Question

Calculate the pH of the solution after the addition of the following amounts of 0.0536 M HNO3 to a 60.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04.

a) 0.00 mL of HNO3; Ph=

b)8.28mL of HNO3; Ph=

c) 42.0ml of HNO3

d) 84.0 mL of HNO3; Ph=

f) 88.5 mL of HNO3; Ph=

Answer 1

millimoles of pyridine = 60 x 0.0750 = 4.5

pKa = 8.04

pKb = 14 - 8.04 = 5.96

Kb = 1.096 x 10^-6

a) 0.00 mL addition of HNO3

pOH = 1/2 [pKb -logC] = 1/2 [5.96 -log0.0750] = 3.54

pH + pOH = 14

pH = 10.46

b) after the addition of 8.28 mL HNO3

millimoles of HNO3 = 8.28 x 0.0536 = 0.4438

millimoles of aziridine = 4.0562

millimoles of salt = 0.4438

pOH = pKb + log [salt / base]

       = 5.96 + log [0.4438 / 4.0562]

       = 5.0

pH = 9.00

c) after the addition of 42 mL HNO3

millimoles of acid = 42 x 0.0536 = 2.25

it is half equivalence point . so

pOH = pKb

pOH = 5.96

pH +pOH =14

pH = 8.04

d) after the addition of 84 mL HNO3

millimoles of HNO3 = 84 x 0.0536 = 4.5

it is equivalence point only salt is formed

salt millimoles = 4.5

salt concentration = millimoles / total volume = 4.5 / (60 + 84) = 0.3125 M

salt is from strong acid weak base so pH <7

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

pH = 7 - 1/2 [5.96 + log 0.3125]

pH = 4.77

e) after the addition of 88.5 mL HNO3

millimoles of HNO3 = 88.5 x 0.0536 = 4.74

here strong acid remians. so

millimoles of HNO3 = 4.74 - 4.5 = 0.2436

[HNO3] = 0.2436 / (88.5 + 60) = 0.00164 M

pH = -log(0.00164)

pH = 2.78