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## Weak acid and base equilibria

# Weak base equilibria

AP.Chem:

SAP‑9 (EU)

, SAP‑9.C (LO)

, SAP‑9.C.3 (EK)

, SAP‑9.C.4 (EK)

, SAP‑9.C.5 (EK)

## Video transcript

- [Instructor] An example of
a weak base is ammonia, NH3, and ammonia will react with water. And in this reaction, water functions as a Bronsted-Lowry acid and donates a proton to ammonia, which functions as a Bronsted-Lowry base and accepts a proton. A proton is H plus. So if you add on H plus onto NH3, you form NH4 plus, which
is the ammonium ion. And if you take away NH plus from H2O, you form the hydroxide
anion, which is OH minus. This reaction comes to an equilibrium, so we have an equilibrium arrow in here, and we can also write an
equilibrium constant expression. So first, we look at our products. We have NH4 plus, and there's a coefficient
of one in front of it, so it'd be the concentration
of NH4 plus raised to the first power times the concentration of the other product, hydroxide, which also has a coefficient of one, so it's the concentration
of hydroxide raised to the first power, divided by the concentration of ammonia, and the coefficient is also one, so that's also raised to the first power. And water is left out of our equilibrium constant expression. The equilibrium constant
K has a subscript of b, which stands for base. And this Kb is called the
base ionization constant, or the base dissociation constant. Ammonia is a weak base, which means it only
partially ionizes in water. And if it only partially
ionizes, at equilibrium, there's going to be a
small amount of products and a large amount of reactants. Therefore, looking at the
equilibrium constant expression, the numerator is gonna
be very small compared to the denominator. So with a small numerator
and a large denominator, that means that the Kb
value for a weak base will be less than one. Let's look at a few
examples of weak bases. Here's the dot structure for ammonia, and notice how ammonia has a nitrogen with a lone pair of electrons that's capable of accepting a proton. A proton is H plus. So if we add an H plus
onto NH3, we form NH4 plus, which is the ammonium ion, and it's the conjugate acid to ammonia. So think about this lone pair of electrons that I circled here picking
up a proton to form this bond. If we replace the hydrogen with what's called a methyl group, CH3, we form a new weak base
called methylamine. Like ammonia, methylamine has a nitrogen with a lone pair of electrons that's capable of accepting a proton. So if methylamine picks up a proton, it forms the methyl ammonium ion. Both ammonia and methylamine are examples of neutral substances. However, a weak base
doesn't have to be neutral. If we look at the
hypochlorite anion, ClO minus, if the hypochlorite
anion picks up a proton, let's say this lone pair of electrons in the oxygen picks up a proton, it would form hypochlorous acid, and hypochlorous is the
conjugate acid to hypochlorite. So these are three examples of weak bases, which we can tell by
looking at their Kb values at 25 degrees Celsius. Notice they all have Kb
values less than one. However, the higher the Kb value, the higher the equilibrium
concentration of hydroxide ion, and therefore the stronger the base. So out of these three week bases, the one with the higher Kb
value is the most basic. So looking at these three values, again, the Kb value for methylamine
is the highest of the three, so it's the strongest base of the three, although all three are
considered to be weak. Let's do a calculation with a weak base. Let's say we have a .14 molar solution, aqueous solution of ammonia
at 25 degrees Celsius, and our goal is to
calculate the concentration of hydroxide ions in solution, and also the pH of the
solution at 25 degrees Celsius. So first, we're gonna calculate the equilibrium concentration
of hydroxide ions, and to help us do that, we're gonna use an ICE table, where I stands for the
initial concentration, C is the change in concentration, and E is the equilibrium concentration. The initial concentration
of ammonia is .14 molar. So we can put that in on our ICE table. And if we assume that ammonia
hasn't ionized at all yet, the initial concentrations of ammonium and hydroxide ions would both be zero. Next, we think about some
of the ammonia ionizing, and we don't know how much, so we write minus x under
ammonia for the change. When NH3 gains a proton,
it turns into NH4 plus, and the mole ratio of ammonia
to ammonium is one to one. Therefore, if we write
minus x for ammonia, we must write plus x for ammonium. And the hydroxide ion also
has a one as a coefficient in a balanced equation. Therefore, it would also
be plus x in our ICE table. So the equilibrium
concentration of ammonia would be .14 four minus x. So let's write that in there, .14 minus x. For ammonium, it would be zero plus x, which is just x, and for hydroxide, it would also be zero
plus x, which is just x. The next step is to write out the equilibrium constant expression for the ionization of ammonia, which we've already talked about. And since the concentrations in our equilibrium constant expression are equilibrium concentrations, we can go ahead and plug in the equilibrium concentrations
of the ammonium ion, the hydroxide ion, and ammonia into our
equilibrium constant expression. We also need to plug in for Kb for ammonia at 25 degrees Celsius. Here's our equilibrium constant expression with everything plugged in. And our goal is to solve for x, because x represented the
equilibrium concentration of hydroxide ions. However, if we solve for x now, we would get a quadratic equation. So to simplify the math, we're gonna use an approximation. We're gonna say that .14 minus x is approximately equal to .14. The reason why we can
make this approximation is because the Kb value
for ammonia is quite small, which means that ammonia
doesn't ionize very much at all, which tells us the
equilibrium concentration of hydroxide ions will
be a very small number. And if x is a really small
number compared to .14, .14 minus x is just
approximately equal to .14. So instead of writing .14 minus x, we can just write in .14. So now we solve for x and we find that x is equal to 1.6 times
10 to the negative third molar, which is the equilibrium
concentration of hydroxide ions. Our goal is to calculate
the pH of our solution. And first, we could find the pOH, because pOH is equal to the negative log of the concentration of hydroxide ions. So the negative log of 1.6
times 10 to the negative third is equal to 2.80. And at 25 degrees Celsius, the pH plus the pOH is equal to 14.00, so we could plug in the
pOH into our equation and solve for the pH, and we would find the pH of
our solution is equal to 11.20. And this makes sense, because
we have a basic solution, so the pH should be greater than seven. Finally, to make sure that
we can use our approximation, let's find the percent
ionization for our weak base. To calculate the percent
ionization of ammonia, we take x, which you could think about as the concentration of
ammonia that ionized, or you could think about it as the equilibrium
concentration of hydroxide ions. And we divide x by the initial
concentration of ammonia and multiply it by 100 to
get a percent ionization. So for this problem, x is equal to 1.6 times 10
to the negative third molar. So we go ahead and plug that in here. So 1.6 times 10 to the
negative third molar, and the initial concentration of ammonia was equal to .14 molar. So molar cancels out and we end up with a percent ionization of 1.1%. Once you calculate the percent ionization, you need to think about the 5% rule. And the 5% rule says, if your percent ionization
is less than 5%, it's okay to use the approximation. So in our case, our percent
ionization was 1.1%, which is less than 5%. So it's okay to say that .14 minus x is approximately equal to .14, because with such a
low percent ionization, x is a really small value compared to .14. However, if you get a percent
ionization greater than 5%, you can't use the approximation and you need to use the quadratic
equation to calculate x. Finally, when we calculated
our equilibrium concentration of hydroxide ions, we only considered the
ionization of ammonia. It's true that the autoionization of water would produce a small amount
of hydroxide ions in solution. However, that concentration
is extremely small compared to the concentration of hydroxide ions from the ionization of ammonia. Therefore, we ignore the
contribution of hydroxide ions from the auto ionization of
water in our calculations.

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