Answer to Question #343398 in General Chemistry for Jiin

Question #343398

A method for the analysis of Ca2+ in water suffers from an interference in the presence of Zn2+. When the concentration of a Ca2+ is 50 times greater than that Zn2+, an analysis for Ca2+ gives a relative error of 2.0%. What is the value of the selectivity coefficient for this method?


1
Expert's answer
2022-05-24T03:17:03-0400

Since only relative concentrations are reported, we can arbitrarily assign absolute concentrations. To make the calculations easy, we will let CCa= 50 (arbitrary units) and CZn = 1. A relative error of 2% means the signal in the presence of Zn2+ is 2% greater than the signal in the absence of Zn2+. Again, we can assign values to make the calculation easier. If the signal for Cu2+ in the absence of Zn2+ is 50 (arbitrary units), then the signal in the presence of Zn2+ is 51. The value of kCa is determined:

"k_{Ca}=\\frac{S_{Ca}}{C_{Ca}}=50\/50=1"

In the presence of Zn2+ the signal is given by

"S_{sample}=51=k_{Ca}C_{Ca}+k_{Zn}C_{Zn}=(50x1)+k_{Zn}x1"

Solving for kZn gives its value as 1. The selectivity coefficient is

"K_{Ca,Zn}=k_{Zn}\/k_{Ca}=1\/1=1"


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