Randy Holmes-Farley
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My Tank Thread
And the answer is...
A.
500 mg/L calcium
19.4 dKH
1286 mg/L magnesium
Here's the rationale:
Totally pure calcium carbonate (CaCO3) has one mole of calcium for each mole of carbonate.
Each mole of carbonate is equivalent to 2 moles of alkalinity (to get to the carbon dioxide endpoint):
CO3-- + 2H+ --> 2H2CO3 ---> H2O + CO2
One mole of calcium weighs 40 grams per mole, so for each 2 mole of alkalinity it has 40 grams of calcium.
Thus the ratio of calcium to calcium to alkalinity in pure calcium carbonate is 40 grams of calcium to 2 moles of alkalinity, which is equivalent to 40 mg of calcium to 2 milliequivalents of alkalinity.
For simplicity, if we think of this as coming from one liter of seawater, then the ratio consumed is 40 mg/L calcium for each 2 meq/L (= 5.6 dKH).
BUT, other ions are incorporated into coral skeletons and coralline algae in a reef tank, especially magnesium. This effect is minor and won't change the answer much, but we should work through it. The amount of magnesium incorporated varies with coral type, ranging from close to 0 to about 4.4% by weight in the calcium carbonate (for coralline algae).
The calcium in calcium carbonate comprises about 40% of the weight of calcium carbonate, so magnesium might range from about 0 to about 11% of the calcium value.
Let's assume the magnesium is about 3% by weight of the calcium carbonate. There must then be less calcium in it, by weight.
Calcium weighs 40 g/mole while magnesium weighs 24.3 g/mole. Since the replacement is 1:1 on an ion (mole) basis, then the calcium declines by more than the 3%. The 3% magnesium in the calcium carbonate is swapping into the crystal in place of 40/24.3 * 3 = 5% (calcium as a percentage of the original total calcium carbonate mass).
So the original 100 grams of this calcium carbonate now must have 35 grams of calcium and 3 grams of magnesium along with the 60 grams of carbonate.
Consequently,our new material has the ratio:
35 mg calcium
3 mg magnesium
60 mg = 2 meq/L = 5.6 dKH of alkalinity
or reduced to 1 meq/L:
17.5 mg calcium
1.5 mg magnesium
30 mg = 1 meq/L = 2.8 dKH of alkalinity
This is the demand ratio that we need to see in the correct answer.
As an aside, when I am quoting the consumption ratio in reef tanks, this result is why I usually say 18-20 ppm calcium for each 1 meq/L (2.8 dKH) of alkalinity, and that magnesium consumption is about 1/10th or less of the calcium consumption.
Back to the original question...
If we check to see the demand ratio in choice A,
500 mg/L calcium
19.4 dKH
1286 mg/L magnesium
we get:
500-420 = 80 mg/L calcium
19.4 - 7 dKH = 12.4 dKH alkalinity
1286-1280 = 6 mg/L magnesium
Adjusting that ratio to 1 meq/L (2.8 dKH) (which means dividing it by 12.4/2.8 = 4.43) we get:
80/4.43 = 18 mg/L calcium
12.4/4.43 = 2.8 dKH of alkalinity (1 meq/L)
6/4.43 = 1.35 mg/L magnesium.
Those values are very close to the theoretical demand ratio we calculated of
17.5 mg calcium
1.5 mg magnesium
30 mg = 1 meq/L = 2.8 dKH of alkalinity
All of the other answer choices are way off of this theoretical ratio in some way.
Choices B and C have far too little alkalinity consumed for the calcium and magnesium consumed. Choice D has too much alkalinity consumed.
Happy Reefing!!
A.
500 mg/L calcium
19.4 dKH
1286 mg/L magnesium
Here's the rationale:
Totally pure calcium carbonate (CaCO3) has one mole of calcium for each mole of carbonate.
Each mole of carbonate is equivalent to 2 moles of alkalinity (to get to the carbon dioxide endpoint):
CO3-- + 2H+ --> 2H2CO3 ---> H2O + CO2
One mole of calcium weighs 40 grams per mole, so for each 2 mole of alkalinity it has 40 grams of calcium.
Thus the ratio of calcium to calcium to alkalinity in pure calcium carbonate is 40 grams of calcium to 2 moles of alkalinity, which is equivalent to 40 mg of calcium to 2 milliequivalents of alkalinity.
For simplicity, if we think of this as coming from one liter of seawater, then the ratio consumed is 40 mg/L calcium for each 2 meq/L (= 5.6 dKH).
BUT, other ions are incorporated into coral skeletons and coralline algae in a reef tank, especially magnesium. This effect is minor and won't change the answer much, but we should work through it. The amount of magnesium incorporated varies with coral type, ranging from close to 0 to about 4.4% by weight in the calcium carbonate (for coralline algae).
The calcium in calcium carbonate comprises about 40% of the weight of calcium carbonate, so magnesium might range from about 0 to about 11% of the calcium value.
Let's assume the magnesium is about 3% by weight of the calcium carbonate. There must then be less calcium in it, by weight.
Calcium weighs 40 g/mole while magnesium weighs 24.3 g/mole. Since the replacement is 1:1 on an ion (mole) basis, then the calcium declines by more than the 3%. The 3% magnesium in the calcium carbonate is swapping into the crystal in place of 40/24.3 * 3 = 5% (calcium as a percentage of the original total calcium carbonate mass).
So the original 100 grams of this calcium carbonate now must have 35 grams of calcium and 3 grams of magnesium along with the 60 grams of carbonate.
Consequently,our new material has the ratio:
35 mg calcium
3 mg magnesium
60 mg = 2 meq/L = 5.6 dKH of alkalinity
or reduced to 1 meq/L:
17.5 mg calcium
1.5 mg magnesium
30 mg = 1 meq/L = 2.8 dKH of alkalinity
This is the demand ratio that we need to see in the correct answer.
As an aside, when I am quoting the consumption ratio in reef tanks, this result is why I usually say 18-20 ppm calcium for each 1 meq/L (2.8 dKH) of alkalinity, and that magnesium consumption is about 1/10th or less of the calcium consumption.
Back to the original question...
If we check to see the demand ratio in choice A,
500 mg/L calcium
19.4 dKH
1286 mg/L magnesium
we get:
500-420 = 80 mg/L calcium
19.4 - 7 dKH = 12.4 dKH alkalinity
1286-1280 = 6 mg/L magnesium
Adjusting that ratio to 1 meq/L (2.8 dKH) (which means dividing it by 12.4/2.8 = 4.43) we get:
80/4.43 = 18 mg/L calcium
12.4/4.43 = 2.8 dKH of alkalinity (1 meq/L)
6/4.43 = 1.35 mg/L magnesium.
Those values are very close to the theoretical demand ratio we calculated of
17.5 mg calcium
1.5 mg magnesium
30 mg = 1 meq/L = 2.8 dKH of alkalinity
All of the other answer choices are way off of this theoretical ratio in some way.
Choices B and C have far too little alkalinity consumed for the calcium and magnesium consumed. Choice D has too much alkalinity consumed.
Happy Reefing!!