How many moles of AgIO3 (Ksp=3.1e-8) will dissolve in one liter of a 1e-5 M solution of NaIO3?
Can anyone explain how to solve this?
this is a
common ion question. the idea is, that only so much AgIO3 can dissolve in water based on the product of the ions. However, if you already have ions in solution, it will shift the equilibrium left, allowing for less AgIO3 to dissolve. Watch how this manifests itself in the journey to the answer.
first recognize that any salt with Na is ALWAYS soluble. so NaIO3 --> Na+ + IO3-
now you know that the molarity of IO3- is 1e-5 M
now you set up ksp formula for AgIO3
AgIO3 (s) --> Ag+ (aq) + IO3- (aq)
ksp = 3.1e-8 = [Ag+][IO3-]
you have 1e-5 of IO3- already in solution, in highschool we leanred about an "ICE CHART" ICE --> Initial, Change (to account for stoichiometric ratios), Equilibrium
AgIO3 (s) --> Ag+ (aq) + IO3- (aq)
I A 0 1e-5
C -x +x +x
E A-x x 1e-5 + x
ksp = 3.1e-8 = [Ag+][IO3-] = (1e-5 +x)(x)
they won't ask you to solve a quadratic.. and so you must assume that x is negligible compared to 1e-5
3.1e-8 = (1e-5)(x)
x = [Ag+] = 3.1e-8/1e-5 M
x =~ 3e-3 M
it is in 1 L, so n = 3e-3 mols. And for every 1 mol of Ag+ produced, you had reacted 1 mol of AgIO3 (stoichiometric ratio).. then 3e-3 mols of AgIO3 dissolved.
hope this is right, and hope this helps.
doesn't Kaplan give explanations to their answers?
