q. a sample of water on analysis was found to

Question

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