Answer:
[tex]\frac{mg}{L}[Carbonate]=0.33X50=16.5\frac{mg}{L} CaCO_{3}[/tex]
[tex]\frac{mg}{L}[BiCarbonate]=1.06X50=53\frac{mg}{L} CaCO_{3}[/tex]
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
[tex]CO_{3}^{-2}=30[/tex]
[tex]HCO_{3}^{-}=61[/tex]
milliequivalents of carbonate will be:
[tex]meq=\frac{10}{30} =0.33[/tex]
millieuivalents of bicarbonate will be:
[tex]meq=\frac{65}{61} =1.06[/tex]
In terms of calcium carbonate, we will multiply the milliequivalents with 50 (the equivalent weight of calcium carbonate).
[tex]\frac{mg}{L}[Carbonate]=0.33X50=16.5\frac{mg}{L} CaCO_{3}[/tex]
[tex]\frac{mg}{L}[BiCarbonate]=1.06X50=53\frac{mg}{L} CaCO_{3}[/tex]
