Question

Determine the lime and soda ash dose, in mg/L as Ca(OH)2 and Na2CO3, to soften the water of a village to a final hardness of 40.0 mg/L as CaCO3. Chemical analysis showed that the water contains Na+ (50 mg/L), Ca2+ (150 mg/L), Mg2+ (30 mg/L), HCO3- (100 mg/L as CaCO3), Cl- (80 mg/L). and CO2 (5.0 mg/L).

Answer 1

Total hardness present = Ca⁺⁺ mg/l(50/20)+ Mg⁺⁺ mg/l(50/12)

= 150*(50/20) + 30*(50/12)= 500 mg/l as CaCO₃

Total hardness to be removed = 500-40 = 460 mg/l as CaCO₃

HCO₃^- as CaCO₃ =100 mg/l

let amount of HCO₃^- Present= x

so, x*50/61 =100 , x = 122 mg/l

Non Carbonate hardness to be removed = 460-100 = 360 mg/l as CaCO₃

Constituents mg/l Eq.mass m.eq/l = (mg/l/Eq.mass)

CO₂   5   22 5/22 =0.227

Ca⁺⁺   150     20 150/20= 7.5

Mg⁺⁺ 30 12 30/12 =2.5

HCO₃^- 122 61 122/61 =2

Requirement of Ca(OH)₂

= [ m.eq/l of CO2 + m.eq/l of HCO₃^- + m.eq/l of Mg⁺⁺ ] * Eq. mass   of Ca(OH)₂

= [0.227 + 2 + 2.5]*37

= 161.20 mg/l

Requirement of Soda Na₂CO₃

Non carbonate Hardness of 100 mg/l as CaCO₃ requires 106 mg/l of Na₂CO₃

Non carbonate Hardness of 360 mg/l as CaCO₃ requires 106/100*360 mg/l of of Na₂CO₃

= 381.6 mg/l