Calculating alkalinity and expressing it in "as CaCO3"by Jim4592 Tags: alkalinity, as caco3, expressing 

#1
Sep2010, 11:13 PM

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 CaCO_{3}. A) 83.00 mg/L Ca^{2+} B) 27.00 mg/L Mg^{2+} C) 48.00 mg/L CO_{2} D) 220.00 mg/L HCO_{3}^{} E) 15.00 mg/L CO_{3}^{2} 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 



#2
Sep2110, 02:59 AM

Admin
P: 22,708

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.




#3
Sep2210, 10:40 AM

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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