Why does adding #"AgNO"_(3("aq"))# to #"I"_(2("s")) rightleftharpoons "I"_(2("aq"))# cause the aqueous iodine to precipitate into a solid?

Question

Henry Antrim · Accepted Answer

Here's what I think.

There are two additional equilibria in question: #"2I"_2"(aq)" ⇌ "I"^"+""(aq)" + "I"_3^"-""(aq)"# and #"I"_3^"-""(aq)" ⇌ "I"_2"(aq)" + "I"^"-""(aq)"# Since #"I"^"-"# is present in an equilibrium concentration, adding #"AgNO"_3# will give a precipitate of silver iodide. #"Ag"^"+""(aq)" + "I"^"-""(aq)" → underbrace("AgI(s)")_color(orange)("bright yellow")# Producing #"AgI"(s)# consumes #"I"^(-)#, which by Le Chatelier's principle induces a shift in the second #"I"_3^(-)# equilibrium towards making more #"I"_2(aq)#. Producing more #"I"_2(aq)# induces its usage in the reverse reaction to form #"I"_2(s)#.

Peyton Adams · Answer

Here's my take on this:

When iodine and water interact, the following equilibrium is reached: #sf(I_(2(s))+H_2O_((l))rightleftharpoonsOI_((aq))^(-)+2H_((aq))^(+)+I_((aq))^(-))# The equilibrium point is located far to the left. A yellow precipitate of silver iodide would form upon the addition of aqueous silver nitrate: #sf(Ag_((aq))^(+)+I_((aq))^(-)rarrAgI_((s)))# Le Chatelier predicted that more solid iodine would dissolve as a result of this shift in the equilibrium position to the right. It is stated in the question that the opposite occurs. I think this hinges on how you interpret exactly what you mean by #sf(I_(2(aq)))#. This is routinely described as "aqueous iodine" or "iodine solution" and is used in volumetric analysis and as a qualitative test for starch. It is actually a solution of iodine in potassium iodide solution due to iodine's low solubility in water. The balance is reached: #sf(I_(2(s))+I_((aq))^(-)rightleftharpoonsI_(3(aq))^-)# The charged #sf(I_3^-)# ion being more soluble such that #sf(K=~400)#. If the question actually means that #sf(I_(2(aq))# is really a solution of iodine in KI solution then adding #sf(Ag_((aq))^+)# would form a precipitate of AgI, thus removing #sf(I_((aq))^-)# from the equilibrium. This would cause the position of equilibrium to shift to the left, producing more solid iodine.

Kai Bowman · Answer

undefined The addition of AgNO3 to I2(aq) forms AgI, a sparingly soluble salt, leading to the precipitation of solid AgI due to its low solubility.