Answer to Question #171735 in Physical Chemistry for Akshat

Question

Calculate the amount of lime (88.3% pure) and soda (99.2% pure) required to soften 24,000L of water per day in a year containing the following: CaCO3 = 1.85 ppm, MgCO3 = 0.42 ppm, MgSO4 = 0,9 ppm, CaSO4 = 0.34 ppm, MgCl2 = 0.76 ppm, NaCl = 2.36 ppm and SiO2 = 2.34 ppm.

Answer 1

For temporary hardness which is determined by the presence of CaCO3 an MgCO3

Volume of water= 24000L

Conc of CaCO3= 1.85ppm= 1.85mg/L

1L contains 1.85mg of CaCO3

24000L will contain 1.85x24000= 44400mg= 44400/1000= 44.4g

Conc of MgCO3= 0.42ppm= 0.42mg/L

1L contains 0.42mg

24000L will contain 0.42 x 24000= 10080mg= 10.08g

Since temporary hardness is soften by slaked lime, we can now set two equations

Ca(OH)2 + Ca(HCO3)2---> 2CaCO3 + 2H2O

From the equation

200g of CaCO3 is precipitated by 74g of Ca(OH)2

44.4g of CaCO3 will be precipitated from 74/200 x 44.4= 16.428g of Ca(OH)2

Also,

Ca(OH)2 + Mg(HCO3)2----> 2MgCO3 + 2H2O

From the equation

168g of MgCO3 is precipitated from 74g of Ca(OH)2

10.08g of MgCO3 will be precipitated from 74/108 x 10.08= 4.44g

Total amount of lime required to remove temporary hardness= 16.428 + 4.44= 20.868g

But the line is only 88.3% pure.

Actual amount= 88.3/100 x 20.868= 18.43g of the lime.

Now, permanent hardness is determined by the presence of CaSO4 and MgSO4.

Mass of CaSO4= 0.34 x 24000= 8160mg= 8.16g

Mass of MgSO4 present= 0.9 x 24000= 21600mg= 21.6g

Since permanent hardness is soften by soda, we can set two equations.

136g of CaSO4 requires 106g of Na2CO3

8.16g of CaSO4 will require 106/136 x 8.16 = 6.36g

Also,

Na2CO3 + MgSO4---> MgCO3 + Na2SO4

120g of MgSO4 requires 106g of Na2CO3

21.6g of MgSO4 will require 106/120 x 21.6= 19.08g

Total amount of soda required= 6.36+ 19.08= 25.44g

But the soda is only 99.2% pure,

Actual amount= 99.2/100 x 28.44= 28.21g of soda.