Question

Calculate the pH of the solution after the addition of each of the given amounts of .0656M HNO3 to a 60mL solution of .075M aziridine. The pKa of aziridinium is 8.04 What is the pH of the solution after the addition of 0mL HNO3? What is the pH of the solution after the addition of 6.6mL HNO3 What is the pH of the solution after the addition of a volume of equal to half the equivalence point volume? What is the pH of the solution after the addition of 64.3mL HNO3 What is the pH of the solution after the addition of a volume of equal to the equivalence point? What is the pH of the solution after the addition of 73.07mL of HNO3?

please answer will rate

Answer 1

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 ) 6.60 mL HNO3 added

millimoles of base aziridine = 0.0750 x 60 = 4.50

millimoles of HNO3 = 6.60 x 0.0656 = 0.433

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

4.50 0.433    0

4.067 0 0.433

pH = pKa + log (4.067 / 0.433)

pOH =8.04 +  log (4.78 / 0.472)

pH = 9.01

c )

at half equivalence point : pOH = pKb

pOH = 5.96

pH = 8.04

d) 64.3mL HNO3 added

millimoles of base aziridine =4.50

millimoles of HNO3 = 4.22

B +    H+ ----------------> BH+

4.50 4.22 0

0.282    0 4.97

it is buffer use above formula

pH = 6.79

e ) at equivalece point

only salt remains

salt concetration = 4.50 / 68.6 + 60 = 0.035 M

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

pH = 7 -1/2 [5.96 + log 0.036]

pH = 4.75

f ) 73 mL HNO3

pH = 2.66