Question

How would you answer this question step by step ?
Calculate the concentration of [CrO4]2- ions in a saturated solution of lead chromate PbCrO4
K(s) [PbCrO4] = 2.5x10^-13

Answer 1

since lead chromate is ionic, it will dissociate in water into the two ions: Pb(+2) and CrO4(-2). Even though it's "insoluble", a very small amount of this salt will dissolve in water, and what does dissolve will dissociate. So what you really have is a system at equilibrium:\[PbCrO_4(s) \rightleftharpoons Pb^{+2}(aq) + CrO_4^{-2}(aq)\]
there will be an equilibrium constant (K), describing this equilibrium:\[K_{SP} = \frac{[Pb^{+2}][CrO_4^{-2}]}{[PbCrO_4(s)]}\] since the concentration of solids never changes, we can ignore the denominator here and just call it \[K_{SP} = [Pb^{+2}][CrO_4^{-2}]\] The value of this equilibrium constant is given: 2.5x10^(-13), we can solve for the concentration of chromate ions.
The concentration of the 2 ions has to be the same, because of the stoichiometric ratio from the balanced equation, so [Pb+2] = [CrO4-2], which we can call some value "x". THis means that\[K_{SP} = [Pb^{+2}][CrO_4^{-2}] = [x]*[x] = 2.5*10^{-13}\] solve for X and you have the concentration of chromate ions (and the lead ions, as well) in a saturated solution of lead (II) chromate.

I get about 5.0x10^(-7)