pH of a Weak Base

pH of .2 M of NH3 (weak base).

We've dealt with the weak acid,
so let's try an example

with the weak base.

Let's say we had ammonia.

That's nitrogen with
three hydrogens.

And it's a weak base because
it likes to accept hydrogen

from water, leaving the water
with just a hydroxide.

So it increases the hydroxide
concentration.

So if you have some ammonia
in an aqueous

solution, plus water.

I'll throw the water in there.

Plus water in an aqueous
solution.

It's a weak base.

So this reaction doesn't go
in just one direction.

It's an equilibrium reaction.

And since this is a weak base,
it-- and this is where the

Bronsted-Lowry definition
really kind of pops out.

Is that it's a proton acceptor
instead of a donor.

So it turns into ammonium,
or an ammonia cation.

Ammonium has another hydrogen
on it, so now

it has another proton.

So it's the plus charge.

An it's an aqueous.

And it took that hydrogen
from the water.

So plus OH minus aqueous.

And remember, if you look at
it from the Bronsted-Lowry

definition , it was
a proton acceptor.

So that made it a base.

Or if you look at the Arrhenius
definition, it

increased the concentration of
OH in the solution, so that

makes it an Arrhenius base.

But anyway, given that
we have-- let me

pick a random number.

Let's say we have 0.2
molar of NH3.

What is going to be the pH?

So what's going to be our pH of
the solution, considering

that it's 0.2 molar of NH3.

So the first thing
we need to do.

We need to figure out the
equilibrium constant for this

base reaction.

And I just went to Wikipedia--
I wanted to say liquidpedia,

I'm talking about
liquids so much.

And equilibrium.

Equipedia.

But I went to Wikipedia, and
they have a little chart for

almost any compound
you look for.

And they give you pKb.

Which is, you see
that p there.

That just means it's the
minus log base 10 of

the equilibrium constant.

And they give that
as being 4.75.

So we can just do a little bit
of math here to solve for the

equilibrium constant.

So let's see.

If we multiply both sides by
negative, you get log base 10

of our equilibrium constant
for this base reaction.

That's why the b is there.

Is equal to minus 4.75,
or 10 to the minus

4.75 should be Kb.

So Kb is equal to 10
to the minus 4.75.

That's not an easy exponent to
figure out in your head, so

I'll bring out the calculator
for that.

So if we take 10 to the 4.75
minus, it equals, let's just

say 1.8 times 10 to
the negative 5.

This is equal to 1.8 times
10 to the minus 5.

So now we can use this
information and we can do a

mathematical thing very
similar to we

did in the last video.

It's going to be hard to
figure out the hydrogen

concentration directly, right?

Because our equilibrium reaction
only has hydroxide.

But if we know the hydroxide
concentration, then we can

back into the hydrogen
concentration, knowing that

this plus the hydrogen
concentration has to equal 10

to the minus 14.

Or if you figure out the pOH,
that plus the pH has to be 14.

And we did that a couple
of videos ago.

So this equilibrium constant
or this formula

would look like this.

1.8 times 10 to the minus 5
will be equal to-- in the

denominator, we have our
concentration of reactants.

And remember, you don't
include the solvent.

So you only include the NH3.

We have 0.2 molars is what we
put in, but some of it, let's

say X of it, is going to be
converted into this stuff on

the right-hand side.

So in the denominator, we're
going to have 0.2 minus

whatever gets converted into
the right-hand side.

And so then in the right-hand
side, we're going to have x of

NH4 and x of OH.

This is the concentration
of ammonia.

And then we have x times x.

This is the concentration of
NH4 plus-- that's a 4.

And then this is the
concentration,

right here, of OH minus.

Right?

And we just solve for x.

Let's do that.

Solve for x.

And once we have x, we know
the concentration of OH.

We'll be able to figure out
the pOH, and then we'll be

able to figure out the pH.

OK.

Multiply this times both
sides of this equation.

And just so you know, that same
simplification step that

we did in the previous thing.

When this is several orders of
magnitude smaller than this

number right here-- I want to
give you-- heuristics are just

kind of rules of thumb
that sometimes work.

Let's just do the quadratic
equation.

But you can kind of think about
sometimes when you can

get rid of that middle term.

But let's just multiply it.

0.2 two times 1.8 is 0.36.

0.36 times 10 to the
minus 5, right?

