posted by Anonymous
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.
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.