Question

A sample water of 100 ml required 12.6 ml of 0.02M EDTA solution with EBT as indicator and 8.4
ml of 0.02 M EDTA for the same volume of water after removing the carbonate hardness. Calculate
the total, permanent hardness in terms of calcium carbonate equivalents.

Answer 1

Explanation:

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Answer 2

Given :

100 ml of water sample = 12.6 ml of 0.02 M EDTA

100 ml of water sample =8.4 ml of 0.02 M EDTA

To Find:

We have to find the total hardness and permanent hardness in terms of calcium carbonate equivalents.

Solution:

It is given that,

 100 ml of water sample = 12.6 ml of 0.02 M EDTA

                                         = of 0.02 EDTA

                                         = 126 ml of 0.02 EDTA

Also, we know that

 Molarity of EDTA = 2×Normality of EDTA

                                = 2× 126 ml of 0.02 EDTA

                                = 252 ml or 0.252 L of 0.02 eq. of

                                = 0.252×0.02×100 g  

Hence total hardness = 0.504g or 504mg of

                                      = 504mg/L or ppm

Now after removing the carbonate hardness

  100 ml of water sample = 8.4 ml of 0.02 M EDTA

                                          =  of 0.02 EDTA

                                          = 84 ml of 0.02 EDTA

Molarity of EDTA = 2× Normality of EDTA

                             = 2×84 ml of 0.02 EDTA

                             = 168 ml or 0.168 L of 0.02 EDTA

                             = 0.168×0.02×100 g of

                             = 0.336 g or 336 mg of  

Hence permanent hardness = 336 mg/L or ppm

∴ The total hardness and permanent hardness is 504mg/L and 336 mg/L