Question

LIUserIUDI Tulured Practice PIUDIM 17.3.3 COUNTS TOWARDS GRADE Calculate pH for a weak base/strong acid titration. Determine the pH during the titration of 33.9 mL of 0.300 M ethylamine (C2H5NH2 , Kb = 4.3x104) by 0.300 M HBr at the following points. (a) Before the addition of any HBr (b) After the addition of 14.7 mL of HBr (c) At the titration midpoint (d) At the equivalence point (e) After adding 52.5 mL of HBr

Answer 1

loro Concontactédon of CzH5NH3 = 0.300M= 0.300 mbele Volusone of C2H5 NH2 Selection = 33.9ml = 33.94 0.0339L Nennbed of moles of CH NH2= 0.300 meelLX 0.0339L =0.01017 mål PKb = -log kb = -log(4.3 x1b') = -hg 4.3 +4 bogio = -0-63+4x1=3.37 les concentration of H2X = 0.300 M = 0.300 mol/L (a) Pefore the addition of any Her Conseloe a hydrolysis racion fema CHENH2 CH - GHS NH2 + H20 Kb Oh + Cats Nth Thereal concentrateen 0.300 change en concentration - final concentration 0-300-x) x kb= OH-IT QHne) - 60)(2) - Br? [242942] O.300-) 0.300-x Since kb is very small hence 0.300-0 ~ 0.300 27 = 0.300 X Kb = 0.300 x 4.3x109 = 4:29x104 26 = 11:29x104 = 1.136x102 Since OH-] = xm thence [OH-] = 1.136X1oM. ..POH = -log[04]=-log (1-136X10=-491-136+ 20.9.10 POH= -0.05+2x/1.95 __.pH = 14.00 - POH= 14.00 -1.95 = 12.05 +2 /PH = 12.051 (b) volumene of her solution Gelded = 14.7 mnd - Home 0067 L Number of hopes of 410730-300 belL x 0.0147L=0.0044/mol Reacheon-Herce)CaN2.607)-> C2H5. NH3 BY 1.0nd of Hbr reacts with 1.0 mol of Ca HSN H₂ Prednces 1.0 most of the N Ha for then 0.00 441 mol of HPor well AMO

pract wilt 0.00441 mol of C2H5.N.Hz.will fonduce 00.044/mol ok (2Hs NH3.for... Nemibes of rules of C2H5NH2. Zemaêning = (0:0/017-0.0044)mol > 0.00576 mol From Hendess.on equalēdu pOH = PKøtleg Comingople to Where (C2H5NH2Por] Call N45] pOH = 3.37 +690.00441 mel = 3.37+loq 0.7636 = 3.37+(-0.1) 3.37 kg 10.00576 het - = 3:26 pH = 14.00 - POH = 14.00-3-26 = 10.74 PH= 10.74] © At the end poênt ... Nunukes of melas of H.Par = $x Number of miles of 52 H&NH2 =X0-01017 mol = 0.005085 mol From reaction, 0.005005 md of Hpor will reace wille 0.005001- mol of C2H5NH2 will produce 0:0050.05 mol of CzHs.Nhz Br. foros Hemelstam egyeleon Nenubes of roles of CzHsnIz remaining = (0.01017-0.00500s T = 0.0050 05 mol Foomas Henderson equation POH= 3.37 +boq (0.0050 85ml) = 3.37 tlag 1 = 3.37+0.00 = 3.37 (PH= 10.63) D) At the equêvlenee poêne Alenubes of moles of HBr = Nemikes of mueles of C2H5. NH2 =0:01017 nol 0.010.17 map of Cats NH.. Well peace wille 0:0/017 mol g HBT well Barebule 0.010.12 mol of C2H5NH2 Pol PH (0.00500shall 14.00 -3.37 - 10.63

Volemme of Her Solutéen meeded to reach the equênalence point = 0.01097 hp = 0.0339190 - 0.30 mall Tofal volume of solution = (0.0339+0:0339) 2= 0.06784 In aquelut Solenteon, CANO][zHs. NH2 Port] 1 0.01017 mol = 0.15 mol/L kar Kw - 1:00x1019 = 0.2228 x 150 Rak 4 :3x10.4. 0.0670L 20 Consédes a hydsoljšis reaction of CHS pt. C2H5NHET +420_KQ >0 H25+ Cats MH2 Thêléal Concentration 0.15 loo Change en concentration - +2 +x Final Concentrate on (0415-X) 26 ka = [H26][ CzHs Nfl2] - (062) = x2 (C2Hs NH3F] (0.15-2) (01/5-2) Since ka is lesy small hole 0.15-x 40.15 - 2 = 0:1 x ka - Ô•/ x 0.2324 x 10 = 0.0348% X! Bºo X=10.03487x17/0 =0:1867x70 = 1.067x106 SenelH20] = xm hauce (H26) = 1.867xcom PH = -log[120] = -ley 1-80 xoʻ= _leg 1-867 +6 Lig 10. PH=- 027+6X = 5.73 P A= 573