In Lab 3, Solubility Rules, students identified ionic salts as soluble or insoluble. The solubility of ionic salts is quantified by an equilibrium constant called Ksp, the solubility product. This Ksp refers to the dissolution reaction of the ionic salts.

(a) Write the dissolution reaction for the ionic solids below. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)

1. BiPO4
BiPO4(s) ? Bi3+(aq) + PO43-(aq)
(it’s wrong)
2. ZnCO3
ZnCO3(s) ? Zn2+(aq) + CO32-(aq)
(it’s wrong)
(b) The concentration of cations and anions in solution can be predicted from the equation

[cation] = sp

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.