Answers
IPOH = 14-PH _ =14-9.95 = 4.0H [OH 7 - 10-poH - 10-4.05 - 8.9x10-5 Ž (18 H₂1 NO H+ + OH (18 H₂, NO₃ + H₂O Initial 0.005 change -8.98 10-5 - Equilibrium 0.005 - (8.9x105) + 8.9x10-5 +8.9x10-5 8.9x10-5 8.9x10-5 (C18 H 12 NO₃ Ht] [oH] [Cis H12 NO3] (8.9x10-5) [0.oos - (8.9x10-57 1.68 10-6 → Answer = Note : 5.80 is the pkb . They have asked for just kb pkb = -log kq = -log (1.6x 10-6) = 5.8
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