I'd like to know what's the difference between, for example,
Sodium (Na) 22.73
&
Sodium (CaCO3) 49.56?
The molecular weight of calcium carbonate is 100. If you put one milligram of calcium carbonate in a flask and add 1 L of deionized water and then bubble CO2 through the water with stirring eventually all the calcium carbonate will dissolved and eventually the pH will reach 8.3. Note that this is the mechanism by which limestone is dissolved naturally i.e. by carbonic acid in subterranean water. As the molecular weight of CaCO3 is 100 it is clear that 100/100 = 1 mmol (millimole) of calcium carbonate has dissolved. As the reaction is
CaCO3 + H2CO3 --> Ca++ + 2HCO3-
it is evident that there is 1 mmol of calcium in this liter of water and 2 mmol of bicarbonate ion. A millimole is 6.023E20 atoms or molecules or ions (or rutabagas or bowling balls...) A milliequivalent is a millimole of charge. Thus there are 2 mEq of calcium charge (we say 2 mEq of calcium understanding that it is the charge we mean) and 2 mEq of bicarbonate ion.
If we are tasked with finding the alkalinity of this water we add a strong acid (e.g. hydrochloric acid) to it until almost all the HCO3- has converted to carbonic acid (and thence to water and CO2). The reaction is
HCl + H2O + HCO3- ---> H3O+ + Cl- + HCO3- --> H2CO3 + Cl- --> CO2 + H2O + Cl-
Thus for each bicarbonate ion 1 hydrogen ion is consumed and as the water contains 2 mmol of HCO3-, 2 mEq of H+ (from 2 mmol of HCl) will have been used). Thus the alkalinity of the water will be expressed as 2 mEq/L. If we now measure the amount of calcium in terms of a molecule which can capture it (EDTA) we would find that, as there are 2 mEq of calcium charges, we would need 2 mEq of capture sites in the EDTA. We say that the 'calcium hardness' is 2 mEq.
So, if 100 mg of CaCO3 is dissolved per liter of water in the natural way (with carbonic acid) its alkalinity is 2 mEq/L and its hardness is 2 mEq/L. In the rest of the world that's what the analysis would show* but in the USA it occurred to some genius that it might be more informative to say that the hardness and alkalinity were both 100 mg/L 'as calcium carbonate' since 100 mg of calcium carbonate leads to 2 mEq/L if, and only if, the acid was carbonic but that's what it usually is in ground water. Thus the 'ppm as CaCO3' is a way of expressing equivalence in which 1 mEq/L is equated to 50 ppm as CaCO3. This has confused more people than Carter's got pills but it is still, though perhaps waning somewhat recently, very much a part of the way water analyses are reported in this country as your report shows.
Now the use of 'as CaCO3' units are usually limited to expressing the equivalent amounts of acid required to reach a standard pH (the alkalinity test) and the equivalence of the amounts of EDTA required to chelate the calcium and magnesium (hardness tests) but there is no reason we cannot specify the concentration of any ion using this terminology. For example your 49.56 Na as CaCO3. We determined above that 1 mEq/L was the same as 50 ppm as CaCO3. Thus 49.56 Na as CaCO3 is 49.56/50 = 0.9912 mEq/L also equal to 0.9912 mml/L as sodium ion has a single charge. Your sodium concentration is then approximately 0.9912*23 mg/L as the molecular weight of sodium is about 23 mg/mmol.
If you followed all the above then enough said. If you didn't (and experience tells me you probably didn't) then you can go back and study it until you do but not everyone concerned about his water needs to be an expert and so you can simply remember that wherever you see 'as CaCO3' or (CaCO3) you should just divide by 50 and multiply by the equivalent weight (the molecular weight divided by the charge) to get the actual concentration in mg/L.
It can get sillier. Bru'n water, a popular and otherwise excellent spreadsheet, sometimes expresses alkalinity not 'as bicarbonate' but just 'bicarbonate' implying in some cases that water containing no bicarbonate what so ever contains some and in other cases that water that actually contains a small positive amount of bicarbonate has a large negative bicarbonate ion content. Once you understand what is going on simply divide the bicarbonate number by 61 (the equivalent weight of bicarbonate) to get the alkalinity in mEq/L and use that or multiply by 50 to get 'ppm as CaCO2' which, while equally silly is at least understood.
At least 6 components are listed in this manner.
Bicarbonate (CaCO3) 344.05
Total Alkalinity (CaCO3) 344.05
Taking our rule to the 344.05 total alkalinity we see that it is 344.05/50 = 6.881 mEq/L. The bicarbonate content is numerically the same and so the bicarbonate ion concentration is 61*6.881 = 419.74 mg/L because the equivalent weight (ans molecular weight) of singly charged bicarbonate is 61.
Sulfate (CaCO3) 46.22
Sulfate (SO4) 44.35
Chloride (CaCO3) 53.25
Chloride (Cl) 37.82
As above divide the CaCO3 numbers by 50 to get mEq/L
Sulfate 46.22/50 = .9244
Chloride 53.25/50 = 1.065
and then multiply by the respective equivalent weights of 48 and 35.15 (approximnately) to get mg/L
*I glossed over the fact that in the rest of the world there were simillarly silly units such as German Degrees (based on the amount of quick lime that lead to the observed alkalinity and hardness), French degrees and others. Most of these have disappeared to be replaced bu mEq/L or mVal/L which is the same thing.
