What is the solubility in moles/liter for lead(II) sulfide at 25 oC

Question

What is the solubility in moles/liter for lead(II) sulfide at 25 ^oC given a K_sp value of 8.8 x 10^-29. 

The answer is 9.38*10^-15 or 9.4*10^-15

Answer 1

This problem asks you to find the solubility of an "insoluble" compound. When you are given a K_sp value, it means that when you put the compound in water, it will just settle to the bottom, but... a tiny tiny amount will dissolve. (If you go into dental school, you may study the solubility of tooth enamel; tooth enamel is "insoluble", but it can gradually dissolve... ever had a tooth cavity?)

To find the solubility of an "insoluble" compound, you write the equilibrium reaction dissociating into its ions and use the R.I.C.E method...

R reaction: PbS_(s) <==> Pb²⁺_(aq) + S^2-_(aq)

I initial: solid 0 moles/L 0 moles/L

C change: +x +x

E equilibrium: x moles/L x moles/L

K_sp = [Pb²⁺] [ S^2- ]

8.8 x 10^-29 = [x] [x]

8.8 x 10^-29. = x²

9.38 x 10^-15 M = x

Answer 2

PbS <-----> Pb²⁺ + S-

Ksp = [Pb²⁺]*[S^-]

8.8 x 10^-29 = x²

x = 9.38 x 10^-15 M