Calculating Water Hardness: Ca2+ & Mg2+

[sdoug041]

Homework Statement

A sample of water is known to have 190 g/m^3 Ca2+ and 84 g/m^3 Mg2+. What is the hardness of the water in mg CaCO3/L?

Homework Equations

Hardness = (Ca2+) + (Mg2+)

The Attempt at a Solution

I need to convert my units into mg CaCO3/L or Normality before summing up the values. The problem is I do not understand how to convert to either.

If I'm converting to mg CaCO3/L do I need a molar mass? And when do I use it?

As for normality I do not understand what a "weight equivalence" is so I can't convert to that either.

Thanks a lot.

 

[nate808]

Hello,

To convert from g/m^3 to mg/L, you need to multiply by 1000 because there are 1000 mg in 1 g. So for the Ca2+ concentration, it would be 190 mg/L and for the Mg2+ concentration, it would be 84 mg/L.

To convert from mg/L to mg CaCO3/L, you need to use the molar mass of CaCO3, which is 100.09 g/mol. This means that for every 100.09 g of CaCO3, there is 1 mol. So to convert from mg to mol, you divide by the molar mass (100.09 g/mol). Then, to convert from mol to mg CaCO3, you multiply by the molar mass (100.09 g/mol). So for the Ca2+ concentration, it would be (190 mg/L) x (1 mol/100.09 g) x (100.09 g/mol) = 190 mg CaCO3/L. And for the Mg2+ concentration, it would be (84 mg/L) x (1 mol/100.09 g) x (100.09 g/mol) = 84 mg CaCO3/L.

To calculate the hardness, you simply add the concentrations of Ca2+ and Mg2+. So in this case, the hardness would be 190 mg CaCO3/L + 84 mg CaCO3/L = 274 mg CaCO3/L.

I hope this helps! Let me know if you have any other questions.