Question

Calculate the pH of the solution after the addition of the following amounts of 0.0577 M HNO3 to a 80.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04

a) 0.00 mL of HNO3 b) 7.72 mL of HNO3 c) Volume of HNO3 equal to half the equivalence point volume d) 99 mL of HNO3

e) Volume of HNO3 equal to the equivalence point f) 107 mL of HNO3

Answer 1

aziridine is C2H5N

pKa Aziridine) = 8.04 so -log Ka = 8.04 Therefore Ka = 10^-8.04

OH - ion conc = (Ka*C)^1/2   = (0.0750*10^-8.04)^1/2   = 2.6154*10^-5

a)

Moles of H+ ion - moles of OH - ion = 0.0577*0.00/1000 - 2.6154*80/1000     (because moles = molarity * volume)

                                = 0 - 0.2090

                              = - o.2092

so OH- ionare in excess , = + .2092

so pOH = = -log (OH-) = - log(0.2092) = 0.6794

pH = 14 - pH = 14 - 0.6794 = 13.32

b)

moles of H+ ion - moles of OH-

   = 0.0577*7.72/1000 - 2.6654*10^-5 * 80/1000

     = 4.454*10^-4 - 2.09232*10^-6

    = 4.433*10^-4 = excess H+ ion conc

pH = - log (H+ ion conc 0 - log(4.433*10^-4) = 3.3533

d)

moles of H+ ions - moles of OH-

    = 0.0577*99/1000 - 2.6154*10^-5*80/1000

   = 5.7123*10^-3 - 2.09*10^-6

so excess of H+ ion conc = 5.71020*10^-3

pH = - log (H+ ion conc = - log(5.71021*10^-3) = =2.2433

f)

moles of H+ ions - moles of OH- ions

   = 0.0577*107/1000 - 2.6154*10^-5 *80/1000

             = 6.1739*10^-3 - 2.09232*10^-6

             = 6.1718*10^-3   = excess H+ ions

so pH = - log H+ ion conc = - log(6.1718*10^-3)   = +2.2096

e)

At equivalence point

pH = 14 + log (Ca*Cb(Kw) /(Ca+Cb)*Ka ) ^1/2

     = 14 + log (0.0577*2.6154*10^-5 * 10^-14 /(0.0577+2.6154*10^-5 ) * 10^-8.04 )^1/2

    =    14 + log (1.509*10^-20/5.2647*10^-10)^1/2 = 14-10.5426   = 3.457

c )

At half equivalence point

   pH = pKa + log base/acid  

But base = acid   = 1 say

so pH = pKa + log 1/1

pKa + log 1 = pKa + 0 = 8.04 +0 = 8.04