# How do I translate my water supply's “Total Hardness” to Mg and Ca or “Effective Hardness”?

My water supply comes through San Diego's Miramar treatment plant, and here's the data sheet. Here's the relevant part:

``````                    Units   Avg     Range
Total Hardness      ppm     214     142 - 243
Total Hardness      gr/Gal  12.2    8.3 - 14.2
``````

And an excerpt from the footnotes:

Hardness is the sum of polyvalent cations present in the water, which is essentially the sum of magnesium and calcium. These cations are usually naturally occurring.

Okay, now, I'm trying to use this scale under "Determining the Beer Styles That Best Suit Your Water" in Ch. 15.3 of John Palmer's How to Brew: What is my Effective Hardness? Is it the same as Total Hardness? (In which case, 214ppm seems very high on this scale—Is it possible my water is that hard?) Or, since the footnotes say Total Hardness is a sum, is my best bet to split 214ppm by some rough, sane-looking proportions like 75ppm Mg and 150ppm Ca, thus crossing over on the scale at about 145ppm Effective Hardness?

By the way, my CaCO3 level is 98.2ppm. Not sure this is relevant to hardness though—I imagine it's the non-alkaline Calcium that hardness is measuring.

I do know a store near me where I can get reverse osmosis water for 35 cents a gallon which is basically nothing. I'm more just curious about tap water chemistry, especially for amateur brewing.

Thanks!

In short, you can't.

If you dig into Palmer's spreadsheet version, you see that effective hardness is equal to:

[ppm Ca / 1.4] + [ppm Mg / 1.7]*

* See cell K15 on the Mash sheet

Not getting too far into detail, this is a simplification of the Kolbach formula that Palmer refers to on the page you linked:

Residual Alkalinity = mEq/L Alkalinity - [(mEq/L Ca)/3.5 + (mEq/L Mg)/7],

used to determine how much the alkalinity is offset by the hardness (more specifically how much is left over, if any). mEq/L means milliequivalents per liter, which is just a way of expressing concentrations of different ions by first converting them to a common unit, similar to using common denominators with fractions.

Here's the catch:

The total hardness of 214 ppm does not directly represent Ca or Mg concentrations, but rather the combination of their equivalent amounts of CaCO3, just as in the equation above. Put another way, it does not mean that the amount of Ca plus the amount of Mg in your water equals 214 ppm (e.g., 150 ppm Ca + 64 ppm Mg), but rather that whatever amount of Ca and Mg there is in your water, together, has an equivalent weight of 214 ppm expressed as CaCO3, telling you nothing about the concentration of either. A bit confusing? For sure.

The reason you can't apply the total hardness as your effective hardness is because Ca and Mg offset the alkalinity to different degrees, as you can see in the equations above. There's just no way to make this formula work without knowing the specific proportions of Ca and Mg. Indeed, this is a very common shortcoming of many municipal water reports, which are definitely not tailored to the brewer's needs.

- edit -

If anyone's interested, here's how the equation in Palmer's spreadsheet is derived:

First, the calcium and magnesium are converted to mEq/L by dividing ppm of each by their respective equivalent weights:

``````1 ppm Calcium / 20
1 ppm Magnesium / 12.2
``````

These numbers are then multiplied by 50, the equivalent weight of CaCO3 so they can be directly compared to alkalinity:

``````Calcium -> [50 / 20]
Magnesium -> [50 / 12.2]
``````

To finish it off, Kolbach's formula is applied to both:

``````Calcium -> [50 / 20] / 3.5 = 0.714
Magnesium -> [50 / 12.2] / 7 = 0.5855
``````

Check that against Palmer's equations:

``````Calcium -> 1 ppm / 1.4 = 0.714, checks out
Magnesium -> 1 ppm / 1.7 = 0.588, close enough!
``````