At 25degreeC, you conduct a titration of 15.00 mL of a 0.0540 M AgNO_3 solution

Question

At 25degreeC, you conduct a titration of 15.00 mL of a 0.0540 M AgNO_3 solution with a 0.0270 M Nal 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 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. 0.600mL 30.00mL 17.10mL 44.90mL.

Explanation / Answer

Given :

T = 25 deg C = 298.15 K

Volume of AgNO3 = 15.00 mL

[AgNO3]= 0.0540 M

[NaI]= 0.0270 M

Reduction potential of saturated calomel electrode = 0.241 V

Ksp of AgI = 8.3 E-17

Solution

We know from the cell

Ecell = EAg/Ag+ - ESCE

Lets use Nernst equation

Ecell = E0cell – (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 0.0540= 0.00081 mol

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

= 0.00081 mol I-

Volume at equivalence point

= moles of I- / molarity of NaI
= 0.00081 mol I- / 0.027 = 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 0.0540 M / (0.015 +0.6 E-3 L )

= (0.030 L – 0.6 E-3 L )x 0.0540 / ( 0.0156 ) L

= 0.1017 M

Ecell = 0.79993 V – (0.0591 V / 1 ) log ( 1 / 0.1017 ) – 0.241 V

= 0.500 V

b) volume = 17.10 mL = 0.0171 L

[Ag+]= (0.030 L – 0.0171 L )x 0.0540 / ( 0.015+0.0171 ) L

= 0.0217 M

Ecell = 0.79993 V – (0.0591 V / 1 ) log ( 1 / 0.0217 ) – 0.241 V

= 0.4606 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

d)

44.90 mL

Ag+ is already converted to Ag(s)

Lets find concentration of I-

[I-] = {[(44.90 – 30.0) x 1E-3 L ]x 0.027 M }/ (0.015 + 0.0449 )L

= 0.0067 M

From Ksp

[Ag+] = ksp / [I-]

= 8.3 E-17 M / 0.0067

= 1.23E-14 M

Lets plug this value in Ecell

Ecell = 0.79993 V – (0.0591 V / 1 ) log ( 1 / 1.23E-14 ) – 0.241 V

= -0.263 V

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