0:01

Well, hopefully, you took the time to look at part a already,

there's going to be some very definite similarities between part a and part b,

and we might gloss over some of the steps because of the similarities.

We have now in a flask the ammonia solution,

and we are still adding into the ammonia the strong acid, HNO3.

So let's begin by writing the reaction.

The strong acid, again, is written as H3O+.

The weak base is NH3.

Because the acid is strong, it's a complete reaction,

a one-way reaction, in which the ammonia accepts a proton and

becomes ammonium, and we're left with water.

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The one-way reaction makes it an ICF table,

in which we know we need to put moles into the table.

So now let's figure out the moles of each substance.

Because they gave me the molarity and the volume of the base, we'll find that first.

The moles of the NH3 would be the molarity of the NH3,

which is 0.350 moles per liter.

Times the volume of that base, which was 0.188 liters.

That will give me a moles of base 0.0658.

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Since they are telling me that the solutions at the equivalence point,

that tells me that the moles of the acid are the same, and we can finish the table.

All of the 0.0658 moles of this and the 0.0658 moles

of that will be reacted to produce 0.0658 moles of the ammonium.

The conjugate acid of ammonia, that consumes these substances,

and we will have 0.0658 moles of that conjugate acid left.

2:12

Let's first determine of these two things the volume of the HNO3 that's needed.

We know moles of the HNO3,

0.0658 moles of HNO3 were placed in the flask.

Well, if that's how many moles we needed, we can go from moles

to liters using the molarity of 0.447 moles per liter.

This will obtain for us the fact that we have

0.147 moles of that acid, HNO3.

Since that acid is in milliliters, let's convert that to milliliters.

That is equal to 147 milliliters of HNO3.

Now we're ready to continue on and figure out the pH of this solution.

To do this, we always follow an ICF table with the ICE table of

ammonium that acid in water.

Whatever sits in this solution is what you'll place and know that it's in water,

and it will reestablish equilibrium with its conjugate.

So the acid will donate to the water to give me the base plus H3O+.

And we know we will put an ICE table together here.

To get an ICE table, we're going to need to know the concentration of ammonium.

So the concentration is moles per liter.

We know the moles of ammonium that we have in solution that was 0.0658.

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And that would give me the molarity of this solution,

which is 0.196, and that would be molarity.

And I can plug that into the table, 0.196.

For this reestablish is equilibrium, those values are 0.

In an ICE table, we will consume some of the reactant and

produce the products in this case.

This will lead me with 0.196 -x, x and x for that table.

Now since this is an ICE table for ammonium, which is a weak acid,

we'll need the Ka value to work the problem.

They gave me the Kb of the base, so

I will take Kw divided by the 1.8 x 10 to the -5.

And I will set that equal to products over reactants

raised to the power of their coefficients.

I can assume that x is very small, and then I can solve for x,

you might want to stop the video and solve for x.

But if you do, the x value is equal to 1.04 x 10 to the -5.

Now I've carried an extra significant figure along there,

I really only know it to 2 because of my Kb value, but I'm carrying extra.

Now x is the H3O+ concentration, so

if I took the negative log of that value, I'd get the pH.

The negative log is equal to 4.98.

So that is the pH at the equivalence point between ammonia and a nitric acid.

And we again see that we should have a pH less than 7 when

the acid is strong and the base is weak.