- #1

Jim4592

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**Calculating alkalinity and expressing it in "as CaCO3"**

## Homework Statement

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-}

## Homework Equations

## 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**