Question

Consider the titration of a 28.0 −mLsample of 0.180 M CH3NH2 with 0.155 M HBr. Determine each of the following.

the inital pH

the volume of added acid required to reach the equivalence point

the pH at 6.0 mL of added acid

the pH at one-half the equivalence point

the pH at the eqivalence point

the pH after adding 6.0 mL of acid beyond the equivalence point

Answer 1

CH3NH2 , Kb = 5.0 x 10^-4

pKb= 3.30

initial pH :

CH3NH3 + H2O -----------------------> CH3NH3+   + OH-

0.180                                                         0                    0 --------------> initial

0.180-x                                                      x                     x ----------------> equilbrium

Kb = x^2 / 0.180 -x

5.0 x 10^-4 = x^2 / 0.180 -x

x^2 + 5.0 x 10^-4 x - 9 x 10^-5 = 0

x = 9.24 x 10^-3

[OH-] = 9.24 x 10^-3 M

pOH = -log ( 9.24 x 10^-3)

pOH = 2.03

pH + pOH = 14

pH = 11.97

the volume of added acid required to reach the equivalence point

at equivalece point : millimoles of acid = millimoles of base

28 x 0.180 = 0.155 x V

V = 32.5 mL

volume of acid = 32.5 mL

the pH at 6.0 mL of added acid

millimoles of acid = 0.155 x 6.0 = 0.93

millimoles of base = 28 x 0.180 = 5.04

CH3NH2 + HBr ---------------------> CH3NH3+Br-

5.04 0.93    0

4.11 0 0.93

pOH = pKb + log (0.93 / 4.11)

pOH = 2.66

pH + pOH = 14

pH = 11..34     

the pH at one-half the equivalence point

at this point

pOH = pKb

pOH = 3.30

pH = 10.70