## Persons stated I need twice my calculated baking soda for Kh ppm

Water analysis, treatment, and mineral recipes for optimum taste and equipment health.
Acavia
In a response to a water discussion, where I stated I use 6.354445mg of baking soda, added to a gallon of distilled water, for each ppm (mg/L) Kh, someone wrote:

"This will give you the ppm of each mineral in solution but most recipes assume ppm CaCO3 equivalence. So to be more accurate you would need to convert using the molar mass of your target compound compared to CaCO3. Also most sources of bicarbonate only dissociate into one equivalent of bicarbonate whereas CaCO3 dissociates into two molar equivalents of bicarbonate so you need twice as much of say sodium or potassium bicarbonate in your calculations to compare it directly to other published recipes if you are making your own solutions."

Why would baking soda be different from CaCO3, needing twice my calculated amount? My calculation matches Barista Hustle and was verified by a knowledgeable posters here. Is this person right - is my Kh half of what I thought it was?

homeburrero
Team HB
You have it right. Your calculation already takes care of the fact that a CaCO3 molecule is chemically equivalent to two NaHCO3 molecules.

Doing a very explicit calculation to convince why:

Given: Molar mass of NaHCO3 is 84 mg/mmol
6.35 mg / 84.0 mg/mmol = 0.0756 mmol of NaHCO3
Given: 1 US gallon = 3.78 liters
0.0756 mmol / 3.78 liters = 0.02 mmol/liter

NaHCO3 is univalent, and 0.02 mmol/L is chemically equivalent to 0.01/L mmol of divalent CaCO3

Given: molar mass of CaCO3 is 100 mg/mmol

0.01 mmol/L * 100 mg/mmol = 1 mg/L as CaCO3
and 1 mg/L is essentially the same as 1 ppm
Pat
nínádiishʼnahgo gohwééh náshdlį́į́h

Acavia (original poster)
Thanks.

Acavia (original poster)
homeburrero wrote:You have it right. Your calculation already takes care of the fact that a CaCO3 molecule is chemically equivalent to two NaHCO3 molecules.

Doing a very explicit calculation to convince why:

Given: Molar mass of NaHCO3 is 84 mg/mmol
6.35 mg / 84.0 mg/mmol = 0.0756 mmol of NaHCO3
Given: 1 US gallon = 3.78 liters
0.0756 mmol / 3.78 liters = 0.02 mmol/liter

NaHCO3 is univalent, and 0.02 mmol/L is chemically equivalent to 0.01/L mmol of divalent CaCO3

Given: molar mass of CaCO3 is 100 mg/mmol

0.01 mmol/L * 100 mg/mmol = 1 mg/L as CaCO3
and 1 mg/L is essentially the same as 1 ppm
I just checked my baking soda and it is Sodium Bicarbonate. So, then I do need to use twice as much?

Acavia (original poster)
Or do I need half as much.

MOL =84

so to hit 60, I would need 60/(100/84) mg per liter. So 50.5mg of Sodium Bicarbonate per liter, which would be 190.5mg per gallon. So, if I use ~381mg per gallon, my Kh is really 120?

Or is NahCO3 half the displacement of CaCO3 as a buffer, so you need twice as much, so 2 x 190mg = ~381mg and that would be the needed added to a gallon to hit ~60Kh?

homeburrero
Team HB
You initially had it figured right for baking soda.

Baking soda, sodium bicarbonate, sodium hydrogen carbonate, NaHCO3 all refer to the same chemical.

Using your already calculated amount, 6.35 * 60 = 381 mg of baking soda per US gallon will give you 60 ppm KH as CaCO3.

And if you want to calculate it out from scratch:

381 mg / 84 mg/mmol = 4.54 mmol
and 4.54 mmol / 3.78 liters = 1.2 mmol/L

1.2 mmol/L of univalent bicarbonate is chemically equivalent to 0.6 mmol/L of divalent carbonate

mass of CaCO3 is 100 mg/mmol and 0.6 x 100 = 60, so your KH is 60 mg/L as CaCO3
Pat
nínádiishʼnahgo gohwééh náshdlį́į́h

Acavia (original poster)
Thanks very much. I am comfortable with my non chemistry-background math on my Gh calculations, but I had taken the Sodium bicarbonate sourced Kh for granted in the amount to use, so it threw me for a loop when that person posted that.