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

Answer:

$Ca^{2+}$= 4.0959 ppm

$Mg^{2+}$= 0.3296 ppm

Explanation:

ppm means milligram of solute per kilogram of solution.

$ppm= \frac{mg of solute}{Kg of solution}$

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:

$100.0 L * \frac{1.001 g }{1 mL } * \frac{1000 mL}{1 L} * \frac{1 kg}{1000 g} =$= 100.1 kg

Now, we have to calculate ppm $Ca^{2+} y Mg^{2+}$

ppm $Ca^{2+}= \frac{0.410 g * (1000 mg/1g)}{100.1 Kg} =4.0959 mg/Kg$

ppm $Mg^{2+}= \frac{0.0330 g * (1000 mg/1g)}{100.1 Kg} =0.3296 mg/Kg$