Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0538 M HNO, to a 70.0 mL solution of 0.0750 M aziridine. The pK, of aziridinium is 8.04. What is the pH of the solution after the addition of 0.00 mL HNO,? pH = What is the pH of the solution after the addition of 5.64 mL HNO,? pH = 9.1889 What is the pH of the solution after the addition of a volume of HNO, equal to half the equivalence point volume? pH = 8.292 What is the pH of the solution after the addition of 92.7 mL HNO,? pH = 6.762 What is the pH of the solution after the addition of a volume of HNO, equal to the equivalence point? pH = What is the pH of the solution after the addition of 101.90 mL HNO,? pH =

Answer 1

pKa = 8.04

pKb = 5.96

molarity of aziridine = 0.0750 M = C

a ) 0.00 mL HNO3 added:

pOH = 1/2 [pKb -logC]

= 1/2 [5.96 - log 0.0750]

= 3.54

pH + pOH =14

pH = 10.46

b ) 5.64 mL HNO3 added

millimoles of base aziridine = 0.0750 x 70 = 5.25

millimoles of HNO3 = 5.64 x 0.0538 = 0.289

B + H+ ----------------> BH+

5.25 0.289    0

4.96 0 0.289

pH = pKa + log (4.96 / 0.289)

pOH =8.04 +  log (4.96 / 0.289)

pH = 9.28

c )

at half equivalence point : pOH = pKb

pOH = 5.96

pH = 8.04

d) 92.7 mL HNO3 added

millimoles of base aziridine = 0.0750 x 70 = 5.25

millimoles of HNO3 = 4.99

B + H+ ----------------> BH+

5.25 4.99 0

0.263    0 4.99

it is buffer use above formula

pH = 6.76

e ) at equivalece point

only salt remains

salt concetration = 5.25 / 97.6 + 70 = 0.0313 M

pH = 7 - 1/2 [pKb + logC]

pH = 7 -1/2 [5.96 + log 0.0313]

pH = 4.77

f ) 101.90 mL HNO3

pH = 2.22