Answer to Question #286034 in Organic Chemistry for Addy

Question

Calculate the temporary and permanent hardness of water sample containing Mg(HCO3)2= 

46.3mg/L, Ca(HCO3)2= 246.2mg/L, MgCl2= 88.5mg/L, CaSO4=243.6mg/L) in terms of 

CaCO3 equivalents.

Answer 1

mole of Ca(HCO3)2=246.2×10^-3/162

= 1.52×10^-3

Mole of CaSO4=243.6×10³/136

= 1.79×10³

Total of Ca = 3.3×10³

​ $mass of CaCO 3 ​ =3.3×10^ {−3} ×100=0.33g$

$ppm (permanent hardness) =\frac{ 6.33}{ 1000} ×10^ 6 =200ppm$