Question

A water sample contains 16.2 mg/L Ca(HCO3)2, 7.3 mg/L Mg(HCO3)2, 9.5 mg/L MgCl2 and 13.6 mg/L CaSO4. Calculate the temporary and permanent hardness of water and what will happen if 10.6 mg/L NaHCO3 is added?​

Answer 1

Explanation:

Solution



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Correct option is

C

150 ppm, 200 ppm

mole of Ca(HCO3)2=162g/mole162×10−3mg=1×10−3moles

Mole of Ca(SO4)=136g/mole136×10−3g=1×103mole

Total mole of Ca=2×10−3mole

mass of CaCO3=2×10−3×100=0.2g

∴ ppm (permanent hardness) =10006.2×106=200ppm

Mole of MgCl2=9595×10−3=1×10−3mole 

Mole Mg (HCg)2=14673×10−3=5×10

Answer 2

Answer:

mole of Ca(HCO3)2=16g/mole16×10−3mg=1×10−3moles

Mole of Ca(SO4)=13g/mole136×10−3g=1×103mole

Total mole of Ca=2×10−3mole

mass of CaCO3=2×10−3×100=0.2g

∴ ppm (permanent hardness) =10006.2×106=200ppm

Mole of MgCl2=9595×10−3=1×10−3mole 

Mole Mg (HCg)2=14673×10−3=5×10−4mole

mole Mg =1.5×