Calculate temporary, permanent and total hardness of water., Chemistry

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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 


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