# Question #f96a9

Feb 25, 2016

Here's what's going on here.

#### Explanation:

The idea here is that silver carbonate, ${\text{Ag"_2"CO}}_{3}$, is insoluble in aqueous solution, which is why a **dynamic equilibrium is established between the solid and the dissolved ions.

${\text{Ag"_2"CO"_text(3(s]) rightleftharpoons color(red)(2)"Ag"_text((aq])^(+) + "CO}}_{\textrm{2 \left(a q\right]}}^{2 -}$

The solubility product constant, ${K}_{s p}$, for this equilibrium looks like this

${K}_{s p} = {\left[{\text{Ag"^(+)]_0^color(red)(2) * ["CO}}_{3}^{2 -}\right]}_{0}$

Here ${\left[{\text{Ag}}^{+}\right]}_{0}$ and ${\left[{\text{CO}}_{3}^{2 -}\right]}_{0}$ represent the equilibrium molarities of the two ions before water is added.

The actual value of the solubility product constant is not important here, all that matters is the fact that you can assume it to be constant, i.e. the temperature of the solution remains unchanged after water is added.

Now, this equilibrium, like any other dynamic equilibrium, is governed by Le Chatelier's Principle, which as you know states that a system at equilibrium will react to a stress applied to the current position of the equilibrium by shifting in such a way as to counteract that stress.

So, what happens when water is added to the solution?

As a first observation, one could say that the volume of the solution increases. Since molarity is

$\textcolor{b l u e}{\text{molarity" = "moles of solute"/"liters of solution}}$

an increase in volume would cause a decrease in concentration.

So, after water is added to the solution, you will have

$\overbrace{\left[\text{Ag"^(+)])^(color(purple)("molarity AFTER adding water")) < overbrace(["Ag"^(+)]_0)^(color(brown)("molarity BEFORE adding water}\right)}$

and

$\overbrace{\left[\text{CO"_3^(2-)])^(color(purple)("molarity AFTER adding water")) < overbrace(["CO"_3^(2-)]_0)^(color(brown)("molarity BEFORE adding water}\right)}$

Now, if the concentrations of the two ions decreased, the product of their new concentrations will be smaller than what you had initially.

${\left[{\text{Ag"^(+)]^color(red)(2)["CO"_3^(2-)] < ["Ag"^(+)]_0^color(red)(2) * ["CO}}_{3}^{2 -}\right]}_{0}$

But remember, this product must be equal to ${K}_{s p}$ again! This means that the equilibrium will shift in such a way as to increase the concentrations of the two ions.

In order for that to happen, more moles of ${\text{Ag}}^{+}$ and of ${\text{CO}}_{3}^{2 -}$ must be released into solution. Simply put, the equilibrium will shift to the right, i.e. more of the solid will be dissolved.

${\text{Ag"_2"CO"_text(3(s]) rightleftharpoons color(red)(2)"Ag"_text((aq])^(+) + "CO}}_{\textrm{2 \left(a q\right]}}^{2 -}$

$\textcolor{w h i t e}{a a} \stackrel{\textcolor{g r e e n}{\rightarrow}}{\textcolor{w h i t e}{a a a a a} \textcolor{red}{\text{Shift to the right}} \textcolor{w h i t e}{a a a a a a a}}$

This means that the number of moles of ${\text{Ag}}^{+}$ will increase as a result of the change in the position of the equilibrium.

Remember, the molar solubility of the compound remains unchanged because an increase in the number of moles of ions comes after an increase in the volume of the solution.