Question

0.28 g of caco3 was dissolved in Hcl and the solution was made to one liter with distilled Water. 100 ml of the above solution required 28 ml of EDTA solution on titration 100 ml of hardwater sample required 33ml of the same EDTA solution on titration. after boiling 100 ml of the water required 10 ml of EDTA solution. calculate temporary hardness, permanent hardness, and total hardness​

Answer 1

Answer:

Explanation:

Given 0.28 g of caco3 was dissolved in Hcl and the solution was made to one liter with distilled Water. 100 ml of the above solution required 28 ml of EDTA solution on titration 100 ml of hardwater sample required 33ml of the same EDTA solution on titration. after boiling 100 ml of the water required 10 ml of EDTA solution. calculate temporary hardness, permanent hardness, and total hardness

Standard hard water of 1000 ml = 0.28 g of CaCO3

Therefore 1 ml of standard hard water will be 0.28 mg of CaCO3

28 ml of EDTA = 100 ml of hard water = 100 x 0.28

Therefore 1 ml of EDTA = 100 x 0.28 x 1/28 = 1mg of CaCO3

Now total hardness  

100 ml of sample requires 33 ml of 0.01 M EDTA

                                           33 x 1 mg

                                             = 33 mg of CaCO3

Now 1000 ml of sample requires = 33 x 1000 / 100

                                                       = 330 mgs of CaCO3

Now total hardness will be 330 ppm

So 100 ml of boiled sample requires 10 ml of 0.01 M EDTA

                                                        = 10 x 1 mg

                                                   = 10 mg

Now boiled sample of 1000 ml requires 10 x 1000 / 100

                                                              = 100 mg of CaCO3

                                                              = 100 ppm

Therefore temporary hardness = total hardness – permanent hardness

                                                    = 330 – 100

                                                  = 230 ppm