Question

Calculate the solubility of silver chloride in 0.17M AgNO3. (Ksp=1.8E-10)

Answer 1

AgNO3 dissociates into Ag+ and NO3-, therefore our Ksp will be: Ksp = [Ag+][NO3-] because AgNO3 is a solid and solids & liquids are never counted in equilibrium problems because their concentrations are always constant.
X moles of the silver chloride will dissociate which means X moles of both Ag+ and NO3- will be in the solution. To do this, I always set up a chart.

AgNO3(s) -> Ag+ NO3-
.17 0 0 Initial concentration
-x +x +x Change in concentration
.17-x x x Equilibrium concentration

Now let's take the equilibrium values and then plug them into our Ksp.
Ksp = 1x10-8
Ksp = [Ag+][NO3-]
1x10-8 = [Ag+][NO3-] *Now recall our equilibrium values.
1x10-8 = (x)(x)
1x10-8 =x^2
x = The square root of 1x10-8 which is 1x10-4