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21 December, 04:50

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.

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  1. 21 December, 05:53
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    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
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