# Calculating alkalinity and expressing it in "as CaCO3"

by Jim4592
Tags: alkalinity, as caco3, expressing
 P: 49 1. The problem statement, all variables and given/known data Convert the following from milligrams per liter as the ion to milligrams per liter as CaCO3. A) 83.00 mg/L Ca2+ B) 27.00 mg/L Mg2+ C) 48.00 mg/L CO2 D) 220.00 mg/L HCO3- E) 15.00 mg/L CO32- 2. Relevant equations 3. The attempt at a solution I think I did the calculations correctly, but there's a slight difference between my answer and the book's answer...I was hoping someone could look and see if I did the steps correctly and if so is it difference negligible? A) (83.00 mg Ca{2+} / L) x (1 mmole Ca{2+} / 40.078 g) x (2 meq Ca{2+} / mmole Ca{2+}) x (100 mg CaCO3 / 2 meq) = 207.096 mg/L as CaCO3, book answer = 207.25 B) (27.00 mg Mg{2+} / L) x (1 mole Mg{2+} / 24.305 g) x (2 meq Mg{2+} / mmole Mg{2+}) x (100 mg CaCO3 / 2 meq) = 111.088 mg/L as CaCO3, book answer = 111.20 C) (48.00 mg CO2 / L) x (1 mmole CO2 / 44.0095 g) x (2 meq CO2 / 1 mmole) x (100 mg CaCO3 / 2 meq) = 109.067 mg/L as CaCO3, book answer = 109.18 D) (220 mg HCO3{-} / L) x (1 mmole HCO3{-} / 61.01684 g) x (1 meq / 1 mmole) x (100 mg CaCO3 / 2 meq) = 180.278 mg/L as CaCO3, book answer = 180.41 E) (15.00 mg CO3 / L) x (1 mmole CO3 / 60.0089 g) x (2 meq / 1 mmole) x (100 mg CaCO3 / 2 meq) = 24.996 mg/L as CaCO3, book answer = 25.02
 Admin P: 21,704 Obviously book uses slightly different molar masses. I wouldn't care too much, differences are below 10-3, that's about as precise as you can determine alkalinity experimentally.
 P: 49 alright thank's borek, I thought I would be able to just write it off but that's the farthest off we've ever been, so I figured I'd ask for a second opinion.

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