I assume you are adding I^- drop wise.
(Ag^+)(I^-) = 8.5E-17
(Tl^+)(I^-) = 5.5E-8
Adding I^- drop wise will ppt AgI when (I^-) = Ksp/(Ag^+) = 8.5E-17/0.035 = 2.41E-15 M
AgI will continue pptng until Ksp for TlI is exceeded. That will be when
(I^- ) = Ksp/(Ag^+) = 5.5E-8/0.035 = 1.57E-6. This is the (I^-) when TlI first ppts. What is the (Ag^+) at that point?It is
(Ag^+) = Ksp/(I^-) = 8.5E-17/1.57E-6 = 5.41E-11. That is what is left of the 0.035 mols Ag+ from AgI initially.
% Ag^+ remaining = (5.41E-11/0.035)*100 = ?
Check these numbers carefully for typos etc. My calculator is on the blink.
The Ksp for AgI is 8.5 x 10-17.
The Ksp for TlI (thallium iodide) is 5.5 x 10-8.
A one litre solution contains 0.035 mol L-1 Ag+ ion and Tl+ ion.
When TlI begins to precipitate, what percentage of Ag+ remains in the solution?
1 answer