Consider the titration of 50.0 mL of 0.20 M NH3 (Kb = 1.8x10^-5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant.

Answer 1

The secret to doing titration problems is to recognize what you have in the solution. In this case, the NH3 is exactly neutralized by the HNO3; therefore, you are at the equivalence point and you have NH4NO3 is solution. How much NH4NO3? That will be 50.0 mL x 0.20 M = 10 millimoles in 100 mL = 0.1 M

Write the hydrolysis equation.
NH4^+ + HOH ==> NH3 + H3O^+
Then do the ICE bit.
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+
Solve for (H3O^+) and convert to pH.

Answer 2

5.13

Answer 3

8.5