydrate for anhydrous, that this solution would now be 20% stronger than the intended solution in Randy's recipe? Memories of high school chemistry remind me that it might not be that simple... would it?

Any help would be much appreciated.

Thanks, and enjoy your day.

 

[disc1]

If it got super super hot then it probably was the anhydrous. No worries about the jug so long as it didn't melt.

Finally, since I used the ratio for dihydrate calcium instead of anhydrous, I'm guessing that my solution is more concentrated. Is it safe to assume that since the recipe calls for a 20% deduction in volume when substituting dihydrate for anhydrous, that this solution would now be 20% stronger than the intended solution in Randy's recipe? Memories of high school chemistry remind me that it might not be that simple... would it?

It's not a chemistry problem. It's a math problem. Unfortunately it's way too early in the morning for me to be doing math.

Basically, 80 is a 20% reduction from 100, but 100 is a 25% increase from 80. Something like that. Math... I think you're over by 25%.

 

[dkeller_nc]

Calcium chloride, anhydrous has a molecular weight of 111 grams/mole; calcium chloride dihydrate has a molecular weight of 147 grams/mole. Calcium has a molecular weight of 40 grams/mole.

So if you used 500 grams of calcium chloride dihydrate to make a solution, you've used 40/147 * 500/147 = 0.93 moles of calcium.

If you used 500 grams of calcium chloride anhydrous to make a solution, you've used 40/111 * 500/111 = 1.62 moles of calcium.

So presuming that the two solutions above had the same volume, your solution would be 1.62/0.93 = 1.74 times more concentrated in the calcium ion than intended.

Keep in mind, however, that it's highly unlikely you'd have a solid that is 100% anhydrous calcium chloride as it's extremely hygroscopic (water loving). Over time, even in a sealed plastic bottle, it will gain moisture, so it would be somewhere in between completely anhydrous calcium chloride and calcium chloride dihydrate.

Given that, I'd simply slow your dosing of the calcium component down by 20-30%, and simply measure your tank's water over a week to determine if you need to make a further dosage adjustment; a slight overdose of the calcium component of a 2-part solution is unlikely to cause problems if it doesn't go on too long.

 

[disc1]

Calcium chloride, anhydrous has a molecular weight of 111 grams/mole; calcium chloride dihydrate has a molecular weight of 147 grams/mole. Calcium has a molecular weight of 40 grams/mole.

So if you used 500 grams of calcium chloride dihydrate to make a solution, you've used 40/147 * 500/147 = 0.93 moles of calcium.

If you used 500 grams of calcium chloride anhydrous to make a solution, you've used 40/111 * 500/111 = 1.62 moles of calcium.

So presuming that the two solutions above had the same volume, your solution would be 1.62/0.93 = 1.74 times more concentrated in the calcium ion than intended.

Keep in mind, however, that it's highly unlikely you'd have a solid that is 100% anhydrous calcium chloride as it's extremely hygroscopic (water loving). Over time, even in a sealed plastic bottle, it will gain moisture, so it would be somewhere in between completely anhydrous calcium chloride and calcium chloride dihydrate.

Given that, I'd simply slow your dosing of the calcium component down by 20-30%, and simply measure your tank's water over a week to determine if you need to make a further dosage adjustment; a slight overdose of the calcium component of a 2-part solution is unlikely to cause problems if it doesn't go on too long.

Your math is way off. If you do the 40/147 thing, then you are just converting weight. That's how you would get from a weight of CaCl2 to a weight of calcium from it.

If you want moles then you have to think the whole molecule, not just the calcium part. Each mole of calcium chloride has one mole of calcium. So you need to be using:

500 / 147 == 3.4 moles of calcium chloride == 3.4 moles of calcium

500 / 111 == 4.5 moles of calcium chloride == 4.5 moles of calcium


So you've got 4.5 / 3.4 == 132% of the calcium you should.

So your dosage should be

100 / 132 == 75% of what it would have been.

So you need to reduce the dosage by 25%. For every 4 ml you were dosing, dose 3 instead.

 

[Sand.man]

Well, math and chem go together like pizza and good times, unfortunately neither of them (math or chem) were my strong suit.

I much appreciate the two of you taking time out to help me with the calculations, the general census seems to be that it is around 25-30% stronger.

Although I agree, overdosing the calcium component isn't likely to cause adverse affects compared to overdosing something like alk, it's still good to know what concentration my solution is at.

Thank you, I really appreciate it. Enjoy your day.

 

[dkeller_nc]

Your math is way off.

Yeah, you're right - same problem; doing math WAY too early with not nearly enough coffee! Of course, yesterday "too early" was 2 p.m. (really late night Friday....)