The reaction ag+(aq) + cl−(aq) ↔ agcl(s) has an equilibrium of 1020 (keq = 1020). if you have a beaker containing 1 liter of water with 0.1 mol of agcl(s) in it along with 10−6 m ag+(aq) and 10−15 m cl−(aq), how does the reaction shift?

When Q is equal the initial concentration of the products / the initial concentration of the reactants. 

so, Q = [Ag]*[Cl-]  and we neglected [AgCl] as it is solid

∴ Q = 10^-6 * 10^-5 

       = 10^-11 

now we will compare the value of Q with the value of Keq:

when Q = Keq so, the system is in equilibrium

and when Q > Keq so, the reaction will go forward (shift to right) to achieve equilibrium.

and when Q< Keq so, the reaction will go left (shift to left) to achieve equilibrium.

when Q = 10^-11 and Keq = 10^20

∴Q< Keq 

and the reaction will shift to left.

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.

Electron equations of the given changes are as follow,a)                              cu      →      cu⁺²  + 2e⁻b)                              mg⁺²  +  2 e⁻      →      mgc)                              cl⁻    →      cl  +  1e⁻d)                              s  +  2 e⁻    →      s²⁻e)                                o²⁻      →      o  +  2 e⁻f)                              p  +  3 e⁻    →      p³⁻g)                              fe²⁺    →      fe³⁺  + 1 e⁻h)                              s²⁻    →      s⁴⁺  +  6 e⁻