Question

The Ksp of AgI is 8.3× 10–17. You titrate 25.00 mL of 0.08970 M NaI with 0.05190 M AgNO3. Calculate pAg after the following volumes of AgNO3 are added:

1) 36.30 mL

2) Ve

3) 47.70 mL

Answer 1

AgNO3 + NaI = AgI + NaNO3 -

Moles of NaI = 25.0 ml X 0.0897 mol/L = 2.2425 m moles NaI

1 ) When 36.3 mL AgNO3 have been added, Volume = 61.3 mL
Moles Ag+ added = 36.3 ml X 0.0.0519 mol/L = 1.88397 m moles
Assume that initially, all of the added Ag+ precipitates. You will be left with a NaI concentration of 3.58 X 10^-3 M
Then, some of the precipitated AgI will dissolve in that solution to give:
Ksp = 8.3 X 10^-17 = [Ag+] [3.58X10^-3]
[Ag+] = 2.318 X 10^-14 M
pAg = 13.63

2) After the addition of equilibrium volume of AgNO3, all of the NaI has been precipitated since you have added an equal number of moles of Ag+ as you initially had of NaI. So,

Ksp = [Ag+][NaI] = 8.3 X 10^-17 M. Here [Ag+] = [NaI] So,
[Ag+]^2 = 8.3 X 10^-17
[Ag+] = 9.11X10^-9
pAg = 8.04

3) When 47.7 mL AgNO3 have been added, Volume = 72.2 mL
Moles Ag+ added = 47.7 ml X 0.0.0519 mol/L = 2.475 m moles
Assume that initially, all of the added NaI precipitates. You will be left with a Ag+ concentration of 2.325X 10^-4 M
Then, some of the precipitated AgI will dissolve in that solution to give:
Ksp = 8.3 X 10^-17 = [NaI] [2.325 x 10^-4]
[NaI] = 3.5698 X 10^-13 M
pNaI = 12.44
pAg = 14 - 12.44 = 1.55