Example #6: A metal cup of surface area 200. cm- needs to be electroplated with silver to a thickness of 0.200 mm. The density of silver is 1.05 x 104 kg m 3 The
mass of a silver ion is 1.79 x 10-25 kg and the charge is the same magnitude as that on
an electron. How long does the cup need to be in the electrolytic tank if a current of
12.5 A is being used?

Ithink is a because if say increased in electronegativity

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.

Answer246 k;