When measured with a F^- ion-selective electrode with a Nernstian response at 25
ID: 519748 • Letter: W
Question
When measured with a F^- ion-selective electrode with a Nernstian response at 25 degree C, the potential due to F^- in unfluoridated groundwater in a nearby city was 40.7 mV more positive than the potential of tap water in a distant city. The distant city maintains its fluoridated water at the recommended level of 1.00 plusminus 0.05 mg F^-/L What is the concentration of F^- in mg/L in groundwater in the nearby city? (Disregard the uncertainty. Assume that the Nernst potential is 0.05916 V.) mg/L The following data were obtained when a Ca^2+ on-selective electrode was immersed in standard solutions whose ionic strength was constant at 2.0 M.Explanation / Answer
The reduction reaction of interest is
F2 (g) + 2 e- -----> 2 F- (g); E0 = standard reduction potential.
Since the concentration of F- is not the standard 1.0 M, we can write for the cell potential as
E1 = E0 – (0.05916 V)*log [F-]2 …..(1)
The concentration of F- in the distant city is 1.00 mg F-/L; the molar mass of F- = 19 g mol-1; therefore, concentration of F- in terms of mol/L = (1.00 mg/L)*(1 g/1000 mg)*(1 mole/19 g) = 5.263*10-5 mol/L. Therefore, [F-] = 5.263*10-5 M.
Given E2 = E1 + 40.7 mV; again, E2 = E0 – (0.05916 V)*log [F-]2 where [F-] is the molar concentration of fluoride in the nearby city. Therefore,
E2 – E1 = -(0.05916 V)*log [F-]2 – [-(0.05916 V)*log (5.263*10-5)2]
===> 40.7 mV = -(0.05916 V)*log [F-]2 - 0.5063 V
===> 0.0407 V = -(0.05916 V)*log [F-]2 - 0.5063 V
===> 0.5470 V= -(0.05916 V)*log [F-]2
===> log [F-]2 = -9.246
===> [F-]2 = 5.675*10-10
===> [F-] = 2.382*10-5
The concentration of F- in the nearby city = 2.382*10-5 M = (2.382*10-5 mol/L)*(19 g/1 mole)*(1000 mg/1 g) = 0.452 mg/L 0.45 mg/L (ans).
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