Titration problem?

Question

What is the pH at the equivalence point in the titration of 10.0 mL of 0.36 M HZ with 0.200 M NaOH? Ka= 6.6 x 10^-7 for HZ

Answer 1

At the equivalence point, all of HZ will be converted to Z- because the protons will be picked up by the equivalent concentration of OH-.  Therefore, .36 M of Z- will dictate the pH.  In order to find the pH, we must first use Kb to find pOH.

Kb = Kw/Ka = 10^-14/6.6E-7 = 1.52E-8

Kb = products/reactants

The chemical equation is as follows:

Z- + H2O ---> HZ + OH-

Kb = [OH-][HZ]/[Z-]

(H2O doesn't need to be in the equation because all pure solids and liquids have a concentration close enough to 1)

Assuming x amount of Z- will react with H20 to form x amount of OH- and x amount of HZ, there will be .36-x Z- remaining.

1.52E-8 = x^2/.36-x

Assuming .36-x = .36 is OK to do because Kb < 10^-4.

1.52E-8 = x^2/.36

1.52E-8(.36) = x^2

x = 7.39E-5

pOH = -log(7.39E-5) = 4.13

pH = 14 - pOH = 14 - 4.13 = 9.87

Intuitively, a pH of 9.87 makes sense.