Lead(II) iodide is considered insoluble in water, which means that the equilibrium that is established when this salt is dissolved in water will lie significantly to the left, i.e. most of the salt will remain undissolved.
#"PbI"_ (color(red)(2)(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + color(red)(2)"I"_ ((aq))^(-)#
The solubility product constant,
#K_(sp) = ["Pb"^(2+)] * ["I"^(-)]^color(red)(2)#
Now, the molar solubility of the salt represents the maximum concentration of lead(II) iodide that dissociates in water to produce a saturated solution.
If you take
The solubility product constant will thus be equal to
#K_(sp) = s * (color(red)(2) * s)^color(red)(2)#
#K_(sp) = 4 s^3#
Rearrange to solve for
#s = root(3)(K_(sp)/4)#
Plug in your value to find
#s = root(3)((8.8 * 10^(-5))/4) = color(darkgreen)(ul(color(black)("0.028 M")))#
The answer is rounded to two sig figs, the number of sig figs you have for the
So, what does this mean?
This tells you that at room temperature, you can only dissolve