Question

You have been tasked with determining the concentration of Ca2+ in an unknown solution of calcium carbonate (CaCO3). You titrate 50.0 ml of the unknown solution with 0.0250 M EDTA. It requires 16.85 ml of EDTA to reach the equivalence point. Determine [Ca2+] of the unknown solution, in ppm. Recall that ppm (parts per million) is equal to mg/L. 2.

If a sample of tap water contains 400 ppm of Ca2+ and 80 ppm of Mg2+, what is the hardness of this water sample in terms of equivalent concentration CaCO3, in mg/L?

Answer 1

1 L of solution will contain 0.008425 moles of calcium ions. THe atomic mass of calcium is 40.1 g/mol.
THe mass of calcium ions is mg.
ppm (parts per million) is equal to mg/L
Hence,

(2)

The molar masses of CaCO3, ion and ion are 100 g/mol, 24.3 g/mol and 40.1 g/mol respectively.
We can say that 24.3 ppm = 100 ppm CaCO3 and 40.1 ppm = 100 ppm CaCO3

So 80 ppm ppm CaCO3
Similalrly 400 ppm CaCO3
Total hardness of the sample = 329.2 + 997.5 = 1326.7 ppm CaCO3
This means that 1 L of water contains 1326.7 mg of CaCO3.
Hence, the hardness of water is 1327 mg/L of CaCO3