Question

a 50 mL sample consisting of 0.15 M nicotine C10H14N2 is tot rated with 0.25 M HBr. The Ka of nicotine is 1.0*10^-6 calculate the pH after adding 5.0 mL of HBr at the equivalence point and after adding 5.0 mL HBr beyond the equivalence point.

Answer 1

millimoles of nicotine = 50 x 0.15 = 7.5

Ka = 1.0 x 10^-6

At the equivalence point :

millimoles of base = millimoles of acid

7.5 = 0.25 x V

V = 30 mL

volume at equivalence point = 30 mL

after the addition of 5.0 mL HBr beyond equivalence point :

beyond the equivalence point :

volume = 30 + 5 = 35

millimoles of HBr = 35 x 0.25 = 8.75

C10H14N2 +   HBr   ---------------->   salt

   7.5               8.75                             0

    0                 1.25                           7.5

here strong acid remains. so

concentration of HBr = 1.25 / (50 + 35) = 0.015 M

pH = -log [H+] = -log [0.015]

      = 1.83

pH = 1.83