The concentration of As(III) in water can be determined by differential pulse po

Question

The concentration of As(III) in water can be determined by differential pulse
polarography in 1 M HCl. The initial potential is set to –0.1 V versus the SCE and
is scanned toward more negative potentials at a rate of 5 mV/s. Reduction of As(III)
to As(0) occurs at a potential of approximately –0.44 V versus the SCE. The peak
currents for a set of standard solutions, which are corrected for the residual current,
are shown in the following table.
[As(III)] (M) ip (A)
1.00 0.298
3.00 0.947
6.00 1.83
9.00 2.72
What is the concentration of As(III) in a sample of water if its peak current is 1.37 A

Explanation / Answer

Ans> The given current is 1.37uA. Now when we increase the concentration from 3 to 6 micromolar the current increases by 0.883 uA. When we increase the concentration further to 9 micromolar then also the current increases by 0.89. So approximately we can say that there is a linearity in the current increase with increasing concentration. Now the rate of current increase is 0.294 uA per 1 micromolar concentration increase.

So the given concentration is 3 + [(6-3)/(1.83-0.947)]*(1.37-0.947) = 3+1.436 = 4.436 micromolar is the concentration of As(III) ions in the water sample.

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