Reef Chemistry Question of the Day #198 Magnesium Consumption

[Randy Holmes-Farley]

Reef Chemistry Question of the Day # 198

In an aquarium where the alkalinity is depleted by 2.8 dKH and the calcium is depleted by 18 ppm each day, what would be the expected average magnesium depletion over a period of ten days?

A. ~0.3 ppm per day
B. ~1.2 ppm per day
C. ~3.9 ppm per day
D. ~6.7 ppm per day

 

[Sabellafella]

If i had to guess on my tanks depletion, it would be inbetween 3.9 and 6.7. But i would expect d

 

[JimWelsh]

I'm going to assume that the depleted alkalinity and calcium are replaced each day by dosing, kalk, CaRX, etc., and that the consumption rate remains constant during those ten days. I know that in your Improved Two-Part article, you assume that magnesium is incorporated at a rate of 2.5% of the skeleton weight, or 6.5% of the calcium portion.

Let's establish that 2.8 dKH = 2.8 / 2.8 = 1 meq/L = 50 PPM CaCO3 = 0.5 moles CO3.

Using the "2.5% of the skeleton" approach, the 0.5 moles of CO3 will weigh 30 mg. Add 18 mg of Ca, and we get 48 mg. That 48 mg is 100% - 2.5% = 97.5% of the total skeleton weight, so the total skeleton weight is going to be 48 / 0.975 = 49.23 mg, of which 1.23 mg is Mg. 1.23 time ten days = 12.3 PPM.

Using the "6.5% of the Ca" approach, the answer is 18 * 6.5% = 1.17 mg times ten days = 11.7 PPM.

Ignoring your assumptions made in another context, and taking the numbers in the question at face value, I see that we have 0.5 moles of CO3, but only 18 / 40 = 0.45 moles of Ca, so the Mg will make up the remaining 0.05 moles. 0.05 moles of Mg weighs 0.05 * 24.3 = 1.215 mg times ten days = 12.15 PPM. This number is 2.47% of the total skeleton weight, and 6.75% of the Ca weight, and, so, is right in line with those other assumptions.

My answer is E: ~12.2 PPM

 

[Randy Holmes-Farley]

Oops, left out two words.

ppm per day

and I had a typo in one answer...

 

[Randy Holmes-Farley]

Sorry for any confusion. I had gone back and forth a few times between giving the answer as daily, or over the ten days total. Hence the confusion in the wording and the answer.

 

[JimWelsh]

Then I'm going with B: ~1.2 PPM per day. And if it weren't for the typo, then I'd have seen the "per day" part in the first place. And I wouldn't have answered so soon, either.

 

[DamianOZ]

My tank currently averages the following consumption
Calcium 23.18ppm / day
Alkalinity 3.32dKH / day
Magnesium 7.74ppm / day

My ration of Alk to Ca consumption isn't 2.8 dKH to 18ppm it is 2.8dKH to 19.54ppm
The expected average magnesium depletion for 2.8 dKH Alkalinity = 6.53ppm
or if calculated at 18 ppm calcium the expected Mg = 6.01ppm

D would be the closest answer.

 

[Zero Nitrates]

B

 

[reef123456]

I would go with D too

 

[DamianOZ]

2.8dKH = 50ppm equivalent CaCO3
The molecular weight of CaCO3 is 100.0869 g/ml and the molecular weight of calcium is 40.078 g/mol, therefore (rounded) Ca ppm as CaCO3 = (2.5*Ca ppm)
The molecular weight of MgCO3 is 84.3139 g/mol and the molecular weight of magnesium is 24.305 g/mol, therefore (rounded) Mg ppm as MgCO3 = (4.1*Mg ppm)

For 2.8dKH Alkalinity (50ppm Alkalinity) we expect 20ppm Ca to be used. If 18ppm Ca is used, Mg is taking the place of 2ppm Ca.
To calculate the magnesium equivalent for 2ppm Ca
2ppm Ca x 2.5 = 5ppm Alk
5ppm Alk / 4.1 = 1.22ppm Mg

Therefore the answer = B



I wander where all my Mg is going

 

[Rybren]

I'd say A.

 

[Randy Holmes-Farley]

And the answer is...B. ~1.2 ppm per day

Folks have shown the math, so I won't bother repeating it.

One reason for this question is for folks to realize that apparent magnesium depletion rates of more than a couple of ppm per day are likely coming from other causes, such as magnesium testing issues, or water changes.

Happy Reefing!