Concentration of indium and cadmium:

Problem:

Indium and Cadmium ions in 0.1 M HCl give peaks at potentials - 0.557 V and - 0.597 V by differential pulse polarographic analysis. When a standard solution containing 0.4 ppm of indium is analyed, it gave peak current (in arbitrary units) corresponding to 100.25 at - 0.557 V and 43.75 at - 0.597 V. A standard solution containing 0.4 ppm of Cd gave i_ps 29.52 at - 0.557 V and 64.8 at - 0.597 V. What is the concentration of indium and cadmium in a sample if the peak current is 190 at & 0.557 V and 120 at & 0.597V.

Answer:

Peak currents in differential pulse polarography are directly proportional to the analytes concentration

i_p = k (analyte ppm)

where k is a constant depending on the analyte and applied potential.

The values of k at potentials -0.557 (E₁) and -0.597 (E₂) for both the metals are given by

K_In.E₁  = 100.25/0.4  =          250.62 ppm^-1

K_In.E₂  = 43.75/0.4    =          109.4 ppm^-1

K_cd.E₁  =29.52/0.4     =          73.8 ppm^-1

K_cd.E₂  =64.8/ 0.4     =           162.0 ppm^-1

Writing the simultaneous equations for the current at the two potentials

iE₁ = 250.62 (Con In) + 73.8 (Con Cd) = 190

iE₂ = 109.4 (Con In) + 162.0 (Con Cd) = 120

On solving these two equations we get the

ConIn = 0.674 ppm

ConCd = 0.285 ppm in the given sample.