Question

Aniline hydrochloride, [C6H5NH3]Cl, is a weak acid. Its conjugate base is the weak base aniline C6H5NH2. Assume 50.0 mL of 0.100 M C6H5NH3 is titrated with 0.125M NaOH. (C6H5NH3)Cl Ka = 2.40 X 10-5. What is the ph at 40.0ml Calculate the pH after 100.0 ml of NaOH has been added

Answer 1

1)

millimoles of C6H5NH3 = 50 x 0.1 = 5

millimoles of NaOH = 0.125 x 40 = 5

C6H5NH3 + NaOH -----------------------> C6H5NH2 + H2O

5                   5                                     0                0

0                  0                                       5

C6H5NH2   concentration = 5 / (50 + 40) = 0.056 M

pOH = 1/2 [pKb - log C]
pOH   = 1/2 [9.38 - log 0.056]

pOH = 5.32

pH + pOH = 14

pH = 8.68

2)

millimoles of C6H5NH3 = 50 x 0.1 = 5

millimoles of NaOH = 0.125 x 100 = 12.5

C6H5NH3 + NaOH -----------------------> C6H5NH2 + H2O

5                 12.5                                    0                0

0                 7.5                                   5

NaOH strong base its concentration = 7.5 / (50 + 100) = 0.05M

pOH = -log OH-
pOH = 1.30

pH + pOH = 14

pH = 12.70