Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0504 M HNO, to a 60.0 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 8.55 mL HNO,? pH = What is the pH of the solution after the addition of a volume of HNO, equal to half the equivalence point volume? pH = What is the pH of the solution after the addition of 85.9 mL HNO,? pH = 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 92.4 mL HNO,? pH =

Answer 1

Given that concentration of HNO3= 0.0504 M concentration of Aziridine (C,HN) = 0.0750 M Volume of C H N = 60.00 mL pka of aziridium ion = 8.04 pKb for aziridine = 14 - pka = 14 – 8.04 = 5.96 Ko of C2H5N = 1.10 x 10-6 a) Before Addition of any acid: CHEN + H20 HN + H2O = 9H,NH* GH_NH + OH- Initial 0.0750 Change Equillibrium 0.0750 - X Kr. =1.10x100 0.0750- x x is small, so 0.0750- x=0.0750 X.X S omen=1.10×100 0.0750 = x² = 8.22x10-8 > x= 2.87x10+ :: [OH]= 2.87x104M pOH = -log(2.87x10+) = 3.54 pH = 14- pOH = 14-3.54 pH =10.46

b) Addition of 8.55 mL of HNO3 : Number of moles of C,H-N = M*V = 0.0750 M* 60.0 mL = 4.5 mmol Number of moles of HNO3 = M*V = 0.0504 M* 8.55 mL = 0.43092 mmol Net moles of C H N = 4.5 -0.43092 = 4.06908 mmol mmol CHEN + HNO3 = C2H5NH+NO; Initial 4.5 Change -0.43092 -0.43092 -0.43092 0.43092 Equillibrium 4.06908 acid)596 +log According to Henderson's equation 0.005949 pOH = PK, +log <= 5.96-0.9751 [base] 0% 0.060198 pOH = 4.9849 pH =14- pOH = 14-4.9849 = 9.02

c) At Half equivalence point: At half equivalence point, pH = pka = 8.04

d) Addition of 85.9 mL of HNO3: Number of moles of C H N = M*V = 0.0750 M* 60.0 mL = 4.5 mmol Number of moles of HNO3 = M*V = 0.0504 M* 85.9 mL = 4.32936 mmol Net moles of C H N = 4.50 – 4.32936 = 0.216 mmol Kh mmol C2H5N + HNO3 = C2H5NH*NO3 Initial 0 0 4.50 - 4.32936 - 4.32936 +4.32936 Change Equillibrium 0.17064 4.32936 = 5.96 +1.404343 According to Henderson's equation 4.32936 pOH = pK, +log bacia = 5.96 +log [base] 2011090.17064 pOH = 7.36 pH =14 - pOH =14 – 7.36 = 6.64

e) At equivalence point Number of moles of C H N = M*V = 0.0750 M* 60.0 mL = 4.5 mmol At the equivalence point, the number of moles of HNO3 is equal to the number of moles of base. So Number of moles of HNO3 = 4.5 mmol Volume of HNO3 needed to reach equivalence point = moles/molarity = 4.5 mmol/0.0504 M = 89.29 mL At this point, there is no C H N is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate acid. moles [C,HNH*NO;-]= Total volume 4.50 mmol 4.50 mmol -=0.030143M 60+89.29 149.29 mL Dissociation of C H NH+ is given by M CHşnci — C2H5N + H+ Initial 0.030143 0 Change - X Equillibrium 0.030143-

A = = -= 10-8.04 0.030143-X x is small, so 0.030143 - x = 0.033578 > 0.030143 12011x10-9 3x² = 2.69452x10-10 >x=1.6415x10-5 ::[H*]=1.6415x10--M pH =- log(1.6415x10-4)=4.78 f) Addition of 92.4 mL of HNO3: Number of moles of C H N = M*V = 0.0750 M* 60.0 mL = 4.5 mmol Number of moles of HNO3 = M*V = 0.0504 M* 92.4 mL = 4.65696 mmol This is after the equivalence point. At this point, the pH of the solution is determined by the amount of unreacted HCl in the solution. Net moles of HNO2 = 4.65696 -4.50 = 0.15696 mmol Total volume of the solution = 60 + 92.4 = 152.4 mL Conc. of HNO3 = [H+] = moles/ volume = 0.15696 mmol / 152.4 mL = 0.00103 M pH =- log [H+] = -log[0.00103] = 2.99