At 25 degree C, you conduct a titration of 15.00 mL of a 0.0460 M AgNO_3 solutio

Question

At 25 degree C, you conduct a titration of 15.00 mL of a 0.0460 M AgNO_3 solution with a 0.0230 M NaI solution within the following cell: Saturated Calomel Electrode || Titration Solution | Ag (s) For the cell as written, what is the voltage after the addition of the following volume of NaI solution? The reduction potential for the saturated calomel electrode is E = 0.241 V. The standard reduction potential for the reaction Ag^+ + e^- rightarrow Ag(s) is E^0 = 0.79993 V. The solubility constant of Agl is K_sp = 8.3 times 10^- 17. a) 0.200 mL b) 19.00 mL c) 30.00 mL. d) 44.10 mL

Explanation / Answer

Using Nernst equation,

E = 0.79993 - 0.0591 log[Ag+] - 0.241

a) 0.2 ml of 0.023 M NaI added

initial moles Ag+ present = 0.046 M x 15 ml = 0.69 mmol

moles I- added = 0.023 M x 0.2 ml = 0.0046 mmol

[Ag+] left in solution = (0.69 - 0.0046) mmol/15.2 ml = 0.0451 M

E = 0.79993 - 0.05916 log(0.0451) - 0.241

   = 0.64 V

b) 19.0 ml of 0.023 M NaI added

initial moles Ag+ present = 0.046 M x 15 ml = 0.69 mmol

moles I- added = 0.023 M x 19 ml = 0.437 mmol

[Ag+] left in solution = (0.69 - 0.437) mmol/34 ml = 0.00744 M

E = 0.79993 - 0.05916 log(0.00744) - 0.241

   = 0.685 V

c) 30 ml of 0.023 M NaI added

initial moles Ag+ present = 0.046 M x 15 ml = 0.69 mmol

moles I- added = 0.023 M x 30 ml = 0.69 mmol

Equivalence point

[Ag+] = sq.rt.(Ksp)

         = sq.rt(8.3 x 10^-17)

        = 9.11 x 10^-9 M

E = 0.79993 - 0.05916 log(9.11 x 10^-9) - 0.241

   = 1.035 V

d) 44.10 ml of 0.023 M NaI added

This is past equivalence point

initial moles Ag+ present = 0.046 M x 15 ml = 0.69 mmol

moles I- added = 0.023 M x 44.10 ml = 1.0143 mmol

[I-] left in solution = (1.0143 - 0.69) mmol/59.1 ml = 5.49 x 10^-3 M

[Ag+] = Ksp/[I-]

          = 8.3 x 10^-17/5.49 x 10^-3

          = 1.51 x 10^-14 M

E = 0.79993 - 0.05916 log(1.51 x 10^-14) - 0.241

   = 1.376 V

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