eston -CHEM 251-Fall17-HAAS Activities and Due Dates Ch15 Electrochemical Probes

Question

eston -CHEM 251-Fall17-HAAS Activities and Due Dates Ch15 Electrochemical Probes 1/30/2017 11:00 PM 64.3/100 Grade Print -Cal ulator Periodic Table stion 14 of 14 Sapling Learning Map db At 25 , you conduct a titration of 15.00 mL of a 0.0260 M AgNO3 solution with a 0.0130 M Nal solution within the following cell: Saturated Calomel Electrode ll 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- Ag(s) is E - 0.79993 V. The solubility constant of Agl is Kup 8.310 a) 0.600 mL c) 30.00 ml Number Number b) 14.80 mL d) 38.20 mlL Number Number Hint OPrevous Grve Up & View Soldon 4Check Answer @Next t] Ex6 about us | careers privacy policyterms of usecontact us

Explanation / Answer

Titration

(a) after 0.6 ml of NaI is added

initial AgNO3 = 0.0260 M x 15 ml = 0.390 mmol

moles NaI added = 0.0130 M x 0.6 ml = 0.0078 mmol

[Ag+] remained = (0.390 - 0.0078) mmol/15.6 ml = 0.0245 M

Using,

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

    = 0.79993 + log(0.0245) - 0.241 = 0.463 V

(b) after 14.80 ml of NaI is added

initial AgNO3 = 0.0260 M x 15 ml = 0.390 mmol

moles NaI added = 0.0130 M x 14.80 ml = 0.1924 mmol

[Ag+] remained = (0.390 - 0.1924)mol/29.80 ml = 0.0066 M

Using,

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

    = 0.79993 + log(0.0066) - 0.241 = 0.430 V

(c) after 30 ml of NaI is added

initial AgNO3 = 0.0260 M x 15 ml = 0.390 mmol

moles NaI added = 0.0130 M x 30 ml = 0.390 mmol

Equivalence point

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

Using,

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

    = 0.79993 + log(0.0245) - 0.241 = 0.083 V

(d) after 38.20 ml of NaI is added

initial AgNO3 = 0.0260 M x 15 ml = 0.390 mmol

moles NaI added = 0.0130 M x 38.20 ml = 0.4966 mmol

[I-] remained = (0.4966 - 0.390) mmol/53.20 ml = 0.002 M

[Ag+] = Ksp/[I-] = 8.3 x 10^-17/0.002 = 4.15 x 10^-14 M

Using,

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

    = 0.79993 + log(4.15 x 10^-14) - 0.241 = -0.233 V

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