0:01

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

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

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

Â We have now in a flask the ammonia solution,

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

Â So let's begin by writing the reaction.

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

Â The weak base is NH3.

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

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

Â becomes ammonium, and we're left with water.

Â 0:49

The one-way reaction makes it an ICF table,

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

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

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

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

Â which is 0.350 moles per liter.

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

Â That will give me a moles of base 0.0658.

Â 1:33

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

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

Â All of the 0.0658 moles of this and the 0.0658 moles

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

Â The conjugate acid of ammonia, that consumes these substances,

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

Â 2:12

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

Â We know moles of the HNO3,

Â 0.0658 moles of HNO3 were placed in the flask.

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

Â to liters using the molarity of 0.447 moles per liter.

Â This will obtain for us the fact that we have

Â 0.147 moles of that acid, HNO3.

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

Â That is equal to 147 milliliters of HNO3.

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

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

Â ammonium that acid in water.

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

Â and it will reestablish equilibrium with its conjugate.

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

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

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

Â So the concentration is moles per liter.

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

Â 4:02

And that would give me the molarity of this solution,

Â which is 0.196, and that would be molarity.

Â And I can plug that into the table, 0.196.

Â For this reestablish is equilibrium, those values are 0.

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

Â produce the products in this case.

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

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

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

Â They gave me the Kb of the base, so

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

Â And I will set that equal to products over reactants

Â raised to the power of their coefficients.

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

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

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

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

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

Â Now x is the H3O+ concentration, so

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

Â The negative log is equal to 4.98.

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

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

Â the acid is strong and the base is weak.

Â