At 25 degree C. you conduct a titration of 15.00 mL of a 0.0320 M AgNO_3 solutio

Question

At 25 degree C. you conduct a titration of 15.00 mL of a 0.0320 M AgNO_3 solution with a 0.0160 M Nal 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 Nal 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. 0.700 mL 30.00 mL 14.20 mL 37.50 mL

Explanation / Answer

Nernst equation to use,

E = [Eo - 0.0592/n logK] - 0.241

    = [0.79993 - 0.0592 log(1/[Ag+])] - 0.241

a) 0.7 ml NaI added

moles of AgNO3 = 0.032 M x 15 ml = 0.48 mmol

moles of NaI = 0.016 M x 0.7 ml = 0.0112 mmol

excess [Ag+] = (0.48 - 0.0112)mmol/15.7 ml = 0.03 M

E = [0.79993 - 0.0592 log(1/0.03)] - 0.241

    = 0.470 V

b) 14.20 ml NaI added

moles of AgNO3 = 0.032 M x 15 ml = 0.48 mmol

moles of NaI = 0.016 M x 14.20 ml = 0.2272 mmol

excess [Ag+] = (0.48 - 0.2272)mmol/29.2 ml = 0.0866 M

E = [0.79993 - 0.0592 log(1/0.0866)] - 0.241

    = 0.437 V

c) 30 ml NaI added

moles of AgNO3 = 0.032 M x 15 ml = 0.48 mmol

moles of NaI = 0.016 M x 30 ml = 0.48 mmol

equivalence point

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

E = [0.79993 - 0.0592 log((1/9.11 x 10^-9)] - 0.241

    = 0.083 V

d) 37.50 ml NaI added

moles of AgNO3 = 0.032 M x 15 ml = 0.48 mmol

moles of NaI = 0.016 M x 37.50 ml = 0.6 mmol

excess [I-] = (0.6 - 0.48)mmol/52.5 ml = 0.0023 M

[Ag+] = Ksp/[I-] = 8.3 x 10^-17/0.0023 = 3.63 x 10^-14 M

E = [0.79993 - 0.0592 log(1/3.63 x 10^-14)] - 0.241

    = -12.88 V

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