Question

2. The pH of a 2.18 M solution of a weak base B is measured to be 11.74 Determine Kb of the base 1.39E-5 correct 0.6/1 (sfig err) Determine K of the weak bases's conjugate acid, HB 7.19E-10 correct 0.6/1 (sfig err) Determine [*] in a 2.00 M solution of the chloride salt, HBCI 0.0000119 M incorrect 0/1 Determine the pH of a 2.00 M solution of the chloride salt, HBCI 4.92 incorrect 0/1

Answer 1

Dar 109 [ht] | Sol:-|| PH of a Solution - LA Inleak base in a dissociate react completely in equilibrium as : solution and does not exists B + Hot BH t + OH for this Ky : reaction : [BHT] [on] [B] [B] = 2.18M [OH tB4 PH= 11.74 given : As pH = -log[htl => 11:14 = -log[Hitl => [ht] = 10116741 = 1.82 X 102 M " At 25° Kia = 1014 => Kw = [Ht] [or] => [011] - 1014 182 X 1012 => Cor = 5.494 X 10 M ER Kp = (5.494 x 103) - 2.46 5 1-384 x 100 M

Page: Date / acid tot in To for a I solution conjugate : HBT + H₂O ² B + H₂O Ka = [B] [ H3 0+1 = [B] [H3O+] [H37 Lor] S 2.18 x 182 x ola f494x103 Tka: 7:22. X1010 ML - ☺ for a Chloride salt HB ce in aqueous solution , the following equilibrium exists : HB C - Hot to Is The equilibrium sign depicts that a salt is a strong electrolyte HBT can undergo hydrolysis forming: - Hot + H₂O Ž H+ + B (H1z0*) from Ka of this reaction is 7.22 X 100 M => Ka = (H301B - 7.22 X10' THRt

Date: x-> degree of Hot + H₂O dissociation. Page: 1 È H₂Ot & B Attro At . equillb C e-ca -owme => [hot] = [B1 => Ka= [H₃O+] ² THBt] TAS HB Ce is a strong electrolyte => [HB u ] - [Hot ] = 2 M => 7:22 x1010 x 2 = [H30*12 [H20(7.) 2 x3 xiolog? => [H30*) = 3.8 xl 5 MI . pH = -log[H3 0+ = -109 [ 3.8 X 6051 het: 4.420.9963:8)