Question

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A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5 mL of 0.0600 M EDTA. Titration of the excess unreacted EDTA required 13.3 mL of 0.0100 M Ca. The Cd2t was displaced from EDTA by the addition of an excess of CN. Titration of the newly freed EDTA required 11.2 mL of 0.0100 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? M Mn2+ concentration: M Cd2 concentration:

Answer 1

EDTA forms 1:1 complex with Cd+2, Mn+2 and Ca+2 So number of moles of Ca+2 or Cd+2 or Mn+2 is equal to the number of moles of EDTA The volume of the sample = 50 mL Let the number of moles of Cd+2 = a the number of moles of Mn+2 = b The molarity of EDTA added initially = 0.0600 M Volume of the EDTA added initially = 56.5 mL Total number of moles of EDTA added = M*V = 0.0600 M* 56.5 mL = 3.39 mmol Number of moles of Ca+2 initially added = M*V = 0.0100 M * 13.3 mL = 0.133 mmol. So the number of moles of EDTA unreacted = 0.133 mmol The number of moles of EDTA reacted with Mn+2 and Cd+2 = 3.39 -0.133 = 3.257 mmol. The sum of the moles of Cd+2 and Mn+2 = a + b = 3.257 The number of mole of Ca+2 after treated with CN-ions = 0.0100 M* 11.2 mL = 0.112 mmol

The number of moles of EDTA released = 0.112 mmol Since the addition of CN-ions form complex with Cd+2, the number of moles of Cd+2 = 0.112 mmol The number of moles of Mn+2 = 3.257 – 0.112 = 3.145 mmol Concentration of Mn+2 = 3.145 mmol / 50.0 mL = 0.0629 | Concentration of Cd+2 = moles / volume = 0.112 mmol / 50.0 mL = 0.00224 M

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We know that Pb+2 forms 1:1 complex with EDTA and BAL moles of EDTA used for 32.25 mL of Pb+2 = M*V = 0.0138 M * 41.89 mL = 0.578082 mmol Molarity of Pb+2 = moles/volume = 0.578082 mmol/ 32.25 mL = 0.017925 M. Number of moles of Pb+2 in aliquot solution = M*V = 0.017925 M* 23.81 mL = 0.426795 mmol moles of EDTA used for excess of Pb+2 = M*V = 0.0138 M * 9.67 mL = 0.133446 mmol Number of moles of Pb+2 react with EDTA ( excess after react with BAL) = 0.133446 mmol Number of moles of Pb+2 that react with BAL = 0.426795 -0.133446 = 0.293349 mmol That is number of moles of BAL = 0.293349 mmol Volume of BAL solution = 11.00 mL Concentration of BAL = moles/volume = 0.293349 mmol/11.00 mL = 0.026668 M Concentration of BAL = 0.02667 M