Consider the titration of 300.0 mL of 0.500 M NH3 (Kb = 1.8 ×10-5) with 0.500 M HNO3. At the stoichiometric point of this titration the pH is?

The first answer is incorrect. At the stoichiometric point, the concentration of the NH3 is almost zero (all of the NH3 has reacted with H+). To a first approximation, it is zero. All that's present is NH4+ (plus the NO3- spectator ions, which we'll ignore). Since NH3 is a weak base, its conjugate (NH4+) is a weak acid, with a Ka of its own.

The reaction involved at this point is:

NH4+ <==> H+ + NH3

Ka = ([H+] * [NH3]) / [NH4+]

However, we're given Kb for NH3, not Ka for NH4+. Fortunately, it's easy to calculate the Ka for NH4+

Ka (NH4+) = Kw / Kb (NH3) = 5.6 x 10^-10

Now, in order to find the pH, we need to find [H+] (we'll use the variable 'x' for simplicity). Realize that whenever an H+ is formed, an NH3 molecule is also formed, so [NH3] also equals 'x'. The [NH4+] was calculated correctly in the previous answer, and is 0.25 M

Thus, Kb = x^2 / 0.25 = 5.6 x 10^-10

x^2 = 0.25 * 5.6 x 10^-10 =1.4 x 10^-10

x = 1.18 x 10^-5

Remember that x stands for [H+], and pH = -log [H+]

pH = -log (1.18 x 10^-5) = 4.93

A couple of comments:

1) Always think back to the problem, to make sure your answer makes sense. At the equivalence point, all of the NH3 has been converted to NH4+. Since NH3 is a weak base, NH4+ is a weak acid. Since you have a solution of a weak acid, the solution must be acidic. If the solution is acidic, the pH must be less than 7.

2) The point at which the pH = pKa, or pOH = pKb occurs at the half-equivalence point, (i.e., when enough titrant has been added to react with half of the acid or base present). In the situation you describe, that would have happened when 150 mL of HNO3 had been added.

3) Remember when I said [NH3] was essentially zero? Look above, and you'll see that it is actually equal to 'x'. Since x = 1.18 x 10^-5, [NH3] is also equal to 1.18 x 10^-5 M (which is pretty close to zero compared to its initial value of 0.3 M)