Carbonate and bicarbonate

 ESTIMATION OF CARBONATE AND BICARBONATE

         To the colorless solution obtained above, add a drop of methyl orange and the solution attains light yellow colour. Titrate this against N/10 H2SO4 till the colour changes to pinkish red. Note down the titre value (‘V1’ ml). This value represents the volume of N/10 H2SO4 required to neutralize the bicarbonate originally present and that from the conversion of carbonate.

         Calculate the separate amounts of carbonate and bicarbonate and express the result as milli equivalent/litre.

CALCULATION

Carbonate

Volume of N/10 H2SO4 required to neutralize half  of the carbonate  = V ml

Therefore volume of N/10 H2SO4 required to

 neutralize the entire carbonate                     =   2 x V = 2V ml.

 1 ml of N/10 H2SO4                                                                 =                0.003 g of CO3

Therefore 2V ml of N/10 H2SO4                                                  =           0.003 x 2V g of CO3

50 ml of water sample contains                                        =          g of CO3

Therefore carbonate content in ppm                                =      0.003 x 2V x 1000 x 1000/50

Milli equivalents of CO3 per litre                      =   ppm / 30

Bicarbonate

Volume of N/10 H2SO4 required to neutralize the

full bicarbonate originally present + the bicarbonate resulted from carbonate

                                                                                            =          V1 ml

Volume of N/10 H2SO4 required neutralizing

half of the carbonate                                                           =          V ml

Therefore volume of N/10 H2SO4 required to

neutralize the bicarbonate originally present                     =          (V1-V) ml

1ml of N/10 H2SO4                                                                =                        0.0061g of HCO3-

Therefore (V-V1) ml of H2SO4 contains            =           0.0061 x (V-V1)g of HCO3-

50ml of water sample contains                                    =

Therefore bicarbonate content in  ppm                      =         0.0061 x (V-V1) x 1000 x 1000/50

Milli equivalents of HCO3- per litre            =         ppm / 61