- [Instructor] Acetic acid is an example of a weak acid
and sodium hydroxide is an example of a strong base.
When acetic acid reacts with sodium hydroxide,
an aqueous solution of sodium acetate is formed
along with water.
Since this reaction is an acid-base neutralization reaction,
and these reactions go to completion,
instead of using an equilibrium arrow,
we simply draw an arrow going to the right.
Next, let's write the overall ionic equation.
This is also called the complete ionic equation.
Let's start with acetic acid.
Acetic acid is a weak acid
that only partly ionizes in solution.
Therefore, an aqueous solution of acetic acid,
most of the acetic acid molecules stay protonated
and don't turn into acetate.
Therefore, we're just gonna write CH3COOH down here.
We're not gonna show acetic acid ionizing
into H+ and the conjugate base.
However, things are different for sodium hydroxide.
Sodium hydroxide is a strong base,
and strong bases dissociate 100% in solution.
So an aqueous solution of sodium hydroxide
consists of sodium cations and hydroxide anions.
Next, let's think about our products.
We have an aqueous solution of sodium acetate.
Sodium acetate is a soluble salt.
Therefore, in aqueous solution,
we would have sodium cations, Na+,
and the acetate anion, CH3COO-.
Because water ionizes only to an extremely small extent,
we don't write it as the ions.
We simply write H2O.
To save some time, I've added in some aqueous subscripts
and a liquid one for water,
some plus signs, and the reaction arrow.
So this balanced equation
represents the complete ionic equation.
Next, we're gonna use the complete ionic equation
to write the net ionic equation
for this weak acid-strong base reaction.
And to do that,
we first need to identify spectator ions.
Remember, spectator ions don't participate in the reaction.
So looking at the complete ionic equation,
there are sodium cations on the left
and sodium cations on the right,
so we can cross out the sodium cation,
that's our spectator ion.
After we take out the sodium cation,
what's left over is our net ionic equation.
So for our net equation,
we have acetic acid plus hydroxide anion,
forms the acetate anion and water.
Now that we have our net ionic equation,
we're gonna think about three different situations.
And the first situation,
we have equal moles of our weak acid and our strong base.
Looking at our balanced net ionic equation,
the mole ratio of our weak acid to our strong base
is one-to-one.
Therefore, if we have equal moles of our weak acid
and our strong base
and the mole ratio is one-to-one,
the weak and the strong base
are going to completely neutralize each other
and form the acetate anion.
So if our goal is to figure out the pH
of the resulting solution,
we don't need to consider acetic acid
and we don't need to consider the hydroxide ions
'cause these have been completely used up.
We need to think about the acetate anion
in aqueous solution.
And in aqueous solution, acetate anions react with water
to form acetic acid and hydroxide ions.
This reaction does come to an equilibrium
and since Kb is less than one,
at equilibrium, there will be a large amount of reactants
and only a small amount of products.
However, since we are increasing the concentration
of hydroxide ions in solution,
the resulting solution will be basic.
Therefore, at 25 degrees Celsius,
the pH of the resulting solution will be greater than seven.
If we wanted to calculate the actual pH,
we would need some more information,
but we would treat it
just like a weak base equilibria problem.
Also note that the hydroxide ions that we wrote down here
are not the same thing as the hydroxide ions
from our net ionic equation.
The hydroxide ions from our net ionic equation
went away completely.
They were neutralized by reacting with the acetic acid.
So these hydroxide ions were generated
from the anion hydrolysis reaction of acetate plus water,
and even though it's only a small amount
or a small concentration of hydroxide ions,
those are the hydroxide ions
that make the pH greater than seven.
For the second situation,
let's think about the weak acid being in excess.
Since the mole ratio of our weak acid to our strong base
is one-to-one,
if we have more of the weak acid
than we do have the strong base,
all of the strong base will be used up,
and we will form the acetate anion.
And when the reaction comes to completion,
there'll be no more hydroxide ion in solution.
So if our goal is to find the pH of the resulting solution,
we don't need to consider the hydroxide ion at all
since it's all been used up.
We need to think about the weak acid
that was in excess, acetic acid,
and we also have to think about the acetate anion
that was formed in the neutralization reaction.
First, let's think about the weak acid that was in excess.
Acetic acid will react with water
to form the hydronium ion, H3O+,
and the acetate anion, CH3COO-.
Because the concentration of hydronium ions
in solution increases,
that makes the solution more acidic.
Therefore, when the weak acid is in excess,
the pH of the resulting solution will be less than seven
at 25 degrees Celsius.
If our goal is to calculate the actual pH
of the resulting solution,
our first thought would be,
this is a weak acid equilibria problem.
And it is, however,
there are two sources of acetate anion.
One source of acetate anion
is from the reaction of acetic acid with water,
and then the other source of the acetate anion
came from our acid-base neutralization reaction.
Therefore, one way to think about this problem is,
it's actually a common-ion effect problem
where the common ion is the acetate anion.
The presence of a common ion
decreases the ionization of acetic acid in water.
However, since the concentration of hydronium ions
will still increase,
the pH of the solution will be less than seven.
Another way to think about this problem is,
when we formed the acetate anion
from the neutralization reaction,
we had some of the weak acid in excess.
And when we have a weak acid and its conjugate base present,
a buffer solution forms.
And so we could also use the Henderson-Hasselbalch equation
to calculate the pH of the resulting solution.
For the third situation,
let's think about the strong base being in excess.
Since the mole ratio of our weak acid
to our strong base is one-to-one,
if we have the strong base in excess,
the strong base is going to neutralize all of the weak acid.
Therefore, when the reaction comes to completion,
there will be no more weak acid present.
So if our goal is to find the pH of the resulting solution,
we would just have to think about the extra hydroxide ions
in solution.
And since there are extra hydroxide ions
from our strong base,
we know at 25 degrees Celsius,
the pH of the solution will be greater than seven,
so the resulting solution will be basic.
If we wanted to calculate the actual pH of the solution,
this would become a strong base pH calculation problem.
And while it is true that the acetate anions
that were produced from the neutralization reaction
will react with water
to also increase the concentration of hydroxide ions,
the concentration of hydroxide ions that we would get
from the anion hydrolysis of the acetate anion
would be negligible to the hydroxide ions
that were left over.
Therefore, we could just consider the strong base
to calculate the pH,
and we don't need to worry about anion hydrolysis
for this particular situation.