# The chemical reaction that represents solid silver chloride dissolving in water is: AgCl (s) rightleftharpoons Ag^+ (aq) + Cl^(-) (aq). K_(sp) = 1.7 xx 10^-10 What would you do to shift this reaction towards solid silver chloride?

Jul 18, 2016

Add a soluble salt that contains silver cations or chloride anions.

#### Explanation:

Like any other dynamic equilibrium, a solubility equilibrium is governed by Le Chatelier's Principle.

As you know, Le Chatelier's Principle states that a system at equilibrium will react to a stress placed on the position of the equilibrium in such a way as to minimize this stress.

In other words, the position of the equilibrium will shift in response to a stress placed on either the reactants', or the products' side.

In this case, the solubility equilibrium for silver chloride, $\text{AgCl}$, looks like this

${\text{AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

Your goal is to find a way to shift this equilibrium to the left, i.e. to find a way to reduce the solubility of the salt.

This can be done by adding a soluble salt that contains silver cation, ${\text{Ag}}^{+}$, or chloride anions, ${\text{Cl}}^{-}$. For example, let's say that you're adding silver nitrate, ${\text{AgNO}}_{3}$, to the solution.

Silver nitrate will dissociate completely in aqueous solution to form

${\text{AgNO"_ (3(aq)) -> color(red)("Ag"_ ((aq))^(+)) + "NO}}_{3 \left(a q\right)}^{-}$

Dissolving this salt into your silver chloride solution will cause the concentration of aqueous silver cations to increase. This will place a stress on the position of the solubility equilibrium.

In response, the equilibrium will react in such a way as to decrease the concentration of silver cations. The only way to do that is to shift to the left, since the reverse reaction

${\text{Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-) -> "AgCl}}_{\left(s\right)}$

consumes silver cations and chloride anions and produces solid silver chloride.

${\text{AgCl"_ ((s)) rightleftharpoons color(red)("Ag"_ ((aq))^(+)) + "Cl}}_{\left(a q\right)}^{-}$

color(white)(a)stackrel(larr)(color(white)(aaacolor(darkgreen)("shift to the left")aaaaa)

The same thing happens if you add, for example, sodium chloride, $\text{NaCl}$, which dissociates completely to produce

"NaCl"_ ((aq)) -> "Na" _ ((aq))^(+) + color(blue)("Cl"_ ((aq))^(-))

The added chloride anions will once again cause the solubility equilibrium to shift to the left. The reverse reaction will be favored, which implies that chloride anions and silver cations will be consumed and solid silver chloride will be produced.

"AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + color(blue)("Cl"_ ((aq))^(-))

color(white)(a)stackrel(larr)(color(white)(aaacolor(darkgreen)("shift to the left")aaaaa)

This decrease in the solubility of a salt as a result of the addition of one its ions is called the common-ion effect and it's a direct consequence of Le Chatelier's Principle.