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.