This is off topic for a swimming pool, but I would like to get the chemistry experts to weight in on this. As part of my work, I am setting up a titration type analyzer which measures total alkalinity in mg/L as calcium carbonate. This analyzer uses a standard alkalinity solution as part of its calibration procedure. For this particular project the calibration standard needs to have 2000 mg/L total alkalinity as calcium carbonate. At first we were looking at using calcium carbonate in DI water for the standard. However calcium carbonate does not have that much solubility in pure water. So we need to use sodium bicarbonate instead. What I am looking at is how much sodium bicarbonate I need to use. As this is a calibration standard fairly high accuracy is needed.
According to the PoolMath calculator I need to use 3.517 grams. According to another thread I found on this site 100 lbs of sodium bicarbonate is equal to 59,57 lbs of calcium carbonate. This gives a value of 3.357 grams
So I tried to perform the the calculations my self: .
The bicarbonate ion has one half the alkalinity as carbonate ion i.e. 1 mole of bicarbonate is equal to 0.5 moles of carbonate.
Calcium carbonate has a mass of 100.0869 g/mol.
Sodium bicarbonate has a mass of 84.0066 g/mol.
So 2000mg of calcium carbonate = 2000 * (87.0066/(0.5*100.0869)) = 3477 mg of sodium bicarbonate
Can you verify this calculation? Now I know the variances are very small, but I want to be as precise as I can.