Asked by Darcy
In a solution of .200M NaI, solid AgI was placed in. Calculate the solubility if AgI (Ksp=1.5*10^-16)
Answers
Answered by
Shenaya
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)
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)
Answered by
Darcy
Thank you so much!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.