A groundwater (GW) was analyzed for ions and the results are reported below: Ca2+ = 200 mg/L (as Ca2+) Mg2+ = 60 mg/L (as Mg2+) Alkalinity = 400 mg CaCO3/L Assume the pH is near neutral a) What is the total hardness of the GW in mg CaCO3/L? b) What is the non-carbonate hardness of the GW in mg CaCO3/L? c) What is the carbonate hardness of the GW in mg CaCO3/L?? d) Would you recommend the GW be treated with a softening process prior to distribution to consumers? Why? List two treatment options.

Answer is given below.

Explanation:

Alkalinity = 400 mg/L in terms of CaCO3  

a.) Total Hardness is defined as the sum of calcium and magnesium iion concentration in terms of CaCO3

Total Hardness = [Ca+2]*{50/20) + [Mg+2]*{50/12) = [200]*{50/20) + [60]*{50/12) = 750 mg/L in terms of CaCO3

b.) Non-Carbonate hardness = Total Hardness - Carbonate Hardness

Now here as the Total Hardness is greater than alkalinity ,Thus

Carbonate Hardness = Alkanlinity = 400 mg/L in terms of CaCO3

Non-Carbonate hardness = 750 - 400 = 350 mg/L in terms of CaCO3

c.) As Total hardness is greater than Alkalinity thus Carbonate hardness is equal to the Alkalinity.

Carbonate hardness = 400 mg/L in terms of CaCO3

d.) Now as the Total hardness of the ground water is more than 180 mg/L,Thus it needs removal of hardness.

There are many treatment options for the removal of hardness:

i) Ion Exchange method

ii) Lime soda method

po-215 i got it right on the quiz

the awnser is group 12