Answer to Question #293359 in Chemical Engineering for wesda

Question

Standard hard water is prepared by dissolving 0.28 gm of calcium carbonate in dilute HCI

and then diluted to 1 liter with distilled water. 100 ml of this standard hard water requires 28 ml of EDTA solution. 100 ml of unknown hard water requires 33 ml of EDTA solution. 100 ml of same water sample after boiling requires 10 ml of EDTA solution. Calculate Total, Permanent and Temporary hardness of water sample

Answer 1

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