Question

A 20.0 mL solution of 0.1004 M KI is titrated with a 0.0845 M solution of AgNO3. What are the voltages after adding 1.0, 15.0, and 25.0 mL of AgNO3? Ksp = 8.3x10^-17

Answer 1

The reaction occurring is

Ag⁺ comes from AgNO₃ and I^- comes from KI.

After adding 1.0 mL of AgNO₃,

Initial I^- = (20 mL)*(0.1004 M) = 2.008 mmol

Initial Ag⁺ = (1 mL)*(0.0845 M) = 0.0845 mmol

In order to find the limiting reagent, we divide the moles/millimoles by the respective stoichiometric coefficient in the balanced equation. The reagent with the lower number is the limiting reagent. Here, the coefficients are both 1 and thus we can directly use the number of millimoles. Ag^₊ is the limiting reagent. The information of limiting reagent is crucial because the other reagent is in excess. Constructing an ICE chart,

	Ag+	I-	AgI
I(Initial)	0.0845	2.008	0
C(Change)	-0.0845	-0.0845	+0.0845
E(Equilibrium)	0	1.9235	0.0845

Concentration of [I^-] = millimoles of I^-/Volume(in mL)

[I^-] = 1.9235/21 = 0.092 M (Note that the total volume after 1 mL addition of AgNO₃ is 21 mL)

K_sp = [Ag⁺][I^-] thus (8.3 X 10-17)/(0.092) = [Ag⁺]

[Ag⁺] = 9.02 X 10^-16 M

From the Nerst equation, E = E^o - (0.0592/n)log(1/[Ag⁺])

Here n = 1 as 1 electron is involved in the process and E^o = 0.799 V that is the reduction potential of Ag.

Substituting the values, E = 0.799 - 0.0592log(1/(9.02 X 10^-16))

E = -0.091 V

After adding 15.0 mL of AgNO₃,

Initial I^- = (20 mL)*(0.1004 M) = 2.008 mmol

Initial Ag⁺ = (15 mL)*(0.0845 M) = 1.2675 mmol

The limiting reagent is Ag⁺. Constructing an ICE chart,

	Ag+	I-	AgI
I	1.2675	2.008	0
C	-1.2675	-1.2675	1.2675
E	0	0.7405	1.2675

[I^-] = 0.7405/35 = 0.0211 M

K_sp = [Ag⁺][I^-] thus (8.3 X 10-17)/(0.0211) = [Ag⁺]

[Ag⁺] = 3.93 X 10^-15 M

From the Nerst equation, E = E^o - (0.0592/n)log(1/[Ag⁺])

E = 0.799 - 0.0592log(1/(3.93 X 10^-15))

E = -0.054 V

After adding 25.0 mL of AgNO₃,

Initial I^- = (20 mL)*(0.1004 M) = 2.008 mmol

Initial Ag⁺ = (25 mL)*(0.0845 M) = 2.1125 mmol

We are beyond the equivalence point as mmol of Ag⁺ have crossed the mmol of I^-.

For the equivalence point, the volume of AgNO₃ added must have been 2.008/0.0845 = 23.76 mL

Excess Ag⁺ = (25.0 - 23.76)*(0.0845) = 0.105 mmol

[Ag⁺] = 0.105/45 = 2.33 X 10^-3 M (Note that the total volume is 45 mL)

From the Nerst equation, E = E^o - (0.0592/n)log(1/[Ag⁺])

E = 0.799 - 0.0592log(1/(2.33 X 10^-3))

E = 0.643 V