Answer to Question #343398 in General Chemistry for Jiin

Question

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?

Answer 1

Since only relative concentrations are reported, we can arbitrarily assign absolute concentrations. To make the calculations easy, we will let C_Ca= 50 (arbitrary units) and C_Zn = 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 k_Ca 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 k_Zn gives its value as 1. The selectivity coefficient is

$K_{Ca,Zn}=k_{Zn}/k_{Ca}=1/1=1$