[fai]

#1
Q: 20.00 mL of 0.0532 M KBr was titrated with 0.0511 M AgNO₃. Calculate pAg⁺ at the following volumes of AgNO₃ added:
a) 20 mL
b) equivalence point
c) 22.60 mL

K_sp of AgBr is 5.0 x 10^-13. So I'm assuming the question is asking for the % moles of individual Ag⁺ ions are in the solution with the K_sp of AgBr considered.


#2
Q: taking into account the true activities, calculate concentration of Tl⁺ in a saturated solution of TlBr in water (K_sp = 3.6x10^-6). (you will have to approximate the concentration of the ions before calculating activity coefficient)

Rough concentration based on Ksp is 0.0019M for Tl⁺ and Br^-
ionic strength = (0.5)(0.0019*1²+0.0019*1²) = ~0.002 M
activity coefficient = 10^{-0.509(12+12)(0.002)1/2} = 0.949

now i'm stuck

[Borek]

#1 - http://www.titrations.info/precipitation-titration-curve-calculation

#2 - so far so good. Use these activity coefficients to calculate again concentrations of Tl⁺ and Br^-. Reepeat whole procedure. Results will converge.

[fai]

Actually activity coefficient should be = 10^{-0.509(12)(0.002)1/2} = 0.949 because it's only one ion.

so then I set:

K_sp = K_sp' [γ_Tl⁺] [γ_Br^-]

(K_sp)/([γ_Tl⁺] [γ_Br^-]) = K_sp' = [Tl⁺][Br^-]

(3.6x10^-6)/(0.949*0.949) = [Tl⁺][Br^-]

correct..?

[Borek]

You have not named your symbols, which makes it difficult to follow, but

(3.6x10^-6)/(0.949*0.949) = [Tl⁺][Br^-]

looks OK. Now, calculate new concentrations of ions in the solution.