Question

a) A 0.1500 M of AgNO₃ solution was employed to titrate a 25.00 mL of 0.1250 M of NaI and 0.2500 M NaCl. Given that Ksp, AgI(s) = 8.3*10^-17, and Ksp, AgCl(s) = 1.8*10^-10, please calculate the concentration of Ag⁺ ion after 6.00 mL of AgNO₃ was added.

b) please calculate the pAg after 100.00 mL of AgNO₃ was added

Answer 1

you would have 0.025 L x 0.1250 mol/L = 0.003125 mol of Nal ions
and 0.025 L * 0.2500 mol/L = 0.00625 mol Nacl ions
then
total volume is 0.050 L after mixing
so
[Nacl] = 0.00625 mol / 0.050 L = 0.125 M
[Nal] = 0.003125 mol / 0.050 L = 0.0625 M

Ksp = [Agl][Agcl]
AgI(s) = 8.3*10^-17, and AgCl(s) = 1.8*10^-10
  8.3*10^-17 x 0.0625 = 81.9375

When 6 mL have been added, Volume = 31 mL
Moles Ag+ added = 0.006 L X 0.100 mol/L = 0.0006 moles
Assume that initially, all of the added Ag+ precipitates. You will be left with a Cl- concentration of 3.5 X 10^-3 M
some of the precipitated AgCl will dissolve in that solution to give:
Ksp = 1.82 X 10^-10 = [Ag+] [3.5X10^-3]
[Ag+] = 5.2 X 10^-8 M
pAg = 7.28