Question

Consider the titration of a 26.0 −mL sample of 0.180 M CH3NH2 with 0.150 M HBr. Determine each of the following.

a) the pH at 4.0 mL of added acid

b) the pH at one-half of the equivalence point

c) the pH at the equivalence point

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

Answer 1

a)

mmoles of CH3NH2 = 26.0 x 0.180 = 4.68

mmoles of HBr = 4.0 x0.150 = 0.6

pKb ofCH3NH2 = 3.30

pKa = 10.70

CH3NH2 +   HBr   -----------> CH3NH3+

4.68             0.6                            0

   4.08              0                           0.6

pH = pKa + log [base / acid]

     = 10.70 + log [4.08 / 0.6]

pH = 11.53

b)

At half - equivalence point :

pOH = pKb

pOH = 3.30

pH = 10.70

c)

At equivalence point :

volume of HBr = 31.2 mL

salt concentration = 4.68 / 31.2 + 26 = 0.0818 M

pH = 7 - 1/2 (pKb + log C)

     = 7 - 1/2 (3.30 + log 0.0818)

pH = 5.89

d)

volume = 37.2 mL

[H+] = 0.0142 M

pH =1.85