Question

At 25 C, you conduct a titration of 15.00 mL of a 0.0400 M AgNO3 solution with a 0.0200 M NaI 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 NaI 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 E0 = 0.79993 V. The solubility constant of AgI is Ksp = 8.3 x 10^-17.

.600 mL = ______ V

10.30 mL= ______ V

30.00 mL= ______V

38.80 mL= ______V

Answer 1

Given :

T = 25 deg C = 298.15 K

Volume of AgNO3 = 15.00 mL

[AgNO3]= 0.0400 M

[NaI]= 0.0200 M

Reduction potential of saturated calomel electrode = 0.241 V

Ksp of AgI = 8.3 E-17

Solution

We know from the cell

Ecell = E_Ag/Ag+ - E_SCE

Lets use Nernst equation

Ecell = E⁰_cell – (0.0591 v / n ) log (1/[Ag+] )

Lets use above given equation and write Ecell of overall reaction

Ecell = 0.79993 V – ( 0.0591 V / 1 ) log ( 1/[Ag+] ) – Esce

Calculation of voltage after addition of following volume.

1). 0.600 mL

Ecell = 0.79993 V – ( 0.0591 V / 1 ) log ( 1/[Ag+] ) – Esce

We have to find volume at equivalence point.

Number of moles of Ag+ = Volume in L x molarity = 0.015 L x 0.0400= 0.0006 mol

Moles of NaI = moles of I - = Moles of Ag+ x 1 mol I- / 1 mol Ag+

= 0.0006 mol I-

Volume at equivalence point

= moles of I- / molarity of NaI
= 0.0006 mol I- / 0.0200 = 0.030 L

= 30.0 mL

There is a reaction between I- and Ag+ to form AgI

The given volume of first addition is 0.600 mL.

There will be Ag+ remains in the solution after addition of 0.600 mL of NaI

Lets find concentration of Ag+

[Ag+] = mol Ag+ / volume of solution

= Volume of Ag+ in L x Molarity of Ag+ / Volume of solution

= (volume at equivalence point – volume of NaI-) x 0.0400 M / (0.015 +0.6 E-3 L )

= (0.030 L – 0.6 E-3 L )x 0.0400 / ( 0.0306 ) L

= 0.961 M

Ecell = 0.79993 V – (0.0591 V / 1 ) log ( 1 / 0.961 ) – 0.241 V

= 0.558 V

b) volume = 10.30 mL = 0.0103 L

[Ag+]= (0.030 L – 0.0103 L )x 0.0400 / ( 0.015+0.0103 ) L

= 0.03115 M

Ecell = 0.79993 V – (0.0591 V / 1 ) log ( 1 / 0.03115 ) – 0.241 V

= 0.47 V

c) volume 30.0 mL

At 30.0 mL

This is equivalence point

Here all the Ag+ are converted to Ag(s)

Lets find concentration of Ag+ by using Ksp

Ksp = [Ag+][I-]= 8.3E-17

Lets find x (molar solubility )

8.3 E-17 = x²

x = sqrt ( 8.3 E-17 )

= 9.11 E-9 M

Ag+ = 9.11 E -9 M

Ecell = 0.79993 V – (0.0591 V / 1 ) log ( 1 / 9.11E-9 ) – 0.241 V

= 0.084 V

c)

38.80 mL

Ag+ is already converted to Ag(s)

Lets find concentration of I-

[I-] = {[(38.80 – 30.0) x 1E-3 L ]x 0.0200 M }/ (0.015 + 0.0388 )L

= 0.0033 M

From Ksp

[Ag+] = ksp / [I-]

= 8.3 E-17 M / 0.0033

= 2.53E-14 M

Lets plug this value in Ecell

Ecell = 0.79993 V – (0.0591 V / 1 ) log ( 1 / 2.53E-14 ) – 0.241 V

= -0.245 V