Map A Sapling Learning macmillan learning At 25 °C, you conduct a titration of 1
ID: 1055746 • Letter: M
Question
Map A Sapling Learning macmillan learning At 25 °C, you conduct a titration of 15.00 mL of a 0.0200 M AgNO3 solution with a 0.0100 M Nal solution within the following ce Saturated Calomel Electrode II Titration Solution I 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 EO 0.79993 V. The solubility constant of Agl is K 8.3 x 10-17 sp c) 30.00 mL a) 0.200 mL Number Number d) 45.00 mL b) 19.50 mL Number Number Previous 3 Give Up & View Solution Check Answer Next DC Exit HintExplanation / Answer
Answer:
For Redox titration involving precipitate formation EMF (E) is given as,
E = E0 + (0.059/n) log(Ksp/[X-].
For the given Redox titration,
# of electrons transfered = 1 and [X-] = [I-],
Hence we have,
E = E0 + (0.059) log(Ksp/[I-]).
We have Ksp for AgI = 8.3 x 10-17.
and E0 = 0.79993 V
Using these values we get,
E = 0.79993 + (0.059) log(8.3 x 10-17/[I-]).
For every given case we have to find out [I-] = ? M/L and then we calculate E = ?
Case-1)
For I- ions, M = 0.01 M and V = 0.2 mL = 0.2 x 10-3 L = 2 x 10-4 L.
[I-] = M x V = 0.01 x 2 x 10-4 = 2 x 10-6.
Using this value in eq. (1),
E = 0.79993 + (0.059) log(8.3 x 10-17/2 x 10-6)
E = 0.79993 + (0.059) log(8.3 x 10-11/2)
E = 0.79993 + (-0.61254)
E = 0.18739 V
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Case-2)
For I- ions, M = 0.01 M and V = 19.50 mL = 19.50 x 10-3 L = 1.95 x 10-2 L.
[I-] = M x V = 0.01 x 1.95 x 10-2 = 1.95 x 10-4.
Using this value in eq. (1),
E = 0.79993 + (0.059) log(8.3 x 10-17/1.95 x 10-4)
E = 0.79993 + (0.059) log(4.26 x 10-13)
E = 0.79993 + (-0.72986)
E = 0.07007 V
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Case-3)
For I- ions, M = 0.01 M and V = 30.0 mL = 30.0 x 10-3 L = 3 x 10-2 L.
[I-] = M x V = 0.01 x 3 x 10-2 = 3 x 10-4.
Using this value in eq. (1),
E = 0.79993 + (0.059) log(8.3 x 10-17/3 x 10-4)
E = 0.79993 + (0.059) log(2.77 x 10-13)
E = 0.05904 V
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Case-4)
For I- ions, M = 0.01 M and V = 45.0 mL = 30.0 x 10-3 L = 4.5 x 10-2 L.
[I-] = M x V = 0.01 x 4.5 x 10-2 = 4.5 x 10-4.
Using this value in eq. (1),
E = 0.79993 + (0.059) log(8.3 x 10-17/4.5 x 10-4)
E = 0.79993 + (0.059) log( 1.84 x 10-13)
E = 0.04855 V
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