a. Kb = 1.8 x 10^-5
b. pH = 7.82
c. pH = 4.82
d. Thymolphtalein is most suitable for this titration because its pKa value is closest to the equivalence point of the titration (4.82).
C6H5NH2 + H2O C6H5NH3+ + OH-
1. Aniline, a weak base, reacts with water according to the rxn above.
a. A sample of aniline is dissolved in water to produce 25.0 ml of a .10 M soln. The pH of soln is 8.82. What is Kb for this rxn?
b. The soln in part b, is titrated with .10 M HCl. Calculate pH of the soln when 5 mL of acid is added.
c. Calculate pH at equivalence point of the titration in part c.
d. The pKA values for several indicators are given below. Which of the indicators is most suitable for this titration. Justify your answer.
Indicator pKa
Erythrosine 3
Litmus 7
thymolphtalein 10
1 answer