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 Â
---------------->Â Â salt
 Â
7.5Â Â Â Â Â Â Â Â Â Â Â Â Â Â
8.75Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
  Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
1.25Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
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
