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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

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# Homework Help: Calculating alkalinity and expressing it in as CaCO3

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