Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0503 M HNO3 0.0503 M HNO3 to a 60.0 mL solution of 0.0750 M 0.0750 M aziridine. The pKa of aziridinium is 8.04

What is the pH of the solution after the addition of 0.00 mL HNO3? 0.00 mL HNO3?

pH=

What is the pH of the solution after the addition of 8.69 mL HNO3? 8.69 mL HNO3?

pH=

What is the pH of the solution after the addition of a volume of HNO3 equal to half the equivalence point volume?

pH=

What is the pH of the solution after the addition of 86.2 mL HNO3? 86.2 mL HNO3?

pH=

What is the pH of the solution after the addition of a volume of HNO3 equal to the equivalence point?

pH=

What is the pH of the solution after the addition of 93.2 mL HNO3?

pH=

Answer 1

pKb = 5.96

A) 0.00 mL HNO3 added :

pOH = 1/2 [pKb -log C]

          = 1/2 [5.96 -log 0.0750 ]

         = 3.54

pH + pOH = 14

pH = 10.46

B) 7.14 mL HNO3 :

mmoles of base = 0.0750 x 60= 4.5

mmoles of H+   = 0.053 x 8.69 = 0.461

B   + H+ ----------------> BH+

4.5    0.461                     0

4.04    0                    0.461

pH = pKa + log (B / BH+)

pH = 8.04 + log (4.04 / 0.461)

pH = 8.98

C) volume of HNO3 at half equaivalence point :

pH = pKa

pH = 8.04

D) 62.7 mL HNO3 :

mmoles of base = 0.0750 x 60= 4.5

mmoles of H+   = 0.053 x 86.2 = 4.57

pH = 3.32

E) volume at equivalece point

pH = 4.77

F) 93.2 mL HNO3 :

pH = 2.54