Uncertain Water Report Interpretation

[MaplePaddle]

I’ve been trying to help myself with understanding my water report for a long time. I THINK I’ve finally figured it out, but terribly uncertain. Bru’n Water indicates I have a balanced water report. I’m having to use the Conversion Calculator to extrapolate some values. I would really appreciate a second opinion.
View attachment IMG_0384.jpg

If my interpretation is correct:

Ca....64.2
Mg....38.9
Na....8.0
HCO3....139.5
CO3....96
SO4....59
Cl....5.3

 

[cire]

Mine is this.

Ca....44
Mg....13.6
Na....8.0
HCO3....136.6
CO3....96
SO4....59
Cl....5.3

Using the following.
2.5 X Ca = calcium hardness.
Total hardness = (2.5 X Ca) + (4.1 X Mg)
Alkalinity as CaCO3 X 1.22 = Bicarbonate

 

[ajdelange]

You have total hardness of 160/50 = 3.2 mEq/L and calcium hardness of 105/50 = 2.1 mEq/L. Given that your iron and manganese are tiny magnesium hardness is 3.2 - 2.1 = 1.1 mEq/L. The equivalent weight of calcium is 20 mg/mEq so you have calcium of 42 mg/L and of magnesium is 12.15 mg/mEq so your magensium content is 12.15*1.1= 13.4 mg/L. Pretty far off on both of those.

Your pH is 7.8. At that pH

r1 = 10^(7.8 - 6.38) = 26.3027
r2 = 10^(7.8 - 10.38) = 0.00263027
f0 = 1/( 1 + r1 + r1*r2) = 0.0365338 is the H2CO3 carbon species fraction
f1 = f0*r1 = 0.960938 is the HCO3- carbon species fraction
f2 = f1*r2 = 0.00252753 is the CO3-- carbon species fraction.

Thus for each mmole of carbon species the charge, at pH 7.8 is -f1 - 2*f2 = - 0.965993 mEq

Assuming your alkalinity test was done by titrating to pH 4.5 (ISO standard) the charge on each mmol of carbo would be -.013. mEq (calculated as above but using 4.5 for the pH). It took 115/50 = 2.30 mEq of protons (acid) to get one liter of your water from pH 7.8 to 4.5 during the alkalinity test. As pure water at pH 4.5 contains 1000*10^-4.5 = 0.0316228 measurement it took 2.30 - 0.03 = 2.27 mEq of acid to change the charge on your carbo species from -0.965993 to -.013 per mmol of carbo (H2CO3 + HCO3- + CO3--). This is 0.965993 -.013 = 0.952993 /mmol. Thus each liter of water must contain 2.27/0.952993 = 2.38 mmol of carbo. f1 is the fraction of this that is bicarbonate and is 0.960938*2.38 = 2.29 mmol. The molecular weight of bicarbonate is 61 mg/mmol so bicarbonate content is 61*2.29 = 139.7 mg/L so you nailed that one.

The fraction of the carbo which is carbonate is f2 = 0.00252753 so carbonate is 0.00252753*2.38 = 0.00601552 mmol/L and with its molecular weight of 60 the ion concentration is at 60*0.00252753*2.38 = 0.36 mg/L

The actual numbers are a little (very little) different from these because sulfate is very slightly alkaline and because of the interractions of ions with each other (this is not an 'ideally dilute' solution).

 

[mabrungard]

Seems like a reasonable interpretation. It's encouraging to see that they are monitoring the most important parameters for mashing pH (alkalinity and Ca hardness) at a weekly rate. The concentrations shown by Cire are more correct...excepting for the carbonate value. The carbonate is actually near zero. At the water pH of over 7, the vast majority of the alkalinity exists as bicarbonate.

 

[Silver_Is_Money]

An easy way to accomplish what A.J. has worked out, and get pretty close overall without much math, is to grab a copy of the free download "Kaiser Water Calculator" spreadsheet, and convert your waters average total hardness and alkalinity to units of GH (total hardness) and KH (alkalinity) by dividing the mg/L (or ppm) for each by 17.848. Simply plug the converted GH and KH values into the Kaiser Water Calculator and you will get:

Ca = 44.8
Mg = 11.6

Not far off from A.J.'s detailed calculations.

