WEBVTT

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

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with the weak base.

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Let's say we had ammonia.

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That's nitrogen with
three hydrogens.

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

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from water, leaving the water
with just a hydroxide.

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So it increases the hydroxide
concentration.

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So if you have some ammonia
in an aqueous

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solution, plus water.

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I'll throw the water in there.

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Plus water in an aqueous
solution.

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It's a weak base.

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

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It's an equilibrium reaction.

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

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Bronsted-Lowry definition
really kind of pops out.

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

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So it turns into ammonium,
or an ammonia cation.

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Ammonium has another hydrogen
on it, so now

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it has another proton.

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So it's the plus charge.

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An it's an aqueous.

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And it took that hydrogen
from the water.

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So plus OH minus aqueous.

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

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definition , it was
a proton acceptor.

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So that made it a base.

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Or if you look at the Arrhenius
definition, it

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increased the concentration of
OH in the solution, so that

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makes it an Arrhenius base.

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But anyway, given that
we have-- let me

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pick a random number.

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Let's say we have 0.2
molar of NH3.

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What is going to be the pH?

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

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that it's 0.2 molar of NH3.

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So the first thing
we need to do.

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We need to figure out the
equilibrium constant for this

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base reaction.

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

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I'm talking about
liquids so much.

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And equilibrium.

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Equipedia.

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

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almost any compound
you look for.

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And they give you pKb.

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Which is, you see
that p there.

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

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the equilibrium constant.

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And they give that
as being 4.75.

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

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equilibrium constant.

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So let's see.

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

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of our equilibrium constant
for this base reaction.

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That's why the b is there.

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Is equal to minus 4.75,
or 10 to the minus

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4.75 should be Kb.

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So Kb is equal to 10
to the minus 4.75.

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

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I'll bring out the calculator
for that.

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

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say 1.8 times 10 to
the negative 5.

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

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

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mathematical thing very
similar to we

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did in the last video.

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

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concentration directly, right?

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Because our equilibrium reaction
only has hydroxide.

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But if we know the hydroxide
concentration, then we can

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back into the hydrogen
concentration, knowing that

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this plus the hydrogen
concentration has to equal 10

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to the minus 14.

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

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And we did that a couple
of videos ago.

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So this equilibrium constant
or this formula

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would look like this.

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

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denominator, we have our
concentration of reactants.

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And remember, you don't
include the solvent.

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So you only include the NH3.

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

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

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the right-hand side.

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

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whatever gets converted into
the right-hand side.

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

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NH4 and x of OH.

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This is the concentration
of ammonia.

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And then we have x times x.

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

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And then this is the
concentration,

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right here, of OH minus.

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Right?

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And we just solve for x.

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Let's do that.

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Solve for x.

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

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

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able to figure out the pH.

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OK.

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Multiply this times both
sides of this equation.

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And just so you know, that same
simplification step that

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we did in the previous thing.

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When this is several orders of
magnitude smaller than this

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

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kind of rules of thumb
that sometimes work.

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Let's just do the quadratic
equation.

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But you can kind of think about
sometimes when you can

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get rid of that middle term.

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But let's just multiply it.

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0.2 two times 1.8 is 0.36.

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0.36 times 10 to the
minus 5, right?

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2 times 1.8 would be
3.6, this is 0.36.

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

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Is equal to that.

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x squared.

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

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

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get 0 is equal to x squared.

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Add this to both sides
of the equation.

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Plus 1.8 times 10 to
the minus 5 x.

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1.8 times 10 to the minus 5.

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Just so you can see the
coefficients separate from the

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x terms.

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Minus 0.36 times 10
to the minus 5.

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So let's solve this.

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

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eliminate this term and then
just figure out the straight

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up square root.

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But we won't do that.

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We'll actually use a
quadratic equation.

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So a is 1.

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b is this.

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That's b.

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And this is c.

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And you just supply than in
the quadratic equation.

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So you get minus b.

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

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Plus or minus.

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

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we'll end up with a negative
concentration.

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

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here-- b squared.

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

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So it's 1.8.

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If you square it, it's 3.24.

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

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

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times c, which is minus.

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

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10 to the minus 5.

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Which is 4 times 0.36
is equal to 1.44.

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

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Now you have 1.44 e minus 5.

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Times 10 to-- let
me write that.

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So this is 1.44.

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And of course all of
this is over 2a.

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So let's see.

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This is my x value.

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My concentration of OH.

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So let's see.

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I have 3.24 times 10
to the minus 10.

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That's that.

