Consider the reaction IO−4(aq)+2H2O(l)⇌H4IO−6(aq);Kc=3.5× 10−2IO4−(aq)+2H2O(l)⇌H4IO6−(aq);Kc= 3.5×10−2 If you start with 25.0 mLmL of a 0.909 MM solution of NaIO4NaIO4, and then dilute it with water to 500.0 mLmL, what is the concentration of H4IO−6H4IO6− at equilibrium?

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.

Answer(28.02g/mol);

The second and last option are the answers