Question

1. A water with a pH of 9.0 contains 10 mg/L CO3-2 and 65 mg/L HCO3-. Calculate the alkalinity of the water expressed as meq/L (milli equivalents per liter) and as mg/L as CaCO3.

Answer 1

Answer:

$\frac{mg}{L}[Carbonate]=0.33X50=16.5\frac{mg}{L} CaCO_{3}$

$\frac{mg}{L}[BiCarbonate]=1.06X50=53\frac{mg}{L} CaCO_{3}$

Explanation:

The milli equivalents of the carbonate and bicarbonate ions can be calculated by multiplying the milligrams of the two with their respective equivalent mass.

The equivalent mass of

$CO_{3}^{-2}=30$

$HCO_{3}^{-}=61$

milliequivalents of carbonate will be:

$meq=\frac{10}{30} =0.33$

millieuivalents of bicarbonate will be:

$meq=\frac{65}{61} =1.06$

In terms of calcium carbonate, we will multiply the milliequivalents with 50 (the equivalent weight of calcium carbonate).

$\frac{mg}{L}[Carbonate]=0.33X50=16.5\frac{mg}{L} CaCO_{3}$

$\frac{mg}{L}[BiCarbonate]=1.06X50=53\frac{mg}{L} CaCO_{3}$