The presence of Pb2 can cause interference in a certain method for the determina

Question

The presence of Pb2 can cause interference in a certain method for the determination of As3 in groundwater. A researcher would like to use this method to determine the amount of As3 in a groundwater sample, but she wants to make sure that the method is sufficiently selective for As3 over Pb2 before she uses the method to test the groundwater sample. The researcher knows that if the concentration of As3 is 200.00 times the concentration of Pb2 , then the relative error is 0.26%. Determine the selectivity coefficient, KAs,Pb, for this method.

Explanation / Answer

For simplicity, we assign absolute concentrations, so that CAs3 = 200 and CPb2 = 1.
consider that having a relative error of + 0.36%, it means the presence of Pb2 is 0.36% higher than the signal in the absence of Pb. If the signal in the absence of Pb2 is 200 then the signal in the presence of zinc is 200.72.

the common formula for these cases is:

kAs=S(As)/C(As)

where:

kAs = 200/200 = 1

In the presence of Pb the signal is given by

Ssamp= K(As)*C(As) + K(Pb)*C(Pb)
we know that Ssamp=200.72, then the selectivity coefficient is:

200.72=(1*200)+ K(Pb)*1

from here K(Pb)=0.72

finally,

K(As,Pb)=K(Pb)/K(As) = 0.72/1 = 0.72

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