Carbonate Hardness = Carbonate Alkalinity based on Source pH?

[Big Monk]

After building my water for a long time, I am entertaining using my municipal water. I have reviewed the last 10 years of WQRs and the values are consistent based on the source, which in my case is 90% from Lake Ontario and 10% from Lake Otisco. I had the water tested a while back and it corresponded almost exactly with the WQR averages for a ten year period. So it's stable:

Ca = 33.9
Mg = 9.2
SO4 = 22.9
Cl = 27.7
Na = 8.9

My question relates to the phenomenon by which Carbonate Hardness = Carbonate Alkalinity for Water pH < 8.5.

Since the CO3 component of the alkalinity will generally be a small percentage of the HCO3 concentration when Water pH is < 8.5, is it fair to say that Carbonate Hardness = Carbonate Alkalinity in that scenario?

If so, it would seem that, for my water at least, HCO3 can be calculated as follows:

Hardness as CaCO3 = (2.497*33.9)+(4.164*9.2) = ~ 123

My local WQR quotes my Water pH as being between 7.1-8.5 (which i'll verify by measuring of course) so let's assume the median of 7.8. Assuming an Alkalinity value equal to the Hardness value:

CO3 = ((123*10^(7.8-10.33))/(1+2*10^(7.8-10.33)))*(60/50) = ~ 0.432

HCO3 = (123/(1+2*10^(7.8-10.33)))*(61/50) = ~ 149.13

CO3 < 0.5% of my water, so Hardness as CaCO3 = ~ Alkalinity as CaCO3, right? Did I make this clear as mud?

In general, my water seems well suited to brewing, with only a modest amount of acid required to neutralize the 149 ppm of HCO3.

 

[ajdelange]

About the best you an do here is add up the charges on the reported ions. You will wind up with a net positive charge of +0.626 mEq/L. You then assume that there are no unreported ions except those responsible for alkalinity (hydroxyl, carbonate and bicarbonate) further assume that the bicarbonate is predominant, multiply 0.626*61 = 38.186 mg/L and declare that to be your bicarbonate content. To estimate the alkalinity you assume that titration to 4.5 will convert all the bicarbonate so that the alkalinity is simply 0.626 mEq/L or 31.3 ppm as CaCO3 (plus 2 ppm for the water itself).

If you want to be more precise than this, which probably isn't justified as you don't know that there isn't some unreported nitrate for example, you would calculate that at pH 7.8 the charge on carbo is -.9664 mEq/mol and that, therefore, you have .626/0.9664 = 0.648 mole of carbo. In the course of doing this calculation (how to do it is in the stickies) you would have found that the carbo is 96.08% bicarbonate ( 0.601 mmol/L = 36.7 mg/L) and 0.26% carbonate (0.1 mg/L) but we don't really care about bicarbonate and carbonate content - just the alkalinity. You would also calculate that, at pH 4.5, the defining pH for alkalinity measurement, the charge per mole is -0.0130 mEq/L. Thus your alkalinity is 0.626*(0.961- 0.013) = 0.593 mEq/L or 29.6 ppm (again add 2 for the water). Indeed the extra effort was not worth it.

Here you have calcium at 33.9 mg/L = 1.69 mEq/L and magnesium at 9.2 mg/L = 0.76 mEq/L for a total of 2.45 mEq/L = 122.5 ppm as CaCO3 so no, your hardness is not equal to your alkalinity. Your carbonate hardness (temporary hardness) is but your total hardness (the rest being permanent hardness) is not.

 

[Big Monk]

About the best you an do here is add up the charges on the reported ions. You will wind up with a net positive charge of +0.626 mEq/L. You then assume that there are no unreported ions except those responsible for alkalinity (hydroxyl, carbonate and bicarbonate) further assume that the bicarbonate is predominant, multiply 0.626*61 = 38.186 mg/L and declare that to be your bicarbonate content. To estimate the alkalinity you assume that titration to 4.5 will convert all the bicarbonate so that the alkalinity is simply 0.626 mEq/L or 31.3 ppm as CaCO3 (plus 2 ppm for the water itself).

If you want to be more precise than this, which probably isn't justified as you don't know that there isn't some unreported nitrate for example, you would calculate that at pH 7.8 the charge on carbo is -.9664 mEq/mol and that, therefore, you have .626/0.9664 = 0.648 mole of carbo. In the course of doing this calculation (how to do it is in the stickies) you would have found that the carbo is 96.08% bicarbonate ( 0.601 mmol/L = 36.7 mg/L) and 0.26% carbonate (0.1 mg/L) but we don't really care about bicarbonate and carbonate content - just the alkalinity. You would also calculate that, at pH 4.5, the defining pH for alkalinity measurement, the charge per mole is -0.0130 mEq/L. Thus your alkalinity is 0.626*(0.961- 0.013) = 0.593 mEq/L or 29.6 ppm (again add 2 for the water). Indeed the extra effort was not worth it.

Here you have calcium at 33.9 mg/L = 1.69 mEq/L and magnesium at 9.2 mg/L = 0.76 mEq/L for a total of 2.45 mEq/L = 122.5 ppm as CaCO3 so no, your hardness is not equal to your alkalinity. Your carbonate hardness (temporary hardness) is but your total hardness (the rest being permanent hardness) is not.

Thanks AJ. As always, a well thought out and presented response.