Question

Consider the titration of 25.00 mL of 0.08313 M KI with 0.05210 M AgNO3. Calculate pAg+ at the following volumes of added AgNO3.

(a)    38.46 mL

(b)    V_e

(c)    43.31 mL

Ve is the volume at the equivalence point and pAg+=-log[Ag+]

Answer 1

Moles of I^- = 0.025L*0.08313mol/L = 2.07 X 10^-3 moles Cl^-

A) When 38.46 mL have been added, Volume = 38.46 mL

Mol Ag⁺ added = 0.03846 L X 0.05210 mol/L = 2 X 10^-3 mol

Assume that initially, all of the added Ag⁺ precipitates.

We will be left with a I^- concentration = 2.07*10^-3 - 2*10^-3

= 0.07 X 10^-3 M

Ksp = 8.3 X 10^-17= [Ag+] [0.07 X 10^-3]

[Ag⁺] = 118.57 X 10^-14 M

pAg⁺ = -log [Ag⁺]

pAg⁺ = 11.92

B) At equivalence point,

Number of moles of I^- = Number of moles of Ag+

So,

Ksp = [Ag⁺][I^-] = 8.3 X 10^-17 M [ Here [Ag+] = [Cl-] ]

[Ag⁺]² = 8.3 X 10^-17

[Ag⁺] = 9.1 X10^-9

pAg⁺ = -log [Ag⁺]

pAg⁺ = 8.040

C) Equivalence point will be at:

Ve = 2.07*10^-3 / 0.05210

= 0.03973 L

= 39.73 mL

At 43.31 i.e. beyond equivalence point,

So the amount past the equivalence point = 43.31 - 39.73 = 3.578 mL

Mol of [Ag+] = 3.578*10^-3*0.05210 = 1.86*10^-4

[Ag⁺] = 1.86*10^-4 mol / (0.025 + 0.04331) L

= 2.72*10^-3 M

pAg⁺ = -log [Ag⁺]

pAg⁺ = 2.56