Edit: Actually it's even more simple than this, as all you really need for the determination of Ca and Mg is to compute and enter into the spreadsheet the GH value (total hardness in units of dKH, or units of German degrees of hardness) for this to work. In this case GH = 160/17.848 = 8.965

 

[ajdelange]

Something is a bit off here. We have calcium hardness of 2.1 mEq/L (105 ppm as CaCO3) which means 1.05 mmol of calcium equivalent to 1.05 mmol of CaO which, as it has molecular weight of 56.0774 grams/mol means 58.8813 mg CaO/L. Dividing that by 10 gives dH = 5.888°.

Converting the hardness to dH by dividing by 17.848 gets the same, or essentially the same, result: 105/17.848 = 5.88301. Working backwards from that number you have 58.8301 mg CaO/L. Dividing that by the molecular weight gives 58.8301/56.0774 = 1.049 and calcium's molecular weight is 40.078 so we should be back to 1.049 *40.078= 42.0418 grams using dH.

So something ain't kosher here.

Beyond that I can't help thinking that calculating 20*calcium_hardness/50 isn't easier than calculating calcium_hardness/17.848 and then inserting that number into a spreadsheet (that apparently is off a bit?). A large part of this is that 20 and 50 I can remember. Besides which dividing by 50 is the same as multiplying by 2 and shifting the decimal point left 2 places. Or do the whole thing by multiplying by 4 and shifting the decimal one place to the left. Remembering 17.848 I'm not so good with.

The alkalinity calculation is certainly trickier but all you need to remember to do that is 6.38 and 10.38. The calculation is very easy to stick into an little Excel (or whatever) spreadsheet of your own. The details are in the Sticky at https://www.homebrewtalk.com/showthread.php?t=473408

Not to mention that it's bad enough that we have North America's calcium carbonate and Martin's bicarbonate as units of alkalinity to deal with but now we want to introduce Germany's calcium oxide. And that Karbonat Hartung (carbonate hardness) is a name for alkalinity drives me nuts too!

 

[ajdelange]

At the water pH of over 7, the vast majority of the alkalinity exists as bicarbonate.

It's at pH < 7 (or really <8) where the majority of the alkalinity is from bicarbonate. In the case of
this water (pH 7.8)

Alkalinity Breakdown mEq/L ppm as CaCO3 %
H+ 0.03 1.48 1.29%
OH- 0.00 0.02 0.02%
Bicarbonate 2.25 112.39 97.92%
Carbonate 0.02 0.77 0.67%
Phosphate 0.00000 0.00 0.00%
Sulfate 0.00238 0.12 0.10%
Silicate 0.00000 0.00 0.00%


Compare to pH 9

Alkalinity Breakdown mEq/L ppm as CaCO3 %
H+ 0.03 1.48 1.29%
OH- 0.01 0.37 0.32%
Bicarbonate 2.04 101.76 88.62%
Carbonate 0.22 11.10 9.67%
Phosphate 0.00000 0.00 0.00%
Sulfate 0.00238 0.12 0.10%
Silicate 0.00000 0.00 0.00%

Now OH- is contributing 0.3% of the total and carbonate 9.67%. Note that at pH 9 this solution is super saturated WRT CaCO3 (but it is at pH 7.8 too - just not so much so).

So at pH < 8 you are generally OK with bicarbonate = 61*alkalinity/50. Above that the fuller calculation is needed (depending on how much of an accuracy freak you are).

 

[MaplePaddle]

Thank you ajdelange, cire, Silver_Is_Money, and mabrungard! Admittedly, I haven’t had the think about chemistry at this level since my early university days. I don’t think I could have gotten there without the guidance. Thank you very much for the help!

I’m happy to see the Mg values are more practical. I also see how my incorrect CO3 would leave and unbalanced report.

Ca...42
Mg...13.4
Na...8.0
HCO3...139.7
CO3....~0
SO4...59
Cl...5.3

I think I can get here again with the next, updated report. Thanks again!