Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0502 M HNO3 to a 70.0 mL solution of 0.0750 M aziridine. The p?a of aziridinium is 8.04.

What is the pH of the solution after the addition of 7.78 mL HNO3?

What is the pH of the solution after the addition of 101 mL HNO3?

What is the pH of the solution after the addition of a volume of HNO3 equal to the equivalence point?

What is the pH of the solution after the addition of 108.89 mL HNO3?

Answer 1

Given that concentration of HNO3= 0.0502 M concentration of Aziridine (CHN) = 0.0750 M Volume of C H N = 70.0 mL pka of aziridiumion = 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: Ko CHEN + H20 HN + H2O = 9H,NH* GH_NH + OH- Initial 0.0750 Change Equillibrium 0.0750 - X Kr. S omen=1.10×100 -=1.10x10 0.0750- x x is small, so 0.0750- x=0.0750 X.X 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 7.78 mL of HNO3: Number of moles of C H N = M*V = 0.0750 M* 70.0 mL = 5.25 mmol Number of moles of HNO3 = M*V = 0.0502 M* 7.78 mL = 0.390556 mmol Net moles of C H N = 5.25 -0.390556 = 4.859444 mmol mmol CHEN + Kb HNO3 = C2H5NH+NO; Initial 4. 50 -0.390556 -0.390556 Change +0.390556 Equillibrium 4.859444 0.390556 According to Henderson's equation pOH = pK, +log acid) = 5.96+log 0:390556 = 5.96–1.0949 [base] pOH = 4.87 pH =14 - POH = 14 -4.87 =9.13 4.859444 pot 4 *POH =14-4

c) Addition of 101 mL of HNO3: Number of moles of C H N = M*V = 0.0750 M* 70.0 mL = 5.25 mmol Number of moles of HNO3 = M*V = 0.0502 M* 101 mL = 5.0702 mmol Net moles of C H N = 5.250 - 5.0702 = 0.216 mmol mmol CH3N + Kb HNO3 = C2H5NH*NO, Initial 0 4.50 - 5.0702 0 - 5.0702 Change +5.0702 Equillibrium 0.1798 5.0702 According to Henderson's equation 5.0702 (acid]_596 +log - 0.1798 = 5.96 +1.450235 pOH = pK, +log = 5.96 +log [base] pOH = 7.41 pH = 14 – POH =14 – 7.41 = 6.59

d) At equivalence point Number of moles of C H N = M*V = 0.0750 M* 70.0 mL = 5.25 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 HNO2 = 5.25 mmol Volume of HNO3 needed to reach equivalence point = moles/molarity = 5.25 mmol/0.0502 M = 104.58 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 5.25 mmol 5.25 mmol [C;H3NH*NO, ]= = 0.0301M Total volume 70+104.58 174.58 mL Dissociation of C H NH+ is given by M CHEN* К. = = C2H5N + H+ Initial 0.0301 Change -X Equillibrium 0.0301 - x

x.x <a -=10-8.04 0.0301-X x is small, so 0.0301- x = 0.0301 x.x 0.0301 = 9.12011*10-4 x = 2.74x10-10 >x=1.66x10 :: [H*]=1.66x10-M pH =- log(1.66x10- ) = 4.78 e) Addition of 108.89 mL of HNO3: Number of moles of C H N = M*V = 0.0750 M* 70.0 mL = 5.25 mmol Number of moles of HNO3 = M*V = 0.0502 M* 108.89 mL = 5.466278 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 HNO3 = 5.466278 -5.250 = 0.21628 mmol Total volume of the solution = 70.0 + 108.89 = 178.89 mL Conc. of HNO3 = [H+] = moles/ volume = 0.21628 mmol / 152.4 mL = 0.001209 M. pH =- log (H+) = -log[0.001209] = 2.92