Question

(5) The following data represent the potentials of a fluoride ion-selective electrode vs SCE for several solutions. All solutions are made up in a way to control their pH and ionic strength, and contain added EDTA. Prepare an excel plot of the measured potential vs pF based on this information and determine the concentration of fluoride in each unknown solution. 5 Solution (M) Potential (mv) 5.0 x 102 -22.4 .0 x 103 36.8 5.0 x 104 96.0 5.0 x 10- 155. 2 Unknown 1 Unknown 1 74.3 Unknown 1 -54.3 90.0

Answer 1

From known solutions the potential (mV) and corresponding pF value calculated from [F-] and plotted in excel :

(mV, [F-] in M, pF) : (-22.4, 0.05, 1.301); ( 36.8, 0.005, 2.301); ( 96, 5E-4, 3.301); (155.2, 5E-5, 4.301)

The graph of potential in mV (X) Vs pF (Y) has been attached here.

After linear fitting the intercept (c) and slope values(m) are 1.679
and 0.0169 respectively has been obtained from excell

Let us calculate concentration of F- in unknown solutions

Unknown solution 1, Potential= 74.3 mV that corresponds to pF can be obtained by using the formula y=mx+c

pF= 0.0169*74.3+1.679= 2.935

therefore concentration F- in unknown solution 1 that exhibit potential 74.3 mV = 1.2X10-3 M

Similarly for unknown solution 2 , pF= 0.0169*190.6+1.679= 4.900

and unknown solution 3 , pF= 0.0169*(-54.3)+1.679= 0.7613

therefore concentration F- in unknown solution 2 that exhibit potential 190.6 mV = 1.3X10-5 M

therefore concentration F- in unknown solution 3 that exhibit potential -54.3 mV = 1.7X10-1 M. The linear plot is given below

The linear fitting shows intercept= 1.679 Slope= 0.0169 pF -40 -20 0 20 40 60 80 100 120 140 160 180 200 Potential (mV)

The linear fitting shows intercept= 1.679 Slope= 0.0169 pF -40 -20 0 20 40 60 80 100 120 140 160 180 200 Potential (mV)