Question

A 75.0-ml volume of 0.200M NH3 (Kb=1.8 x 10^-5) is titrated with 0.500M NHO3 . Calculate the pH after the addition of 13.0mL of NH3.

Answer 1

This forms a buffer.
meq of NH3=75*0.2=15
meq of HNO3=0.5*13=6.5
meq of NH3 left=8.5.
meq of NH4+ = 6.5
By Henderson's equation
pOH= pKb + log (base/salt)
pOH=5-log1.8+log85/65
pOH=4.86
pH=14-4.86=9.14