Question

a 25.0 ml solution of 0.1000M benzylamine (pkb= 4.67) was titrated with 0.1000M HBr. Calculate the pH of solution at the following volumes of added acid: 0.00ml, 5.00ml, 12.50ml, 25.00ml and 30.00ml

Answer 1

a. before addition of HBr

   pH = 14 - 1/2(pkb-logC)

   pkb = 4.67

   C = concentration of benzylamine = 0.1 M

pH = 14 - 1/2(4.67-log0.1)

    = 11.16

b) no of mol of benzylamine = 0.1*25 = 2.5 mmol

    no of mol of HBr = 5*0.1 = 0.5 mmol

   pH = 14 - (pkb+log(acid/base))

      = 14 - (4.67+log(0.5/(2.5-0.5))

      = 9.93
c) no of mol of benzylamine = 0.1*25 = 2.5 mmol

    no of mol of HBr = 12.5*0.1 = 1.25 mmol

   pH = 14 - (pkb+log(acid/base))

      = 14 - (4.67+log(1.25/(2.5-1.25))

      = 9.33

   d) no of mol of benzylamine = 0.1*25 = 2.5 mmol

    no of mol of HBr = 25*0.1 = 2.5 mmol

   now the system reaches to equivalence point.

concentration of salt = 2.5/(25+25) = 0.05 M

pH = 7-1/2(pkb+logC)

     = 7-1/2(4.67+log0.05)

     = 5.31

   e) concentration of excess HBr = (30-25)*0.1/(30+25) = 0.0091 M

      pH = -log(H+)

         = -log0.0091

         = 2.041