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

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So that's this whole thing
under the radical.

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And I want to take the
square root of that.

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And so that is to
the 0.5 power.

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So I get 0.00379.

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So I'll switch colors.

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

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plus 0.003794.

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All of that over 2.

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Do the math.

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

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right here.

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I have this value.

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I'm just subtracting this.

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Minus 1.8 e 5 negative
is equal to that.

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This is the whole numerator.

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And now I need to just
divide it by 2.

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Divided by 2 is equal
to 0.001.

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Let me write that.

00:10:04.430 --> 00:10:05.910
So x.

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So I'll switch colors
arbitrarily again.

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

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forth and so on.

00:10:19.690 --> 00:10:21.740
But if you remember from
our original equation.

00:10:21.740 --> 00:10:22.490
What was x?

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It was what's both the ammonium
concentration and the

00:10:26.760 --> 00:10:27.855
hydroxide concentration.

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We care about the hydroxide
concentration.

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So this is equal to my
concentration of hydroxide.

00:10:36.060 --> 00:10:40.580
Now if I want to figure out my
pOH, I just take the minus log

00:10:40.580 --> 00:10:44.610
of this number, which
is equal to--

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So let's just take
the log of it.

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

00:10:49.180 --> 00:10:55.180
So it's 2.72.

00:10:55.180 --> 00:11:00.420
And now if we want to figure out
the pH, my concentration

00:11:00.420 --> 00:11:03.370
of hydrogen ions-- just
remember, when you're in an

00:11:03.370 --> 00:11:10.740
aqueous solution at 25 degrees
Celsius, your pK of water is

00:11:10.740 --> 00:11:15.870
equal to your pOH
plus your pH.

00:11:15.870 --> 00:11:19.730
This at 25 degrees is 14.

00:11:19.730 --> 00:11:22.815
Because you have 10 to the minus
14 molar concentration--

00:11:22.815 --> 00:11:24.580
well no, actually, I don't
want to go into that.

00:11:24.580 --> 00:11:26.720
You have 10 to the minus
7 of each of these.

00:11:26.720 --> 00:11:28.510
But anyway, this is equilibrium
constant for the

00:11:28.510 --> 00:11:30.310
disassociation of water.

00:11:30.310 --> 00:11:37.170
This, when water's neutral is 7
or a concentration of OH of

00:11:37.170 --> 00:11:38.280
10 to the minus 7.

00:11:38.280 --> 00:11:39.760
We can take the minus
log, this becomes 7.

00:11:39.760 --> 00:11:44.070
But now we know we have a much
higher concentration of OH.

00:11:44.070 --> 00:11:45.080
2.72.

00:11:45.080 --> 00:11:47.550
Remember, that minus log
kind of flips it.

00:11:47.550 --> 00:11:52.670
So a lower pOH means a higher
concentration of pOH.

00:11:52.670 --> 00:11:53.230
Right?

00:11:53.230 --> 00:11:57.200
And a lower pOH, if this
is lower, right?

00:11:57.200 --> 00:11:58.480
This is a lower pOH.

00:11:58.480 --> 00:12:00.290
That means your pH is higher.

00:12:00.290 --> 00:12:03.570


00:12:03.570 --> 00:12:04.870
So what is your pH
going to be?

00:12:04.870 --> 00:12:15.024
So your pH is going to be
equal to 14 minus 2.72.

00:12:15.024 --> 00:12:21.710
So let me do the minus
plus 14 is equal to--

00:12:21.710 --> 00:12:23.570
let's just say 11.3.

00:12:23.570 --> 00:12:26.650
So your pH is equal to 11.3.

00:12:26.650 --> 00:12:30.850
Which makes sense, because we
said this was a weak base.

00:12:30.850 --> 00:12:33.680
Ammonia is a weak base.

00:12:33.680 --> 00:12:35.070
So it's basic.

00:12:35.070 --> 00:12:39.120
So it should increase your
pH above the neutral 7.

00:12:39.120 --> 00:12:42.805
So the pH should be greater than
7, but as you compare it

00:12:42.805 --> 00:12:45.280
to some of the strong bases
before that took our pH when

00:12:45.280 --> 00:12:49.410
you added a molar to 14, this
took our pH-- although we only

00:12:49.410 --> 00:12:54.010
did add 0.2 molar
of it to 11.3.

00:12:54.010 --> 00:12:56.970
Anyway, this is more of a math
problem than chemistry, but

00:12:56.970 --> 00:12:59.800
hopefully it clarified
a few things as well.