Which would better conduct electricity: a 2.00M solution of ammonium phosphate or a 3.00M lead (II) nitrate solution? Support your answer with a calculation.

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na

Ametal with a large number of free flowing electrons most likely has good conductivity

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.

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