At 25 °C, you conduct a titration of 15.00 mL of a 0.0300 M AgNO3 solution with

Question

At 25 °C, you conduct a titration of 15.00 mL of a 0.0300 M AgNO3 solution with a 0.0150 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- ---> Ag(s)

is E0 = 0.79993 V. The solubility constant of AgI is Ksp = 8.3 × 10-17.

a) .600 ml

b)12.10 ml

c)30.00 ml

d)39.40 ml

Explanation / Answer

Titration,

Ag+ + I- ---> AgI

a) 0.6 ml of 0.015 M NaI added

initial Ag+ = 0.03 M x 15 ml = 0.45 mmol

NaI added = 0.015 M x 0.6 ml = 0.009 mmol

[Ag+] remains = (0.45 - 0.009) mmol/15.6 ml = 0.028 M

E = (Eo + 0.0592 log[Ag+]) - 0.241

   = [0.79993 + 0.0592 log(0.028)] - 0.241

   = 0.47 V

b) 12.10 ml of 0.015 M NaI added

initial Ag+ = 0.03 M x 15 ml = 0.45 mmol

NaI added = 0.015 M x 12.10 ml = 0.1815 mmol

[Ag+] remains = (0.45 - 0.1815) mmol/27.1 ml = 0.0099 M

E = (Eo + 0.0592 log[Ag+]) - 0.241

   = [0.79993 + 0.0592 log(0.0099)] - 0.241

   = 0.44 V

c) a) 30 ml of 0.015 M NaI added

initial Ag+ = 0.03 M x 15 ml = 0.45 mmol

NaI added = 0.015 M x 30 ml = 0.45 mmol

Equivalence point

[Ag+] = sq.rt.(Ksp) = sq.rt.(8.3 x 10^-17) = 9.11 x 10^-9 M

E = (Eo + 0.0592 log[Ag+]) - 0.241

   = [0.79993 + 0.0592 log(9.11 x 10^-9)] - 0.241

   = 0.083 V

d) 39.40 ml of 0.015 M NaI added

Past equivalence point

initial Ag+ = 0.03 M x 15 ml = 0.45 mmol

NaI added = 0.015 M x 39.40 ml = 0.591 mmol

[Ag+] = Ksp/[I-] = 8.3 x 10^-17/2.6 x 10^-3 = 3.19 x 10^-14 M

E = (Eo + 0.0592 log[Ag+]) - 0.241

   = [0.79993 + 0.0592 log(3.19 x 10^-14)] - 0.241

   = -0.240 V

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