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

`(mg)/(L)[Carbonate]=0.33X50=16.5(mg)/(L) CaCO_(3)`

`(mg)/(L)[BiCarbonate]=1.06X50=53(mg)/(L) CaCO_(3)`

Step-by-step 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=(10)/(30) =0.33`

millieuivalents of bicarbonate will be:

`meq=(65)/(61) =1.06`

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

`(mg)/(L)[Carbonate]=0.33X50=16.5(mg)/(L) CaCO_(3)`

`(mg)/(L)[BiCarbonate]=1.06X50=53(mg)/(L) CaCO_(3)`