Question

Cd2+ and Pb2+ ions in a polluted water sample are measured by complexation. In a first titration, 50.00 mL of water at pH 10 is titrated by 40.09 mL of 0.004870 M EDTA. In a second titration, a 75.00 mL aliquot of water is first treated with an excess of KCN to mask the cadmium in Cd(CN)4 2- before being titrated by 31.44 mL of the same EDTA solution. Calculate the concentrations of Cd2+ and Pb2+ in the polluted water sample. Cd2+ + 4 CN- <===> Cd(CN)4 Kf = 1.0 x 10^19 Cd2+ + Y4- <===> CdY2- Kf = 2.9 x 10^16 Pb2+ + Y4- <===> PbY2- Kf = 1.1 x 10^18 aY4- = 0.36

Answer 1

We know that

Cd2+ + 4 CN- <===> Cd(CN)4

Kf = 1.0 x 10^19

Cd2+ + Y4- <===> CdY2-

Kf = 2.9 x 10^16 = [CdY^2-] / Cd²⁺] [Y^4-]αY4-

2.9 x 10^16 X αY4- = [CdY^2-] / Cd²⁺] [Y^4-]

Similarly for Pb+2

Pb²⁺ + Y4- ==> PbY2-

Kf = 1.1 x 10^18 = [PbY^2-] / Pb²⁺] [Y^4-]αY4-

1.1 x 10^18 X αY4- = [PbY^2-] / Pb²⁺] [Y^4-]

Now

The moles of EDTA used for complete complexation of both the ions = Molarity X volume = 0.00487 X 40.09 / 1000 = 0.000195 moles

Moles of EDTA used after masking the Cd+2 ion = 0.004870 X 31.44 / 1000 = 0.000153

So these are moles of EDTA used against Pb+2 , so moles of Pb+2 present = 0.000153

So moles of Cd+2 = Moles of EDTA used agsint both the ions - Moles of EDTA used agasinst Pb+2 only

Moles of Cd+2 = 0.000195 - 0.000153 = 0.000042 moles

Concentration of Pb+2 = Moles / Volume in litres = 0.000153 X 1000 / 50 = 0.00306 M

Concentration of Cd+2 = Moles / Volume in litres = 0.000042 X 1000 / 50 = 0.00084 M