# Question #62820

Sep 14, 2015

The percent ionization actually increases.

#### Explanation:

Acetic acid, $\text{CH"""_3"COOH}$, is a weak acid, which means that it does not ionize completely in aqueous solution.

The equilibrium that is established in an acetic acid solution looks like this

${\text{CH"""_3"COOH"_text((aq]) + "H"_2"O"_text((aq]) rightleftharpoons "CH"""_3"COO"_text((aq])^(-) + "H"_3"O}}_{\textrm{\left(a q\right]}}^{+}$

Water is a reactant in this equilibrium reaction, which means that adding water, which is what you essentially do when you dilute something, will affect the position of the equilibrium.

According to Le Chatelier's Principle, adding more water molecules will disturb the equilibrium. In order to counteract this disturbance, the equilibrium will shift in such a way as to reduce the number of water molecules.

This of course means that the equilibrium will shift right, which in turn means that more acetic acid molecules will be ionized.

Therefore, diluting the initial solution by doubling its volume will increase the degree of ionization of the acid.

However, the pH of the solution will actually increase, despite the fact that more acetic acid molecules have donated their acidic proton.

This happens because of the significant increase in volume in comparison to the relatively small increase in percent ionization.

As you dilute the solution more and more, the pH of the solution will come closer and closer to that of pure water.

Now, the reaction quotient tells you the ratio between the concentrations of the product and reactants at a given moment in the evolution of the reaction.

At equilibrium, the reaction quotient is always equal to the equilibrium constant at a given temperature and pressure.

${Q}_{c} = {K}_{c} \to$ at equilibrium

This means that as long as ${K}_{c}$ is constant, any position of the equilibrium will be characterized by ${Q}_{c} = {K}_{c}$.

So, as long as equilibrium is reached, ${Q}_{c}$ will always be constant, provided that ${K}_{c}$ is constant as well.

Another way to look at this is to take into account that dilution decreases the concentration of all the species in the solution (except that of water).

This means that the reaction quotient

${Q}_{c} = \left(\left[\text{CH"_3"COO"^(-)] * ["H"_3"O"^(+)])/(["CH"_3"COOH}\right]\right) \to$ after dilution

will be smaller than ${K}_{c}$, since the numberator will decrease faster than the denominator (remember, all concentrations are smaller than they were before the dilution).

As a result, the equilibrium will shift to the right, which means that more acid molecules will be ionized.