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What is the total hardness of sample of water has the following impurities in mg/I.
Ca(HCO3)2 = 162 CaCI2 = 22.2
MgCI2 = 95 NaCI = 20
1 Answer
written 3.9 years ago by |
Given:
Ca(HCO3)2=162CaCl2=22.2MgCl2=95NaCl=20
Solution:
Equivalents of Calcium Carbonate=Mass of hardness producing substance×50Chemical equivalent of hardness producing substance
1.
Ca(HCO3)2=162×50162.11=49.966 mg/L
2.
CaCl2=22.2×50110.98=10 mg/L
3.
MgCl2=95×5095.21=49.889 mg/L
4.
NaCl=20×5040=25 mg/L
Total Hardness=49.97+10+49.89+25=134.86 mg/L