Re adding a shot of concentrate with suspended calcim carbonate"
When I add one shot of it to 3 liters in a gallon jug it dissolves completely. I never get sediment on the bottom of the jug. When adding it directly to the reservoir I do that before putting the machine to bed, so it has overnight for the suspended particles to dissolve fully. Keep in mind this is a fairly low concentration - about 20 mg/L calcium as CaCO3.millmountain wrote:Agitation will distribute the particles in suspension, but if the calcium carbonate remains undissolved, surely it isn't helpful?
It's more complex than that when dissolving in water where it reacts with dissolved CO2 to become calcium bicarbonate, Ca²⁺ + 2HCO₃⁻. At 25C and with water at equilibrium with CO2 in the air, just over 50 mg/L of CaCO3 will fully dissolve in water.millmountain wrote:Wikipedia says the solubility of CaCO3 in water is 13 mg/l at 25°C, and of MgCO3 is 139 mg/l at 25°C and 60 mg/l at 100°C.
I just use just Bulk Supplements MgCO3, which it turns out is actually dypingite - 4MgCO₃ * Mg(OH)₂* 5H₂O. Once dissolved you get magnesium bicarbonate same as you would if you were to use simple magnesium carbonate. (The magnesium hydroxide ends up as magnesium bicarbonate -- Mg(OH)₂ + 2H₂CO₃ → Mg(HCO₃)₂ + 2H₂O .) Molar mass for dypingite is 486 g/mol which is 97g per mole of Mg. For MgCO3 it would be 83g per mole of Mg, so you do need to allow for that for accurate recipe calculations.millmountain wrote:I'd rather find a decent source of MgCO3, but after some web-search grinding, Bulk Supplements is the only thing that came up as available. Shipping is pretty steep for me. Maybe none of it matters if the Mg concentrate will just saturate?
I looked at those references and now understand the reason for that odd calculation:
Those methods assume you have mass measurements for the major ions, Ca, Mg, SO4, Cl, K, Na, but not for the HCO3 ion. They use the factor 0.6 * alkalinity as a way to estimate the mass of the HCO3 ion. When you are making water with just NaHCO3 or KHCO3 you know the mass of the HCO3 ion and don't need to estimate it from the alkalinity.millmountain wrote:I wonder if this isn't the reason for reformulating carbonates and bicarbonates as CaCO3 equivalents with a 0.6 factor.
[To be continued - I clicked submit when I was only halfway thru my response to Mike's last post]