At 25 °C, you conduct a titration of 15.00 mL of a 0.0240 M AgNO3 solution with

Question

At 25 °C, you conduct a titration of 15.00 mL of a 0.0240 M AgNO3 solution with a 0.0120 M Nal solution within the following cell: Saturated Calomel Electrode I| 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 E9-079993 V. The solubility constant of Agl is Ksp-8.3 x 10, ) 0.500 mL c) 30.00 mL Number Number b) 16.70 mL d) 39.20 mL Number Number

Explanation / Answer

Titration

Nernst equation,

Ecell = Eo - 0.0592/n logQ

For the given cell,

Ecell = 0.79993 + 0.0592 log[Ag+] - 0.241

a) 0.5 ml NaI added

initial Ag+ = 0.024 M x 15 ml = 0.360 mmol

I- added = 0.012 M x 0.5 ml = 0.006 mmol

[Ag+] remaining = 0.354 mmol/15.5 ml = 0.023 M

Ecell = 0.79993 - 0.0592 log(0.023) - 0.241

         = 0.463 V

b) 16.70 ml

initial Ag+ = 0.024 M x 15 ml = 0.360 mmol

I- added = 0.012 M x 16.70 ml = 0.2004 mmol

[Ag+] remaining = 0.1596 mmol/31.7 ml = 0.005 M

Ecell = 0.79993 - 0.0592 log(0.005) - 0.241

         = 0.423 V

c) 30 ml

initial Ag+ = 0.024 M x 15 ml = 0.360 mmol

I- added = 0.012 M x 30 ml = 0.360 mmol

Equivalence point

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

Ecell = 0.79993 - 0.0592 log(9.11 x 10^-9) - 0.241

         = 0.083 V

d) 39.20 ml

initial Ag+ = 0.024 M x 15 ml = 0.360 mmol

I- added = 0.012 M x 39.20 ml = 0.4704 mmol

[I-] remaining = 0.1104 mmol/54.2 ml = 0.00204 M

[Ag+] = Ksp/[I-] = 8.3 x 10^-17/0.00204 = 4.1 x 10^-14 M

Ecell = 0.79993 - 0.0592 log(4.1 x 10^-14) - 0.241

         = -0.234 V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.