Hi Matthew Gilchrist, I'm wondering if you just theoretically want to know how to calculate this or if you experimentally want to make such a solution. The problem I see is that the solubility of CaCO3 is only ca. 14 mg / L at ambient temperature. 😳
This looks an awful lot like a homework problem especially since, as Frank T. Edelmann pointed out, you can't have that much CaCO3 dissolved in water; at least assuming ambient conditions (and no acids either!)... Since it has been a few months since you asked the question I feel okay answering it though. I hope this helps.
Lets go ahead and get some of the math out of the way. You know that you (in theory) have 1 g of Ca dissolved in 1 L of water. So, in 0.5 L (I assume by Ml you mean mL) of water, you would have 0.5 g of Ca. Very simple math. You just split things in two there is all...
At this point, we know how much Ca we have (0.5 g in 500 mL of water). What we need to figure out at this point is how much CaCO3 it would take to get that amount of Ca. Personally, I find it helpful at this point (and at most points) to think in terms of one of my favorite things: pizza. Imagine that CaCO3 is a pizza and that Ca is just a slice of the pizza. What the molar masses of Ca and CaCO3 gives us is a tool or a way to relate the size of the slice to the size of the pizza (and vice versa). What do I mean by a tool?: We know that because there is a constant and proportional relationship between the size of the pizza (CaCO3 = 100.1) and the size of the slice (Ca = 40.1) we can use the weight of the slice is 0.5 g to determine the weight of the pizza. This is what atomic masses allows us to do. So, to determine how much the pizza must weigh given the weight of the slice, we do this arithmetic: pizza atomic mass/slice atomic mass * mass of the slice = mass of the pizza. Translating from food talk, this is: 100.1/40.1 * 0.5 g= ~1.25 g CaCO3.
As demonstrated by Frank T. Edelmann the solubility of Ca in ambient water is 14 mg/L and based on the simplified explanation by Eron R Raines you may need 1.25 g of CaCO3 to reach the 1 g/L concentration of Ca. In this situation, you should be aware of the fact that you may not have a 1 g/L Ca solution by adding 1.25 g of pure carbonate calcium to the water, unless you could dissolve all the CaCO3 by the means of acidified compositions (like injecting CO2 gas into the water). Otherwise, you will have a 14 mg/L Ca solution, even after adding 1.25 g of pure calcium carbonate to 500 mL of water.
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