Question

The Ksp of AgI is 8.3× 10–17. You titrate 25.00 mL of 0.08870 M NaI with 0.05050 M AgNO3. Calculate pAg after the following volumes of AgNO3 are added: (a) 37.20 ml (b) Veq (c) 47.20 ml

Answer 1

(a) 37.20 ml

molarity of NaI in the solution = millimoles of NaI / total volume

                                             = 25 x 0.08870 / 25 + 37.20

                                             = 0.03565 M

molarity of AgNO3 in the solution = 0.05050 x 37.20 / 25 + 37.20

                                                      = 0.03020 M

NaI           + AgNO3 -------------------> AgI + NaNO3

0.03565     0.03020                          

AgNO3 completely consumed

remaining NaI = 0.03565 -0.03020 = 5.45 x 10^-3 M

AgI ---------------------> Ag+ + I-

Ksp = [Ag+][I-]

8.3 x 10^-17 = [Ag+] (5.45 x 10^-3)

[Ag+] = 1.52 x 10^-14 M

pAg = -log [Ag+]

pAg = 13.8

b) Veq

Ksp = 8.3 x 10^-17 = S^2

S = [Ag+] = 9.11 x 10^-9

pAg = 8.04

c) 47.20 mL

molarity of AgNO3 = 0.0330 M

[I]= 2.64 x 10^-3 M

[Ag+] = 3.15 x 10^-14 M

pAg = 13.5