The room-temperature electrical conductivity of a semiconductor specimen is 2.8 × 10 4 (ω∙m)–1. if the electron concentration is 2.9 × 1022 m–3 and electron and hole mobilities are 0.14 and 0.023 m2/v∙s, respectively, calculate the hole concentration. (a) 1.24 × 1024 m–3 (b) 7.42 × 1024 m–3 (c) 7.60 × 1024 m–3 (d) 7.78 × 1024 m–3

The correct answer is d

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.