Question

Calculate the pH of the solution after the addition of the following amounts of 0.0656 M HNO3 to a 50.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04. a) Volume of Hno3 equal to half the equivalence point volume b)53.7mL of HNO3 c) Volume of Hno3 equal to the equivalence point

Answer 1

pkb of aziridine = 14-pka = 14-8.04 =5.96

we have Henderson equation   pOH = pkb + log [aziridinium ]/[aziridine]

a) At half equivalence point pOH = pkb , hence pOH = 5.96 , pH = 14-5.96 = 8.04

b) Moles of H+ added = M x V = 0.0656 x 53.7/1000 = 0.003523

Moles of azridine)= M x V = 0.075 x 50/1000 = 0.00375

Now Azridine reacts with H+ to form Aziridinium

hene azridne moles after H+ reaction = 0.00375-0.003523= 0.000227

Moles of aziridinium = moles of H+ added = 0.003523

Now pOH = 5.96 +log ( 0.003523/0.000227)    ( vol term of concentration cancells out from numberator and denominator)

pOH = 7.15

c) at equivalnece point H+ moles = aziridine moles = 0.00375

hence azridinium moles formed = 0.00375

Now we have reverse equilibrium

Aziridinium <---> Azridine + H+

Ka = 10^ -8.04 = 9.12 x 10^-9

Molarity of H+ = Moles of H+ / vol   hence 0.0656 = 0.00375/vol of H+

vol of H+ = 0.05716 L , total vol of solution = 0.05716+0.05= 0.10716 L

Now at equilibrium [Azirdinium] = ( 0.00375-X) / 0.10716 , [H+] =[Aziridjne]= X/0.10716

Ka = [H+] [Aziridien]/[Aziridinium]

9.12 x 10^-9 = ( X/0.10716)^2 / ( 0.00375-X) /0.10716

9.773 x 10^-10 = X^2 / ( 0.00375-X)

X = 1.9 x 10^ -6 , [H+] = X/0.10716 = ( 1.9x10^-6)/ 0.10716 = 1.7865 x 10^ -5

pH = -log [H+] =-log ( 1.7865 x 10^-5) = 4.75