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A lithium ion-selective electrode gave the potentials given below for the following standard solutions of LICI and two samples of unknown concentration: Solution (ALLI+) Potential vs. SCE, mV 0.100 M +1.0 0.050 M -30.0 0.010 M -60.0 0.001 M -138.0 Unknown 1 -48.5 Unknown 2 -75.3 (a) Construct a calibration curve of potential versus log aLi1 and determine if the electrode follows the Nernst equation. (b) Use a linear least-squares procedure to determine the concentrations of the two unknowns.

A lithium ion-selective electrode gave the potentials given below for the following standard solutions of LICI and two samples of unknown concentration: Solution (ALLI+) Potential vs. SCE, mV 0.100 M +1.0 0.050 M -30.0 0.010 M -60.0 0.001 M -138.0 Unknown 1 -48.5 Unknown 2 -75.3 (a) Construct a calibration curve of potential versus log aLi1 and determine if the electrode follows the Nernst equation. (b) Use a linear least-squares procedure to determine the concentrations of the two unknowns.

Chemistry
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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Chapter 21, Problem 28QP Problem A lithium ion-selective electrode gave the potentials given below for the following standard solutions of LICI and two samples of unknown concentration: Solution (ali+) Potential vs. SCE, mV 0.100 M +1.0 0.050 M -30.0 0.010 M -60.0 0.001 M -138.0 Unknown 1 -48.5 Unknown 2 -75.3 (a) Construct a calibration curve of potential versus log alLi1 and determine if the electrode follows the Nernst equation. (b) Use a linear least-squares procedure to determine the concentrations of the two unknowns Step-by-step solution Step 1 of 5 (a) The data given is used in the following spreadsheet to plot the calibration curve: potential vs SCH, mV aLM logLt 1 0.1 -1 -30 0.05 -1.301 -60 0.01 -2 -138 0.001 -3 So, the calibration curve of the electrode potential versus log a, as shown below: Plot of potential vs SCE and logali* logaLi y = 66.64x + 64.885 -4 -3 -2 -1 20 -20 -40 -60 potential vs SCE, mV -80 -100 -120 -140 -160 Step 2 of 5 Nernst equation for Lit /Li couple is written as follows: E = E° +59.2 log a ..... (1) Here, E is electrode potential of the cell, fº is standard electrode potential of the cell, and a, is activity of the lithium ions. With relevance to general equation: y =c+mx (where y, c, m, and x are dummy variables and m signifies slope of the equation) Thus, the slope for Equation (1) must be equal to 59.2. On the other hand, actual equation as shown above in the calibration curve is v = 66.64x + 64.885 and the slope is 66.64. Therefore, the electrode does not follow the Nernst equation as the slope of the calibration curve is 66.64. Comments (1) Anonymous for OWLV2 it does follow Step 3 of 5 (b) Determination of the concentration of the two unknowns using linear least-squares procedure. This is done by simply following the given rules: • Go to data • Go to data analysis • Choose regression as the function • Input values of a,, in the option input Y range • Input values of potential vs. SCE in the option input X range Then values the potential of unknown 1 is put in the following expression: Y = 0.077499 +0.000656×X (as derived in the excel sheet) Here 0.077499 is taken from coefficient of the intercept and 0.000656 is taken from coefficient of the X variable 1 of the ANOVA table shown in the following excel sheet. • The steps where repeated for unknown 2. Step 4 of 5 The spreadsheet is as follows: Step 5 of 5 potenial vs SCE, mVaytM 1 0.1 -30 0.05 -60 0.01 -138 0.001 -48.5 0.045665084 Y-007799-0.000656*X -75.3 0.028074266 SUMMARY OUTPUT legression Statstics 0.866330384 R Square 0.750528335 Adjusted R Square 0.25792502 Standard Eror 0.027631085 Observatons. 4. ANOVA Sgiftcame F 1 004593796 0.0045938 6.01694252 0.133669616 SS MS F Regressin Residual 2 0.001526954 0.0007635 Total 3 0.00612075 Coefficients Standard Error tStat P-value Lower 95% Uppr 95% | Lower 95.0% Upper 95.0% Intercept X Varable 1 0.077499215 0.020529703 3.7749798 0.06355638 -001083297 0.165831399 -0.01083297 0.165831399 0.000656374 0.000267586 2.4529457 0.13366962 -0.000494956 0.001807703 -0000494956 0.001807703 Therefore, the concentrations of Lit for two unknowns are 0.0457 M and 0.0281 M Comments (1) Anonymous incorrect values
