Q. A sample of water on analysis was found to contain the following analytical data:
Mg (HCO3)3 = 16.8mg/L, MgCl2 = 19mg/L,
Mg (NO3)2 = 29.6 ppm, CaCO3 = 20 ppm,
MgSO4 = 24.0 mg/L and KOH = 1.9 ppm,
Calculate the temporary, permanent and total hardness of water sample.
ans.
|
S. No.
|
Substances
|
Amount of substances
|
Multiplication factor
|
CaCO3 equivalent
|
|
1.
|
Mg (HCO3)2
|
16.8mg/L
|
100/146
|
100*16.8/146=11.50ppm
|
|
2.
|
MgCl2
|
19mg/L
|
100/95
|
100*19/95 = 20 ppm
|
|
3.
|
Mg (NO3)2
|
29.6 ppm
|
100/148
|
100*29.6/148= 20 ppm
|
|
4.
|
CaCO3
|
20 ppm
|
100/100
|
100*20/100= 20 ppm
|
|
5.
|
MgSO4
|
24mg/L
|
100/120
|
100*24/120= 20 ppm
|
|
6.
|
KOH
|
1.9 ppm
|
-
|
-
|
KOH does not cause hardness
Temporary hardness due to = [Mg (HCO3)2 + CaCO3] ppm
= 11.50 +20 ppm = 31.50 ppm
Permanent hardness due to MgCl2, Mg (NO3)2 and MgSO4
= [MgCl2 + Mg (NO3)2 + MgSO4] ppm = [20+20+20] ppm = 60 ppm
Total hardness = Temporary hardness + Permanent hardness
= 31.50 +60 ppm = 91.50 ppm