Question

Consider the titration of 25.00 mL of 0.1250 M AgNO3 with 0.05012 M KI. ksp = AgI(s) = 8.3*10-17.

1)  For the same titration as shown in Question 2, please calculate the pAg value after 90.00 mL of KI is added.

2)  For the same titration as shown in Question 2, please calculate the volume of KI that is required to reach a pAg of 2.00.

Answer 1

1)

Moles of KI = 0.025 L X 0.05012 mol/L = 1.25 X 10^-3 moles Kl

When 90 mL have been added, Volume = 115 mL
Moles Ag+ added = 0.090 L X 0.1250 mol/L = 11.25 X 10^-3 moles


Assume that initially, all of the added Ag+ precipitates. You will be left with a Kl concentration of 3.5 X 10^-3 M
Then, some of the precipitated Agl will dissolve in that solution to give:
Ksp = 8.3 X 10^-17 = [Ag+] [3.5X10^-3]
[Ag+] = 3.2 X 10^-8 M
pAg = 5.28

After the addition of 25 mL of AgNO3, all of the Kl- has been precipitated since you have added an equal number of moles of Ag+ as you initially had of Kl-. So,

Ksp = [Ag+][Kl-] = 8.3 X 10^-17 M.

Here [Ag+] = [Kl-] So,
[Ag+]² = 8.3 X 10^-17
[Ag+] = 9.11043358e-9
pAg = 6.87