Question

The molar solubility of silver iodide in a 0.186 M potassium iodide solution is __________ M.

Answer 1

.....AgI ==> Ag^+ + I^-
I....solid...0......0
C....solid...x......x
E....solid...x......x

....KI ==> K^+ + I^- (100% ionized)
I..0.186...0.....0
C.-0.186..0.186..0.186
E...0.....0.186..0.186

Ksp = (Ag^+)(I^-)
Ksp = look up
(Ag^+) = x from above
(I^-) = 0.186 + x. 0.186 from the KI and x from the AgI.
Solve for x.

Answer 2

0.186M(KI) => 0.186(K^+) + 0.186(I^-)
------- AgI <=> Ag^+ + I^-
C(eq) ----------x-----0.186M

Ksp = [Ag^+][I^-] = (x)(0.186M)
Ksp(AgI) = 8.3x10^-17 (fm table of Ksp values) = (x)(0.186M)

x = Solubility in presence of 0.186M(KI) = (8.3x10^-17/0.186)M = 4.5x10^-16M

In pure water Solubility = Sqr-Root(Ksp) = 9.1x10^-9M

Here's one for the ole gray matter... What's the solubility of AgI in the presence of 1.0M NH3? (Kf(Ag(NH3)2^+) = 1.7x10^+7, Ksp = 8.3x10^-17). Check out Complex-Ion equilibrium.