Question

Calculate the pH of the solution after the addition of the following amounts of 0.0515 M HNO3 to a 60.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04.

e) Volume of HNO3 equal to the equivalence point (have to calculate the volume of HNO3 this is not given in the question)

f) 90.5 mL of HNO3

Answer 1

e) pkb of aziridine = 14-8.04 = 5.96

No of mol of waqq taken = 0.075*60 = 4.5 mmol

   No of mol of HNO3 added = 4.5 mmol (equivalence point)

volume of HNO3 must add = 4.5/0.0515 = 87.4 ml

concentration of salt(aziridine-HNO3) = 4.5/(87.4+60) = 0.0305 M

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

   pkb = 5.96

C = 0.0305 M

pH = 7-1/2(5.96+log0.0305)

    = 4.78

f) excess volume of HNO3 = 90.5-87.4 = 3.1 ml

    concentration of excess HNO3 = 3.1*0.0515/(90.5+60)

                                 = 0.00106 M

    pH = - log[H+]

       = - log0.00106

       = 2.975