Question

2. Use the following data to answer the questions about the scenario below: * Ka of HNHOH' is 7.09x109 * The pH of a 0.750 M NaSCH; solution is 8.31 * The % dissociation in a 0.333 M CHN solution is 7.15x10 % A 0.50 L solution was prepared to contain 0.500 M each of NH,OH, NaSCH; and CsHșN. a. Write the hydrolysis reaction for each base in the mixture. (6 marks) b. Determine Kb for each reaction in part a. Attach each Kb value to the appropriate reaction. (3 marks) c. Determine the pH of the mixture. (3 marks)

Answer 1

a. Solution: 0.5 M of NaSCH NaSCH 0.5M → Nat + CH S 0.5M 0.5M It comes CH SH (weak acid) It hydrolyze CH,SH + HO CH,S + H2O Acts as a base 0.5 M of CHN (weak base) It hydrolyze CHEN + H2O CH,NH+ + + HOT A 0.5 M of NH,OH (weak base) It hydrolyze NH,OH+H,0 A HNH,OH*

b. Starting from: CH S + Na+ CH, SNa 0.75 0.75 CH,S +H20 CH,SH + HO 0.75-x х х Calculate the concentration of HO pH = 8.31 ---->pOH = 14-8.31 = 5.69 ----> Calculate the concentration of HO = 10 -5.69 = 2.04x10CM [CH,SH] = [HO-] = 2.04x10-6 M = x [CH, S ] = 0.75 M - 2.04x10M = 0.7499 M [CH,SH] [HO] Ko CH,S (2.04x106)2 0.7499 K = 5.56x10-12 Starting from: 0.33 M of CHN % a = 7.15x103 CHEN + H2O 0.33 -a CHÚNH". + HO a a [C,H,NH*] % a = x 100 [C5H5N] [CH NH+] = % a x [CH,N] /100 = 7.15x103 x 0.33 M/ 100 2.36x10M [CH NH] = [H0] = a---> a= 2.36x105 M [CH3NH+] = [H0] = 2.36x105 M [CH,N] = 0.33 M - 2.36x105M = 0.3299 M К. : [C,H,NH'][HQ] [C,H,N] (2.36x105)? Ko = 0.3299 K = 1.7x10° For NH ZOH: HNH,OH, K = 7.09x10° K = k/K HNH,OH + H2O NH,OH + H2O* K = 7.09x109 K = K/K K = 10-14/7.09x109 = 1.41x10

х х Starting from: NaSCH3, 0.5 M, K7 =5.56x10-12 CH,S" +H20 CH,SH + HO 0.5-x [CH,SH][H0] Kg = [CH,S] x? Ko = 0.5-X X? 0.5- X = 0.5 5.56x10-12 0.5 - X x 5.56x10-12 0.5 5.56x10-12(0.5) = x 1.7x106 = x [H0] = 1.7x106 pOH = -log 1.7x10-6 pOH = 5.77 pH = 14 - 5.77 pH = 8.23 - - [H] = 108.23 = 5.9x10CM Starting from: 0.5 M of CHEN K = K = = 1.7x109 CHEN + H20 CHÚNH+ + + HO 0.5 - X [C,H,NH*][HO] [CH.N] х х Ko 0.5 -X x2 0.5 -X = 0.5 1.7x10º = 0.5 1.7x109 (0.5) = x 2.9x105 = x -----> [H0] = 2.9x10M POH = -log 2.9x105 pOH = 4.54 ---> pH = 14 - 4.54 = 9.5 х х For NH,OH: -----> [H] = 109.5 = 3.16x10-10 0.5 M Ky = 1.41x106 NH,OH + H20 HNH,OH+ + HO 0.5 - X [HNH,OH'][HQ] Ko [NH,OH] x Ko = 0.5 -X x? 0.5 -X = 0.5 1.41x106 0.5 1.41x10° (0.5) = x 8.4x104 pOH = -log 8.4x104 pOH = 3.07 ---> pH = 14 - 3.07 = 9.93 = X ---> [HO] = 8.4x104 M ----> [H] = 10 -9.93 = 1.17x101 [H] **] total = 5.9x109M +3.16x10-10 M + 1.17x10-10 M [H]. = 5.33x10 Stotal pH = -log [total pH = -log5.33x10 pH = 8.27