Transcribed Image Text:Chapter 21, Problem 28QP Problem A lithium ion-selective electrode gave the potentials given below for the following standard solutions of LICI and two samples of unknown concentration: Solution (ali+) Potential vs. SCE, mV 0.100 M +1.0 0.050 M -30.0 0.010 M -60.0 0.001 M -138.0 Unknown 1 -48.5 Unknown 2 -75.3 (a) Construct a calibration curve of potential versus log alLi1 and determine if the electrode follows the Nernst equation. (b) Use a linear least-squares procedure to determine the concentrations of the two unknowns Step-by-step solution Step 1 of 5 (a) The data given is used in the following spreadsheet to plot the calibration curve: potential vs SCH, mV aLM logLt 1 0.1 -1 -30 0.05 -1.301 -60 0.01 -2 -138 0.001 -3 So, the calibration curve of the electrode potential versus log a, as shown below: Plot of potential vs SCE and logali* logaLi y = 66.64x + 64.885 -4 -3 -2 -1 20 -20 -40 -60 potential vs SCE, mV -80 -100 -120 -140 -160 Step 2 of 5 Nernst equation for Lit /Li couple is written as follows: E = E° +59.2 log a ..... (1) Here, E is electrode potential of the cell, fº is standard electrode potential of the cell, and a, is activity of the lithium ions. With relevance to general equation: y =c+mx (where y, c, m, and x are dummy variables and m signifies slope of the equation) Thus, the slope for Equation (1) must be equal to 59.2. On the other hand, actual equation as shown above in the calibration curve is v = 66.64x + 64.885 and the slope is 66.64. Therefore, the electrode does not follow the Nernst equation as the slope of the calibration curve is 66.64. Comments (1) Anonymous for OWLV2 it does follow Step 3 of 5 (b) Determination of the concentration of the two unknowns using linear least-squares procedure. This is done by simply following the given rules: • Go to data • Go to data analysis • Choose regression as the function • Input values of a,, in the option input Y range • Input values of potential vs. SCE in the option input X range Then values the potential of unknown 1 is put in the following expression: Y = 0.077499 +0.000656×X (as derived in the excel sheet) Here 0.077499 is taken from coefficient of the intercept and 0.000656 is taken from coefficient of the X variable 1 of the ANOVA table shown in the following excel sheet. • The steps where repeated for unknown 2. Step 4 of 5 The spreadsheet is as follows: Step 5 of 5 potenial vs SCE, mVaytM 1 0.1 -30 0.05 -60 0.01 -138 0.001 -48.5 0.045665084 Y-007799-0.000656*X -75.3 0.028074266 SUMMARY OUTPUT legression Statstics 0.866330384 R Square 0.750528335 Adjusted R Square 0.25792502 Standard Eror 0.027631085 Observatons. 4. ANOVA Sgiftcame F 1 004593796 0.0045938 6.01694252 0.133669616 SS MS F Regressin Residual 2 0.001526954 0.0007635 Total 3 0.00612075 Coefficients Standard Error tStat P-value Lower 95% Uppr 95% | Lower 95.0% Upper 95.0% Intercept X Varable 1 0.077499215 0.020529703 3.7749798 0.06355638 -001083297 0.165831399 -0.01083297 0.165831399 0.000656374 0.000267586 2.4529457 0.13366962 -0.000494956 0.001807703 -0000494956 0.001807703 Therefore, the concentrations of Lit for two unknowns are 0.0457 M and 0.0281 M Comments (1) Anonymous incorrect values
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