Determine how many moles of ba(oh)2 are required to completely neutralize 0.117 mole of hc2h3o2.

HC₂H₃O₂ (CH₃COOH) - acetic acid

2CH₃COOH + Ba(OH)₂ > (CH₃COO)₂Ba + 2H₂O
2 mol              1 mol

2 mol CH₃COOH 1 mol Ba(OH)₂
0,117 mol CH₃COOH X
X = 0,117/2
X = 0,0585 mol  Ba(OH)₂

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.