Atomic absorption spectroscopy was used to determine the amount of aluminum in a sample of sunscreen. A standard solution of 1.00 ng Al /L was prepared and measured seven times to determine the detection and quantitation limits of the spectrometer. The standard measurements were: 0.72, 0.97, 1.40 , 1.21 , 0.79 , 1.17 , and 1.08 ng Al /L. Calculate the detection and quantitation limits for this instrument. detection limit: ng/L quantitation limit: ng/L

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.