1. Experiment 5 Determining the Ksp of AgCl using Potentiometric Titrations

2. Adding together redox reactions Ag + + e- ↔ Ag (s) E 0 1 AgCl (s) + e- ↔ Ag (s) + Cl - E 0 2

3. Ag + + e- ↔ Ag (s) E 0 + Ag (s) + Cl - ↔ AgCl (s) + e- E 0 - __________________________________ Ag + + Cl - ↔ AgCl (s) E 0 cell = E 0 + - E 0 -

4.  G 0 = -nFE 0 cell = -RT ln(K sp ) K sp = e -  G0/RT = e (nF/RT)E0cell = e E0cell/59.16

5. Adding together redox reactions Ag + + e- ↔ Ag (s) E 0 1 AgCl (s) + e- ↔ Ag (s) + Cl - E 0 2

6. Experimental Setup Voltmeter Double junction Ag/AgCl Reference Electrode Ag bullet Electrode [Ag +] = 0.10 0.1 M KCl in buret

7. Potentiometric titration Use very low current electrochemical cell to monitor the course of a redox tiratration The reactions at the cathode do not change the bulk concentrations of reagents

8. Experiment: Part 1 Add 0.0500 M AgNO 3 to 100 mL of 0.500 M KNO 3. Titration reaction; none. This is not a titration Cathode reaction: Ag + + e- ↔ Ag (s) Cell Voltage increases because the Ag + ion concentration increases Ecell = E 0 (Ag + /Ag) – 59.16log(1/[Ag + ]) - E ref

9. Part 2 Add a diluted solution of KCl Titration reaction; Ag + + Cl - → AgCl(s) Equivalence point reached when just enough Cl - reacts with all the Ag + Cathode reaction before Equiv. point: Ag + + e- ↔ Ag (s) Cell Voltage decreases because the Ag + ion concentration decreases Ecell = E 0 (Ag + /Ag) – 59.16log(1/[Ag + ]) - E ref

10. At the Equivalence point A sharp drop in cell voltage indicates when the equivalence point is reached The equivalence point is used to determine the concentration of the diluted KCl solution

11. After the Equivalence Point In bulk: Adding excess Cl - Reaction at cathode; AgCl (s) + e- ↔ Ag (s) + Cl - Cell Voltage decreases because the Cl - ion concentration increases Ecell = E 0 (AgCl/Ag) – 59.16log([Cl - ]) - E ref

12. Recap Before Equivalence point: Ecell = E 0 (Ag + /Ag) – 59.16log(1/[Ag + ]) – E ref Plot Ecell vs. log(1/[Ag + ]) and intercept = E 0 (Ag + /Ag) - E ref

13. Recap After Equivalence point: Ecell = E 0 (Ag/AgCl) – 59.16log[Cl - ] – E ref Plot Ecell vs. log [Cl - ] and intercept = E 0 (Ag/AgCl) - E ref

14. intercept 1 = E 0 (Ag + /Ag) - E ref intercept 2 = E 0 (AgCl/Ag) – E ref Int 1 – Int 2 = E 0 (Ag + /Ag) - E 0 (AgCl/Ag) K sp = e (Int1 –Int 2)/59.16