Calcium and magnesium

ESTIMATION OF CALCIUM AND MAGNESIUM

Principle

         Calcium and magnesium get complexed by EDTA in the order Ca first and Mg afterwards. In this experiment Ca is estimated first by using an indicator mureoxide at pH 12 in the presence of sodium hydroxide. Then Ca + Mg are estimated by using an indicator Eriochrome Black-T at pH 10 in the presence of ammonium chloride and ammonium hydroxide buffer solution.

Reagents

•   0.02 N EDTA

•   10 % Sodium hydroxide

•   Ammonium chloride – Ammonium hydroxide buffer

•   Mureoxide indicator

•   Eriochrome Black-T indicator

Procedure

Calcium

Pipette out 10 ml of irrigation water sample into a porcelain basin. Add 10 % Sodium hydroxide solution drop by drop to neutralize the acidity if any (use litmus paper) and add another 5 ml excess to maintain the pH at 12. Add a pinch (50 mg) of mureoxide indicator and titrate with 0.02N EDTA till the colour changes from pinkish red to purple or violet.

Calcium + Magnesium

Pipette out 10 ml of irrigation water sample into porcelain basin. Add buffer solution drop by drop to neutralize the acidity if any (use litmus paper) and 5 ml excess to maintain the pH at 10.  Add 2-3 drops of Eriochrome Black-T indicator solution and titrate with 0.02N EDTA till the colour changes from purple red to sky blue.

Calculation

Volume of 0.02N EDTA used in the titration

for Calcium + Magnesium                                                      =    X’ ml

Volume of 0.02 N EDTA used for Calcium                             =    ‘Y’ml

Therefore volume of 0.02N EDTA used for

Magnesium alone                                                                  =    (X-Y) ml

1ml of 0.02N EDTA                                                                =                   0.0004 g of Ca

Therefore Y ml of 0.02N EDTA                                              =                0.0004 x Yg of Ca

This is present in 10ml of water sample

 Therefore in 106 ml of water sample (ppm)   =        0.0004 x Y/10 x 106

In terms of milli equivalent per litre                      =           ppm/20

1ml of 0.02N EDTA                                              =          0.00024 g of Mg

Therefore (X-Y) ml of 0.02 N EDTA                      =         0.00024 x (X-Y)g of Mg.

This is present in 10ml of water sample

 Therefore 106 ml of water sample (ppm)  =      0.00024 (X-Y)/10 x 106

In terms of milli equivalent per liter                        =           ppm/12