How to modify prepared Seawater chemistry so that Mg/Ca ratio = 1?

Question

I prepared 1 liter of artificial sea water by mixing 1 liter of pure water and 35 grams of "Instant Ocean" salt. According to the recipe, salinity = 35000 ppm, Magnesium (Mg) = 1320 ppm and Calcium (Ca) = 400 ppm. My objective is to modify this sea water so that the Mg/Ca ratio equals 1. Therefore, my first step was to figure out what is the Mg/Ca ratio of the current solution, then the second step is to modify the ratio so it equals 1. My approach was: first part: Mg = 1320 ppm or 1320 mg/Ca Ca = 400 ppm or 400 mg/L Molarity of Mg = 1320/(1000 * 24.305) = 0.054 moles/L Molarity of Ca = 400/(1000* 40.078) = 0.0099 moles/ L Mg/Ca = 0.054/0.0099 = 5.7 Second part: for Mg/Ca = 1, then 0.054/Ca = 1 Ca = 0.054 0.054 = ppm *0.001/40.078 ppm = 2164 or mg/L So i have to add 2164 - 400 = 1764 mg to 1 Liter of water of the solution to get Mg/Ca = 1 My questions are: 1. How i the rationale of my approach(i.e. is this the correct approach? 2. Is it correct to assume that ppm = mg/L, considering that we are dealing with Saline water not Pure so 1 Kg doesnt equal 1 Liter. what is the alternative way?

Answer 1

Do you want the Mg/Ca ratio to be 1 mole/1 mole, or 1 ppm to 1ppm?  There's a big difference, I'm sure you know.  Apparently, from you calculations, you want them to be equal on a mole basis.  Not sure of your 2nd part calculations, but I would do it this way.

0.054 moles/L Mg and 0.0099 moles/L Ca.

To get Ca to 0.054, you need 0.054 - 0.0099 = 0.0441 moles Ca

0.0441 moles Ca x 40.078 g Ca/mole = 1.767 grams of Ca (which is pretty much what you calculated).

So, to answer your questions:

(1) your rationale seems sound

(2) yes, it is correct to assume that ppm = mg/L. The difference in mass and volume is going to be so negligible that the degree of accuracy is probably beyond the limits of the equipment you use to measure such quantities.  Besides, ppm is generally meant to be mass/volume and not mass/mass when it comes to solutions.