Question

Wastewater from a cement factory contains 0.410 g of Ca2+ ion and 0.0330 g of Mg2+ ion per 100.0 L of solution. The solution density is 1.001 g/mL. Calculate the Ca2+ and Mg2+ concentrations in ppm (by mass).

Answer 1

`Ca^(2+)`= 4.0959 ppm

`Mg^(2+)`= 0.3296 ppm

Step-by-step explanation:

ppm means milligram of solute per kilogram of solution.

`ppm= (mg of solute)/(Kg of solution)`

Solutes:
`Ca^(2+) y Mg^(2+)`

Mass
`Ca^(2+)` = 0.410g

Mass
`Mg^(2+)` = 0.0330g

Solution: 100.0 L, whose density is 1.001 g/mL

First, let us to calculate mass of solution:

`100.0 L * (1.001 g )/(1 mL ) * (1000 mL)/(1 L) * (1 kg)/(1000 g) =`= 100.1 kg

Now, we have to calculate ppm
`Ca^(2+) y Mg^(2+)`

ppm
`Ca^(2+)= (0.410 g * (1000 mg/1g))/(100.1 Kg) =4.0959 mg/Kg`

ppm
`Mg^(2+)= (0.0330 g * (1000 mg/1g))/(100.1 Kg) =0.3296 mg/Kg`