- [Instructor] Ammonia is an example of a weak base.
and hydrochloric acid is an example of a strong acid.
Ammonia reacts with hydrochloric acid
to form an aqueous solution of ammonium chloride.
And because this is an acid-base neutralization reaction,
there's only a single arrow going to the right,
indicating the reaction goes to completion.
Next, let's write the overall or complete ionic equation.
Let's start with ammonia.
Ammonia is a weak base,
and weak bases only partly ionize in aqueous solution.
Therefore, since weak bases only partly ionize,
we're not gonna show this as an ion.
We're simply gonna write NH3 in our equation.
However, for hydrochloric acid,
hydrochloric acid is a strong acid,
and strong acids ionize 100%.
Therefore, an aqueous solution,
we need to show this as the ions,
so H plus and Cl minus.
Ammonium chloride is a soluble salt,
therefore, an aqueous solution,
we show it as the ions.
So ammonium chloride consists of the ammonium ion,
NH4 plus, and the chloride anion, Cl minus.
To save some time,
I've drawn in the aqueous subscripts,
and also put in the reaction arrow and a plus sign.
So this represents the overall,
or the complete ionic equation.
And we can use the complete ionic equation
to find the net ionic equation
for this weak base, strong acid reaction.
To do that, we first need to identify these spectator ions.
And remember, these are the ions that do not take part
in the chemical reaction.
Since there's a chloride anion on the left side
and on the right side,
the chloride anion is the spectator ion for this reaction.
And once we take out our spectator ion,
we're left with our net ionic equation,
which is aqueous ammonia plus H plus yields NH4 plus.
So this is one way to write our net ionic equation.
However, remember that H plus and H3O plus
are used interchangeably in chemistry.
Therefore, another way to write the net ionic equation
is to show aqueous ammonia plus the hydronium ion,
H3O plus, yields the ammonium ion, NH4 plus, plus water.
Now that we have our net ionic equation,
we're gonna consider three different situations.
In the first situation,
we have equal moles of our weak base and strong acid.
Looking at our net ionic equation,
the mole ratio of ammonia to hydronium ion is one to one.
Therefore, if we have equal moles of our weak base
and strong acid,
the weak base and strong acid
will completely neutralize each other
and produce the ammonium ion NH4 plus.
So if our goal is to figure out the pH
of the resulting solution,
we don't need to consider the weak base,
or this strong acid.
We need to think about the ammonium cation
in aqueous solution.
And in solution,
the ammonium cation acts as a weak acid
and donates a proton to water
to form the hydronium ion, H3O plus, and aqueous ammonia.
The ammonium cation, NH4 plus, is a weak acid.
Therefore, the Ka value is less than one.
And since Ka is less than one at equilibrium,
there are mostly reactants and not very many products.
However, the concentration of hydronium ions in solution
is increased, and therefore,
the resulting solution will be acidic.
So the resulting solution will be slightly acidic.
And at 25 degrees Celsius,
the pH of the solution will be less than seven.
If we wanted to calculate the actual pH,
we would treat this like a weak acid equilibrium problem.
Also, it's important to emphasize that the hydronium ions
that gave the resulting solution a pH less than seven
came from the reaction of the ammonium cation with water.
The hydronium ions did not come from the strong acid.
All of those hydronium ions were used up
in the acid-base neutralization reaction.
For the second situation,
we have more of the weak base than the strong acid,
therefore, we have the weak base in excess.
And because the mole ratio of the weak base
to the strong acid is one to one,
if we have more of the weak base than the strong acid,
all of the strong acid will be used up.
So when the reaction goes to completion,
we'll have ammonium cations in solution,
and we'll also have some leftover ammonia.
So after the neutralization reaction is complete
and all the H3O plus is used up,
we'll have some leftover ammonia.
That ammonia will react with water
to form hydroxide anions and NH4 plus.
Because the concentration of hydroxide ions in solution
has increased at 25 degrees Celsius,
the resulting solution will be basic
and the pH will be greater than seven.
If we wanted to calculate the actual pH,
we would treat this like a weak base equilibria problem.
However, we have two sources for the ammonium cation.
One source is from ammonia reacting with water
to form NH4 plus,
and the other source came from the neutralization reaction.
So actually, this would be a common-ion effect problem.
The other way to calculate the pH of this solution
is to realize that ammonium NH4 plus is a weak acid,
and ammonia NH3 is its conjugate base,
therefore, if we have similar amounts of a weak acid
and its conjugate base,
we have a buffer solution and we could calculate
the pH using the Henderson-Hasselbalch equation.
For our third situation,
let's say we have the strong acid in excess.
Since the mole ratio of weak base to strong acid
is one to one,
if we have more of the strong acid than the weak base,
all of the weak base will be used up
and we'll have some strong acid in excess.
Therefore, there'll be a concentration of hydronium ions
in solution, which would make the resulting solution acidic.
So at 25 degrees Celsius, the pH would be less than seven.
We could calculate the actual pH of the resulting solution
by doing a strong acid pH calculation problem.
And while it's true that the ammonium cation
can function as a weak acid
and also increase the concentration of hydronium ions,
it's such a small increase compared to
the hydronium ions we have in solution from our strong acid
that we don't need to worry about the contribution
of the ammonium cations.
We can just treat this
like a strong acid pH calculation problem.