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.