A sample of water is found to contain the following analytical data in mgs/lit.Mg(HCO3)2=14.6;MgCl2=9.5;MgSO4=6.0;Ca(HCO3)2=16.2.calculate the temporary and permanent hardness of the sample of water.
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Answered by
2
Answer:
We can calculate the hardness of given water sample as:
(i) Ca(HCO
3
)
2
CaCO3≡CaCO
3
162ppm≡100ppm
∴81ppm≡50ppm
(ii) Mg(HCO
3
)
2
≡CaCO
3
146 ppm ≡ 100 ppm
∴73ppm≡50ppm
(iii) CaSO
4
≡CaCO
3
136ppm≡100ppm
∴68ppm≡50ppm
(iv) MgSO
4
≡CaCO
3
120ppm≡100ppm
∴60ppm≡50ppm
Total degree of hardness of water =50×4=200ppm
Answered by
1
Answer: Temporary hardness = 30.8mg/lit
Permanent hardness = 15.5mg/lit
Explanation:
Temporary hardness causing Impurities = Mg(HCO3 )2, Ca(HCO3 )2
Permanent hardness causing Impurities = MgCl2,MgSO4
Temporary hardness = 14.6+16.2 = 30.8mg/lit
Permanent hardness = 9.5+6 = 15.5mg/lit
Total hardness in water = Temporary hardness + Permanent Hardness
=30.8+15.5
Total hardness in water = 46.3mg/lit
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