Titration curve(Xác định pAg+ trong quá trình chuẩn độ)

AgI  Ag+ + IKsp = 8.3 x 10^-17
(1) 25.00 mL of 0.100M I- titrated with 0.05M Ag+
Moles of I- = 0.1 x 0.025 = 0.0025 mol
**When we do not add more Ag+
Ksp = [Ag+][I-] = 8.3 x 10^-17
 [Ag+] = 8.3 x 10^-16
 pAg+ = 15.08
**When we add 10.00 ml of Ag+
 Moles of Ag+ added = 0.05 x 0.01 = 0.005 mol
Moles of I- = Original moles of I- - Moles of Ag+ added
= 0.0025 – 0.0005 = 0.002 mol
The volume after adding = 25 + 10 = 35 ml = 0.035 L
 [I-] =
So : [Ag+] =
 pAg+ = -log[Ag+] = 14.84
In the equivalence point
[Ag+] = [I-] = x
Ksp = [Ag+][I-] = x^2 = 8.3 x 10^-17
 x = 9.1 x 10^-9
 pAg+ = 8.04
After the equivalence point


There is no I- to consume Ag+
If we add more 10ml of Ag+
Moles of Ag+ = 0.01 x 0.05 = 0.0005 mol
 [Ag+] = 5.9

x 10^-3
 pAg+ = 2.23