An EDTA standard was made at 22.5 °C by dissolving 0.6206 g (uncorrected mass) o

Question

An EDTA standard was made at 22.5 °C by dissolving 0.6206 g (uncorrected mass) of Na2H2EDTA·2H2O in a 499.859 mL volumetric flask. Several 9.9873 mL aliquots of unknown were titrated for total divalent metal ions, and required an average of 23.197 mL of EDTA solution to reach the end point. After precipitation of Mg2+ from several additional 9.9873 mL aliquots of unknown, an average of 15.660 mL of titrant was required to reach the end point. All blanks were zero. Determine the total hardness in mg/L as CaCO3, given that the molar mass of CaCO3 = 100.0868 g/mol. Finally, calculate the concentrations of Ca2+ and Mg2+ individually, in mM. Be sure to explicitly show the mole ratio in your sample stoichiometry calculations— do not treat this like a dilution problem. [Note: you do not have to show how to calculate the density of water at 22.5 °C, the buoyancy correction factor for EDTA, or the molar mass of CaCO3, but you do have to show how to use each of these quantities in the sample problems.]

[EDTA standard (corrected to 20 °C): 3.34072 mM]

Explanation / Answer

The concentration of EDTA = Mass of EDTA / equilvalent weight of Na2EDTA X volume in L

Normality = 0.6206 / 186.12 X 0.499 = 0.00668 Normal

The volume of EDTA used = 23.197 mL

Equivalents of EDTA used = Volume X normality = 23.197 X 0.00668 = 0.155 milliequivalents

So equivalents of Mg+2 = 0.155 milliequilvalents

So concentration of Mg+2 = Milliequilvalents / volume used = 0.155 / 9.9873 = 0.0155

so concentration of Mg+2 in terms of CaCO3 in g/ L = 0.0155 X 50 g / L = 0.775 g / L = 775 ppm

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