6) A 0.160 mol sample of PCls(g) is put in a rigid 1.0 L container at an unknown temperature. After a period of time, the system comes to equilibrium. A graph
showing the concentration of PCls(g) as a function of time has been plotted
below.
PCls(g) = PC13(g) + Cl2(g)
0.160
Concentration (mol/L)
0.093
[PCI, ]
I
Time
a. What was the concentration of PCls when the system established
equilibrium?
b. Find the concentrations of PCl3 and Cl2 at equilibrium. (Hint: Use an ICE
chart with the initial and equilibrium concentrations of PCI).
c. On the above graph, sketch curves that show how the concentrations of
PClz and Cl, change over time.
d. Calculate the value of the equilibrium constant K for the reaction at this
temperature.

The answer is not c

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.