Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0561 M HNO, to a 80.0 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 8.67 mL HNO,? pH= 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 102 mL HNO,? pH = What is the pH of the solution after the addition of a volume of HNO3 equal to the equivalence point? pH= What is the pH of the solution after the addition of 112 mL HNO,? pH =

Answer 1

75 (M) CHH concentration of azi ridine = 0.075 (M) pka of aziridinium is 8.04 рКА - РК - 14 th (Aziriding) so, pkb of aziridine is (11-8.04) = 5.96 pkb of aziridine is 5.96 tter addition of 0.00 me HNo means at starting sie before starting of adding NORO ANO. Aziridine is weak base. Se=cone, of a ziridine So, pot = 2 pkb - & lege t or, pot e sigb - - log (6.075) ar POH = (2,98 + 0.5624) = 3.5424 so poH - 3.5424 We know that pH + pot = 14 so pH = 14-POH - 14-3.5424 ar pH = 10.4576 So, Initial pH is 10. 4576