Question

100 ml of standard hard water containing 1 mg of pure CaCO3 per ml consumed 20 ml EDTA. 50 ml of water sample consumed 20 ml of same EDTA solution in another titration Calculate total hardness in ppm, °Cl and ° Fr.​

Answer 1

2000 ppm  

Explanation:

1 ml of standard hard water = 1 mg of CaCO₃ eq.

∴ 100 ml SHW = 100 mg of CaCO₃ eq.

   100 ml of SHW = 20 ml of EDTA solution

   20 ml of EDTA = 100 mg of CaCO₃ eq.

∴ 1 ml of EDTA = 100/20

                          = 5 mg of CaCO₃ eq.

50 ml of water sample = 20 ml of EDTA solution

∴ 1000 ml of water sample = 1000×20/50

                                               = 400 ml of EDTA solution

1 ml of EDTA = 5 mg of CaCO₃ eq.

∴ 400 ml of EDTA = 5 × 400

                                = 2000 mg of CaCO₃ eq.

∴ total haedness of water = 2000 mg/L

  water sample = 2000 ppm