Adding iodide ion will precipitate the minute amount of silver ion as AgI and more importantly, shift the equilibrium to the right producing more silver ion and chloride ion. The silver ion reacts with iodide ion to produce the more insoluble silver iodide, thus reducing the concentration of iodide ion.
AgCl(s) + I- --> AgI(s) + Cl-
The resulting iodide ion concentration is the square root of the Ksp of AgI, as shown below.
If the addition of the AgCl(s) doesn't change the volme of the liquid, then the concentration of I{-} ions will be the same as the concentration of KI, 0.15 M.
Answers & Comments
Silver iodide is about 2 million times more insoluble than silver chloride and so the silver chloride will dissolve and form silver iodide.
AgCl(s) <==> Ag+ + Cl- .............. Ksp = 1.8x10^-10
AgI(s) <==> Ag+ + I- ................... Ksp = 8.5x10^-17
Adding iodide ion will precipitate the minute amount of silver ion as AgI and more importantly, shift the equilibrium to the right producing more silver ion and chloride ion. The silver ion reacts with iodide ion to produce the more insoluble silver iodide, thus reducing the concentration of iodide ion.
AgCl(s) + I- --> AgI(s) + Cl-
The resulting iodide ion concentration is the square root of the Ksp of AgI, as shown below.
AgI(s) <==> Ag+ + I- .............. Ksp = 8.5x10^-17
Ksp = [Ag+][I-]
8.5x10^-17 = x²
x = 9.2x10^-9
[I-] = 9.2x10^-9M
If the addition of the AgCl(s) doesn't change the volme of the liquid, then the concentration of I{-} ions will be the same as the concentration of KI, 0.15 M.
Idk