Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0676 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 6.99 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 62.9 mL HNO,? pH =
please explain as best as possible.

Answer 1

60 me solution of o.o75 (m) aziridine (Az) Number of mols & 6020.075 x 103 mole of Az 4.5 mmdes of Az. Az + H20 F ARH' + OH- = [H2H] [04-) » [OH-]² = ko [hz] Giver, [AZ]= 0.075 (M) Pkw = 14-pka = 14-8.04 - 5.96 Ko = [az] = lo 5.96 ③ Before addition of HNO3 : [OH-] = (108.96 0.075)? - 2.868x104 CM) - pot = log, 50H-] = 3.54 pH = 14-pOH = 14-3.54 * 10:46 7 After addition of 6.99 m2 0.0676(M) HNO 3 : AZ + H+ - Azh* According to Henderson-Hasselbalch equation, pon - photo log [A24 Number of mores troz added = 6.99*0.0676x10 more = 0.473 mmole of HNO3 So, the number of AzHt formed = 0.473 mmola the number of Az after reaction = (4.5-0.473) mmile = 4.028 minore pOH = 5.96+ log 24.028713 - 5.96-0.93 25.03 P+ 14 - -- 14 -5:03 = 891 va total volume

- 3 ② At half equivalence point, [AZH+] = [AZ ] [2] S so, poh = pkb = 5.96 pH = 14-5.96=8.04 - ② After addition of 62.9 me HNO3: number of moles of Hoog added = 62.9*0:0676 *10 mile = 4.252 mimores of two so, the moles of Azht produced = 4,252 mmoles - unreacted after addition of Horong = (4.5-4.252) mundes = 0.248 mmoler. so, pot pke, log (4.252/V) = 5186 +1.234 = 7.184 pret thang To.2481V) 1 PH-14-70194 = 6.81 (6 At equivalence point, (EP), so to reach the volume of Hwroz needed - the number of mores of az S urember of acid = number of base. the EP, 1. 4.5 mmoles mane 66.568 me 0.0676 mmousine Total rolume of the reaction misture = 66.568+ 60 ml = 126.568 me. AZH ) = 4.5/126.568 (M) 20.0356 CM) [141] = CKQX CAZH+))? - (1780"X 0.0356)2 ~ 1.8x105 pH = -los, CHA) = 4.74

6 After addition of 71.01 me HNO₃ : 3 - urember of moles of Horo₂ =71.014 0.0676 mmolers con 4,8 mmoles of Hiros rrumber of tt present = (4.8-4.5) immons R = 0.3 mmoles of Hi. Total volume = (71.01 +60) ml = 131.01 ml. [Ft +) = 632.) (m) = 0.00229 pH = -log, (tt) = 2.64 .