**Question:**Given the following selectivity coefficients for a K⁺ Ion Selective Electrode (ISE), which of the below ions would cause the greatest interference to this ISE?- \( K_{\text{K}^+, \text{Na}^+} = 1 \times 10^{-5} \)- \( K_{\text{K}^+, \text{Rb}^+} = 2.8 \)- \( K_{\text{K}^+, \text{Cs}^+} = 0.44 \)**Options:**- ⦿ A. K⁺- ⦿ B. Na⁺- ⦿ C. Rb⁺- ⦿ D. Cs⁺**Answer Explanation:**When evaluating interference in an ion selective electrode (ISE), the selectivity coefficient (\( K_{\text{K}^+, \text{ion}^+} \)) indicates the electrode’s preference for the ion compared to the primary ion, K⁺ in this case. A higher value of the selectivity coefficient reflects greater interference. Given the coefficients:- \( K_{\text{K}^+, \text{Na}^+} = 1 \times 10^{-5} \) means Na⁺ has very low interference.- \( K_{\text{K}^+, \text{Rb}^+} = 2.8 \) implies Rb⁺ shows significant interference.- \( K_{\text{K}^+, \text{Cs}^+} = 0.44 \) suggests moderate interference from Cs⁺.Therefore, option C (Rb⁺) would cause the greatest interference.

Here we are asked to select the ion which cause the greatest interference.

Given the following selectivity coefficients for a K* ISE, which of the below ions would cause the greatest interference to this ISE? KK+, Na+ = 1 x 10-5 KK,Rb+= 2.8 KK+,Cs+ = 0.44 ● . A. K+ B. Nat OC. Rb+ OD. Cs+

Given the following selectivity coefficients for a K* ISE, which of the below ions would cause the greatest interference to this ISE? KK+, Na+ = 1 x 10-5 KK,Rb+= 2.8 KK+,Cs+ = 0.44 ● . A. K+ B. Nat OC. Rb+ OD. Cs+

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

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Chemistry Question
Explain (with the aid of one or more equations) why a LOWER selectivity coefficient for an interfering ion means an ISE is actually MORE selective for the target ion.