A complexation titration was performed to determine the concentration of Cd2+ in

Question

A complexation titration was performed to determine the concentration of Cd2+ in a sample. A 50 mL aliquot of the Cd2+ sample was taken and titrated with 0.002 M EDTA at a constant pH of 10. It was found that 30 mL of EDTA were required. No auxiliary complexing agent was used. The formation constant for the CdY= complex is 3 × 1016. The 4 at pH = 10 is 0.35. The atomic weight of Cd is 112.

a) What is the concentration of Cd2+ in the sample?

b) How many grams of Cd were in the original aliquot?

c) What are the concentrations of Cd2+, CdY2-, and Y4- after 15 mL of EDTA was added?

d) What fraction of the CdY2- dissociates under the solution conditions after 15 mL of EDTA was added?

Explanation / Answer

1)   

as the no of moles of EDTA = no of moles of Cd

molarity * volume (EDTA) =   molarity * volume   of Cd

0.002 * 30 = 50 * molarity of Cd

molarity of Cd = 0.002 * 30 / 50 = 0.0012 M

concentration of Cd2+ in the sample =  0.0012 M

2) as no of moles = molarity * volume = 0.0012 M *(0.050)= 0.00006

given mass = no of moles * molar mass = 0.00006 * 112.41 = 0.00674 g

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