Sure.
If you or mother nature dissolves 100 mg of calcium carbonate (which is 1 mmol because the molecular weight of calcium carbonate is 100) in 1 L of water using carbon dioxide as the acid the following reaction takes place: CaCO3 + H2O + CO3 --> Ca++ + 2HCO3-
Thus the 1 mmol]L of calcium carbonate give rise to 2 mmol/L of bicarbonate a 1 mmol/L of calcium ion. But there are 2 mmol/L of negative charge from the bicarbonate ions and 2 mmol/L positive charge from the calcium ions. Moles of charge are called equivalents so 100 mg/L gives 2 mEq/L each of positive and negative charge.
Alkalinity is measured by supplying hydrogen ions (from acid) to the sample until all the bicarbonate is gone:
2H+ + 2HCO3- --> 2CO2.
Thus it takes 2 mEq/L hydrogen ions to "neutralize" the 2 mEq/L bicarbonate that came from the 100 mg/L calcium carbonate with which we started. We could say that the alkalinity is 2 mEq/L and in fact that is how it is commonly stated in Europe. In the US the idea is to specify the alkalinity in terms of how much calcium carbonate it would take to produce that alkalinity were it dissolved in the natural (CO2) way. Thus 2 mEq/L ~ 100 mg/L and so, in the US, we take the alkalinity in mEq/L and mulitply by 50. Then we say the resulting number is the "alkalinity as calcium carbonate".
Now the calcium ion from 100 mg/L also contributes 2 mEq/L positive charge so we have the same correspondence. Ca++ at 2 mEq/L ~ 100 mg/L so we multiply calcium ion equivalence by 50 as well. Since calcium and magnesium ions both react with bicarbonate in the same way we also multiply magnesium equivalence by 50 and call the result "magnesium hardness as CaCO3".
So if you see Total Hardness (CaCO3) = 7.3 mg/L you know that there are 7.3/50 mEq/L of positive charge attached to calcium and magnesium ions but we can't tell how much is attached to each unless we are given the hardness of one or the other. If we were told that the magnesium hardness was 2.3 mg/L as CaCO3 we could subtract that from 7.3 and determine that the calcium hardness was 5.0 mg/L as CaCO3. This is equivalent to 5/50 = 0.1 mEq/L and, as calcium has an equivalent weight of 20 mg/mEq, the calcium ion concentration is 0.1*20 = 2 mg/L "as the ion".
We said that alkalinity is determined by adding acid to the water sample until all the bicarbonate is gone. That isn't quite true. It is done until the pH reaches an "end point" pH which varies depending on analyst but which is, in brewing, generally 4.3. At this pH nearly, but not quite all, the HCO3- has been converted to CO2 and left the sample. The alkalinity is formally defined as the number of mL of 0.1 N acid which must be added to a 100 mL sample of the water to bring it's pH to 4.3. Sometimes they stop first at pH 8.3 and note the number of mL required to get to 8.3. Obviously, this is 0 if the starting pH of the sample is less than 8.3. As 8.3 is the pH at which phenolpthalein changes color this first number is often referred to as the "P alkalinity". For sample pH < 8.3 P alkalinity is 0.
there are people who understand the details and other's who can distill the details down to something I use in practice. 
