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

Jun 11, 2017

Here's what I think.

#### Explanation:

There are two more equilibria involved:

$\text{2I"_2"(aq)" ⇌ "I"^"+""(aq)" + "I"_3^"-""(aq)}$

and

$\text{I"_3^"-""(aq)" ⇌ "I"_2"(aq)" + "I"^"-""(aq)}$

Since $\text{I"^"-}$ is present in an equilibrium concentration, adding ${\text{AgNO}}_{3}$ will give a precipitate of silver iodide.

"Ag"^"+""(aq)" + "I"^"-""(aq)" → underbrace("AgI(s)")_color(orange)("bright yellow")

Producing $\text{AgI} \left(s\right)$ consumes ${\text{I}}^{-}$, which by Le Chatelier's principle induces a shift in the second ${\text{I}}_{3}^{-}$ equilibrium towards making more ${\text{I}}_{2} \left(a q\right)$.

Producing more ${\text{I}}_{2} \left(a q\right)$ induces its usage in the reverse reaction to form ${\text{I}}_{2} \left(s\right)$.

Jun 25, 2017

Here's my take on this:

#### Explanation:

Iodine reacts with water and the following equilibrium is established:

$\textsf{{I}_{2 \left(s\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s O {I}_{\left(a q\right)}^{-} + 2 {H}_{\left(a q\right)}^{+} + {I}_{\left(a q\right)}^{-}}$

The position of equilibrium lies well to the left.

Adding aqueous silver nitrate would cause a yellow precipitate of silver iodide to form:

$\textsf{A {g}_{\left(a q\right)}^{+} + {I}_{\left(a q\right)}^{-} \rightarrow A g {I}_{\left(s\right)}}$

On this basis, according to Le Chatelier, you would expect the position of equilibrium to be shifted to the right, causing more solid iodine to dissolve.

The question states that the opposite takes place.

I think this hinges on how you interpret exactly what you mean by $\textsf{{I}_{2 \left(a q\right)}}$. This is routinely described as "aqueous iodine" or "iodine solution" and is used in volumetric analysis and as a qualitative test for starch.

In reality, because of the low solubility of iodine in water, it is actually a solution of iodine in potassium iodide solution.

The equilibrium is established:

$\textsf{{I}_{2 \left(s\right)} + {I}_{\left(a q\right)}^{-} r i g h t \le f t h a r p \infty n s {I}_{3 \left(a q\right)}^{-}}$

The charged $\textsf{{I}_{3}^{-}}$ 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 $\textsf{A {g}_{\left(a q\right)}^{+}}$ would form a precipitate of AgI, thus removing $\textsf{{I}_{\left(a q\right)}^{-}}$ from the equilibrium. This would cause the position of equilibrium to shift to the left, producing more solid iodine.