An analytical chemist is titrating 220.3 mL of a 0.3600 M solution of diethylamine ((C2H5)2NH) with a 0.2600 M solution of HNO3. The pKb of diethylamine is 2.89. Calculate the pH of the base solution after the chemist has added 125.1 mL of the HNO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added. Round your answer to 2 decimal places.

Solution:

millimoles of (C2H5)2NH (base) = 220.3 x 0.3600 = 79.308

millimoles of HNO3 (acid) = 125.1 x 0.2600 = 32.526

(C2H5)2NH + HNO3 > (C2H5)2NH2+NO3- (salt)

79.308           32.526             0    > before reaction

46.782             0                  32.526     > after reaction

in the solution only base and salt remained . so it can form basic buffer

For Basic buffer  

Henderson-Hasselbalch equation

pOH = pKb + log [salt/base]

pOH = 2.89 + log ( 32.526 / 46.782)

pOH = 2.73

pH + pOH = 14

pH = 14-pOH

pH = 11.27

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .