Question

Consider the titration of a 26.0 −mL sample of 0.180 M CH3NH2 with 0.155 M HBr. Determine each of the following.

a) the pH at 5.0 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 5.0 mL of acid beyond the equivalence point

Answer 1

millimoles of CH3NH2 = 26 x 0.180 = 4.68

pKb of CH3NH2 = 3.30

a) the pH at 5.0 mL of added acid

millimoles of HBr = 0.155 x 5 = 0.775

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

4.68           0.775                            0

3.905           0                                 0.775

pOH = pKb + log [CH3NH3Br / CH3NH2]

pOH = 3.30 + log (0.775 / 3.905)

pOH =2.60

pH + pOH = 14

pH = 11.40

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

here pOH = pKb = 3.30

pH + pOH = 14

pH = 10.70

c) the pH at the equivalence point

here HBr volume = 4.68 / 0.155 = 30.2 mL

here only salt remains = salt concentration = 4.68 / (26 + 30.2) = 0.0833 M

pH = 7 - 1/2 [pKb + log C]

pH = 7 - 1/2 [3.30 + log 0.0833]

pH = 5.89

d) the pH after adding 5.0 mL of acid beyond the equivalence point

total volume = 30.2 + 5 =35.2 mL

[H+] = 0.155 x 35.2 - 4.68 / (26 + 35.2) = 0.0127 M

pH = -log [H+]

pH = -log (0.0127)

pH = 1.90