Question

Calculate the pH of the solution after the addition of the following amounts of 0.0525 M HNO_3 to a 80.0 mL solution of 0.0750 M aziridine. The pK_a of aziridinium is 8.04. a) 0.00 mL of HNO_3 pH = 12.9 b) 6.44 mL of HNO_3 pH = 9.26 c) Volume of HNO_3 equal to half the equivalence point volume pH = 8.04 d) 111 mL of NHO_3 pH = 6.57 e) Volume of HNO_3 equal to the equivalence point pH = 4.77 f) 118 mL of HNO_3 pH = 3.0

Answer 1

ta 8.okq .04 Ha A) 0.00mL HND3 0n dlintm) po 3.SL

-0434 Fina 6"419 mON .661 76. 냐u 0.066 Molu

03 m mole 0.34 T6..44 = 3.933 X10+3 ー le b pb 5.96

03 O-O 66 u me HNg 여 hf HN0.0S2SXS7.142 milli 4 21 Fl 2,4 9.21 990.9 9.11 o.aa 0.029 M 137.14 37.142 0,022 Baut (0,4リ

173 1a1l ial Co.03053 0 Voluome iレ 1 Cy 0.0525 ем ‘ 9 428

, 030 0.030 DL 0.030-ㅈ 4NO2 H시 0-145 mmole !ท 1 Z 0:19. 9.rs χΙδ 4 198