Answer to Question #300056 in Organic Chemistry for Abhishek

Question

Calculate temporary and total hardness of water sample of water 

containing: Mg(HCO3)2=9.3 mg/l, Ca(HCO3)2=17.4 mg/l, MgCl2=8.7 mg/l and 

CaSO4= 12.6 mg/l.

Answer 1

Temporary hardness causing impurities = Mg(HCO3)2 and Ca(HCO3)2

Permanent hardness causing impurities = MgCl2 and CaSO4

Mole of Ca(HCO3)2 is;

(17.4×10^-3)mg ÷ (17.4g/mole)

=1 × 10^-3 mole

Mole of Ca(SO4) is;

(12.6×10^-3)mg ÷ (12.6g/mole)

= 1 × 10^-3 mole

We know that total mole of Ca is;

= 2 × 10^-3 mole

Which means mass of CaCO3 is;

= 2 × 10^-3 × 100 = 0.2g

Therefore, permanent hardness is;

= (0.2 × 10^6) ÷ 1000 = 200ppm

Permanent hardness;

= 200ppm