Question

i Review | Constants | Periodic Table Calculate the percent ionization of hydrazoic acid (HN3) in solutions of each of the following concentrations (Ka is given in Appendix D in the textbook). Part A 0.424 M. Express your answer using two significant figures. IVO A o 2 ? ΑΣΦ Submit Request Answer

Ka= 1.9*10^-5

Answer 1

- N3 + H₂O A) N3h + H2O 0.424 M (initial) 0.424 a 0.424 d 0.424 (1-0) (at equilibrium) Ka = 10.4248) (0.424x) = 0.42442 0.424(1-x) 1-a21 (since & is very small for weak acid) Giron: Ka = 1.9x105 1.98 105 0.424 az1.9X100% = 0.448x1014 0.424 * = (0-448x1024%2 = 0.669 x 1012 % imization = 8 x 100% = 0.669 x 10 x 100 % - 0.669 % = 0.67%

B) Similarly: Ka=0.10002 1.9x105 = 0.10002 az IAX Low = 19 x 10? 0. leo. a = (1.981004)% = 1.38*10-2 % ionization = xx 100% = 1.38x10²x 100% = 1.38% . = 1.4% c) Similarly: ka = 4.18x10222 1.98105 4.18% 10%? x² = 1.9x105 = 4.55x100 4.18x102 x = (4.55% 104)% = 2.13 X102 % ionization = x 160% = 2.13* 10 x 100% = 2.13% = 2.1 %