Answer to Question #163753 in Organic Chemistry for Wincy Mispa

Question

Calculate the total, temporary and permanent hardness of water sample having following analysis: Mg(HCO3)2 = 73mg/l, Ca(HCO3)2 = 162mg/l, CaSO4 = 136mg/l, MgCl2 = 95 mg/l, CaCl2 = 111 mg/l and NaCl = 100 mg/l.

Answer 1

mole of Ca(HCO3)2 = (162 g/mole)/162×10-³mg ,= 1×10−3 moles

Mole of Ca(SO)4 = (136g/mole)/(136×10-3g) = 1×10-3 mole

Total mole of Ca=2×10 −3 mole

mass of CaCO3 = 2×10 −3 ×100=0.2g

∴ ppm (permanent hardness) = 6.2/1000 × 10^6 =200ppm

Mole of MgCl 2 = 95 ×10 −3g = 1×10 −3 mole

Mole Mg (HCO3) 2 = 146/73×10 −3 = 5 ×10-4 mole

mole Mg =1.5×10 −4 mole

Mole CaCO 3v= 1.5×10 −3

mass =1.5×10 −3

=0.150g

ppm (temporary hardness) = 100× 0.150×106=150ppm