Answer to Question #179067 in Inorganic Chemistry for Sky

Question

 A sample of water on analysis has been found to contain the following in ppm Ca(HCO3)2 = 10.5; Mg(HCO3)2 = 12.5; CaSO4=7.5; MgSO4 = 2.6; CaCl2 = 8.2; Calculate temporary, permanent and total hardness of the water sample in degree Clark and degree French. (Atomic masses Mg = 24; and Ca = 40).

Answer 1

Temporary hardness due to Mg(HCO3) and Can(HCO3)

8.562 + 6.481 = 15.043 mg/L

In degree french = 15.043 × 0.1°fr = 1.5043 °fr

In degree Clark = 15.043 × 0.07° Clark= 1.05301°clark

Permanent hardness due to MgSO4, CaSO4 and CaCl2 = 5.1 + 7.387 + 2.167 = 15.069 mg/L

In degree french = 15.069 × 0.1= 1.5069 °fr

In degree Clark = 15.069 × 0.07 = 1.05483°Clark

Total hardness = 15.043 + 15.069

= 30.112

In degree french = 30.112 × 0.1 = 3.0112°fr

In degree Clark = 30.112× 0.07 = 2.1078°Clark