At 25degree C, you conduct a titration of 15.00 mL of a 0.0220 M AgNO_3 solution

Question

At 25degree C, you conduct a titration of 15.00 mL of a 0.0220 M AgNO_3 solution with a 0.0110 M Nal solution within the following cell: Saturated Calomel Electrode || Tration 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. a) 0.900 mL b) 18.10 mL c) 30.00 mL d) 44.60 mL

Explanation / Answer

Reduction half reaction:

Ag+ (aq) + e- -> Ag (s)

E0red = 0.79993 V

Oxidation half reaction:

Hg (l) + Cl- -> ½ Hg2Cl2(s) + e- ->

E0ox = - 0.241 V

E0cell = E0red + E0ox - 0.05916 log (1/ [Ag+])

E0cell = 0.79993 – 0.241 + 0.05916 log ([Ag+])

= 0.55893 + 0.05916 log ([Ag+])

a)

Let x mmoles of AgI precipitate.

mmoles of Ag+ in solution = 15 mL * 0.022 M – x

= 0.33 – x

mmoles of I- in solution = 0.9 * 0.011 – x

= 0.0099 – x

Solution volume = 15 mL + 0.9 mL = 15.9 mL

Ksp = 8.3 x 10-17

(0.33 – x) (0.0099 – x) = Ksp = 8.3 x 10-17

Solving we get x = 0.0099

[Ag+] = (0.33 – x) / 15.9 M = 0.02 M

E0cell = 0.55893 + 0.05916 log (0.02)

= 0.458 V

b)

Let x mmoles of AgI precipitate.

mmoles of Ag+ in solution = 15 mL * 0.022 M – x

= 0.33 – x

mmoles of I- in solution = 18.1 * 0.011 – x

= 0.1991 – x

Solution volume = 15 mL + 18.1 mL = 33.1 mL

Ksp = 8.3 x 10-17

(0.33 – x) (0.1991 – x) = Ksp = 8.3 x 10-17

Solving we get x = 0.1991

[Ag+] = (0.33 – x) / 33.1 M = 0.003955 M

E0cell = 0.55893 + 0.05916 log (0.003955)

= 0.417 V

c)

Let x mmoles of AgI precipitate.

mmoles of Ag+ in solution = 15 mL * 0.022 M – x

= 0.33 – x

mmoles of I- in solution = 30 * 0.011 – x

= 0.33 – x

Solution volume = 15 mL + 30 mL = 45 mL

Ksp = 8.3 x 10-17

(0.33 – x) (0.33 – x) = Ksp = 8.3 x 10-17

Solving we get x = 0.32999959

[Ag+] = (0.33 – x) / 33.1 M = 9.1 x 10-9 M

E0cell = 0.55893 + 0.05916 log (9.1 x 10-9)

= 0.083 V

d)

Let x mmoles of AgI precipitate.

mmoles of Ag+ in solution = 15 mL * 0.022 M – x

= 0.33 – x

mmoles of I- in solution = 44.6 * 0.011 – x

= 0.4906 – x

Solution volume = 15 mL + 44.6 mL = 59.6 mL

Ksp = 8.3 x 10-17

(0.33 – x) (0.4906 – x) = Ksp = 8.3 x 10-17

Solving we get x = 0.32999999999816

[Ag+] = (0.33 – x) / 33.1 M = 3.09 x 10-14 M

E0cell = 0.55893 + 0.05916 log (3.09 x 10-14)

= - 0.240 V

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