In a solution of .200M NaI, solid AgI was placed in. Calculate the solubility if AgI (Ksp=1.5*10^-16)

Answer 1

Assuming complete dissociation of NaI as it is an ionic compound,

NaI --> Na^+ + I^-
x x (moldm^-3)
and,
AgI <==> Ag^+ + I^-
y y (moldm^-3)

y=solubility of AgI

So the total [I^-]=(x+y) moldm^-3

But, Ksp AgI = [Ag^+][I^-]=y(x+y) mol^2dm^-6 = 1.5*10^-16 mol^2dm^-6

Hope this helps..

(Please note that I haven't mentioned the state of the above compounds)

Answer 2

Thank you so much!