2 times 1.8 would be
3.6, this is 0.36.

Minus 1.8 times 10 to the
minus 5 x, right?

Is equal to that.

x squared.

Let's put everything on the
same side of the equation.

I'm going to move all of these
the right-hand side, so you

get 0 is equal to x squared.

Add this to both sides
of the equation.

Plus 1.8 times 10 to
the minus 5 x.

1.8 times 10 to the minus 5.

Just so you can see the
coefficients separate from the

x terms.

Minus 0.36 times 10
to the minus 5.

So let's solve this.

And once again, if you wanted
to kind of do it, you could

eliminate this term and then
just figure out the straight

up square root.

But we won't do that.

We'll actually use a
quadratic equation.

So a is 1.

b is this.

That's b.

And this is c.

And you just supply than in
the quadratic equation.

So you get minus b.

So you minus 1.8 times 10
to the minus 5 power.

Plus or minus.

We'll only have to do the plus
because if we do the minus,

we'll end up with a negative
concentration.

So plus, the square root-- we
have to do a lot of math

here-- b squared.

So it's 1.8 times 10
to the negative 5.

So it's 1.8.

If you square it, it's 3.24.

So it's 3.24 times-- if you
square 10 to the minus 5-- 10

to the minus 10 minus 4
times a, which is 1,

times c, which is minus.

So it's 4 times-- the minuses
cancel out-- times 0.36 times

10 to the minus 5.

Which is 4 times 0.36
is equal to 1.44.

I should have been able
to do that in my head.

Now you have 1.44 e minus 5.

Times 10 to-- let
me write that.

So this is 1.44.

And of course all of
this is over 2a.

So let's see.

This is my x value.

My concentration of OH.

So let's see.

I have 3.24 times 10
to the minus 10.

That's that.

Plus 1.44 times 10 to the minus
5 is equal to that.

So that's this whole thing
under the radical.

And I want to take the
square root of that.

And so that is to
the 0.5 power.

So I get 0.00379.

So I'll switch colors.

So I get x is equal to a minus
1.8 times 10 to the minus 5

plus 0.003794.

All of that over 2.

Do the math.

So to that I'm going to subtract
minus this point

right here.

I have this value.

I'm just subtracting this.

Minus 1.8 e 5 negative
is equal to that.

This is the whole numerator.

And now I need to just
divide it by 2.

Divided by 2 is equal
to 0.001.

Let me write that.

So x.

So I'll switch colors
arbitrarily again.

x is equal to 0.001888-- I mean,
then there's a 3 and so

forth and so on.

But if you remember from
our original equation.

What was x?

It was what's both the ammonium
concentration and the

hydroxide concentration.

We care about the hydroxide
concentration.

So this is equal to my
concentration of hydroxide.

Now if I want to figure out my
pOH, I just take the minus log

of this number, which
is equal to--

So let's just take
the log of it.

The log is that, and then I
take the minus of that.

So it's 2.72.

And now if we want to figure out
the pH, my concentration

of hydrogen ions-- just
remember, when you're in an

aqueous solution at 25 degrees
Celsius, your pK of water is

equal to your pOH
plus your pH.

This at 25 degrees is 14.

Because you have 10 to the minus
14 molar concentration--

well no, actually, I don't
want to go into that.

You have 10 to the minus
7 of each of these.

But anyway, this is equilibrium
constant for the

disassociation of water.

This, when water's neutral is 7
or a concentration of OH of

10 to the minus 7.

We can take the minus
log, this becomes 7.

But now we know we have a much
higher concentration of OH.

2.72.

Remember, that minus log
kind of flips it.

So a lower pOH means a higher
concentration of pOH.

Right?

And a lower pOH, if this
is lower, right?

This is a lower pOH.

That means your pH is higher.

So what is your pH
going to be?

So your pH is going to be
equal to 14 minus 2.72.

So let me do the minus
plus 14 is equal to--

let's just say 11.3.

So your pH is equal to 11.3.

Which makes sense, because we
said this was a weak base.

Ammonia is a weak base.

So it's basic.

So it should increase your
pH above the neutral 7.

So the pH should be greater than
7, but as you compare it

to some of the strong bases
before that took our pH when

you added a molar to 14, this
took our pH-- although we only

did add 0.2 molar
of it to 11.3.

Anyway, this is more of a math
problem than chemistry, but

hopefully it clarified
a few things as well.