Question

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.160 M pyridine, CsHsN(aq) with 0.160 M HBr(aq): Number (a) before addition of any HBr Number (b) after addition of 12.5 mL of HBr Number (c) after addition of 24.0 mL of HBr Number (d) after addition of 25.0 mL of HBr Number (e) after addition of 29.0 mL of HBrIL

Answer 1

Given that concentration of HBr= 0.160 M concentration of C5H5N=0.160 M Volume of C5H5N=25.0 mL Kb of C5H5N = 1.7 x 10-9 a) Before Addition of any acid: OH Initial Change Equillibrium 0.160 0 0 + X + X 0.160 - X x-x =1.7×10 ,-0.160-х x is small, so 0.160-x-0.160 x-x =1.7x10 ー9 0.160 → x-2.72x10-10 x-1.65 x 10-5 :. [OH 1 1.65x10 M p0H--log ( 1 .65 x 10-5 )-4-78 pH = 14-478-922

b) Addition of 12.5 mL of HBr: Number of moles of C5H5N-M*V = 0.160 M* 25.0 mL = 4.00 mmol Number of moles of HBr = M*V = 0.160 M* 12.5 mL = 2.00 mmol Net moles of CH5N 4.00 2.00 2.00 mmol his is the mid point mmol CSHsN Initial Change Equillibrium 4.00 -2.00 2.00 0 -2.00 0 0 + 2.00 2.00 net moles 2.00 ol 2.00 mmol 0.053333M Total volume25+12.5 net moles 37.5 mL 2.00 mmol 2.00 mmol 0.053333M Tota l volume 25+12.5 37.5 mL According to Henderson's equation [acid] - 8.77+0 [base] basel og(.7x10)+ log 0.053333 pOH =8.77 pH 14 pOH -14-8.77 - 5.23

c) Addition of 24.0 mL of HBr Number of moles of C5H5N = M*V = 0.160 M* 25.0 mL = 4.00 mmol Number of moles of HBr M*V 0,160 M 24.0 mL 3,84 mmol Net moles of CH5N 4.00 3.84 0.16 mmol mmol C5H5N Initial Change Equillibrium 4.00 -3.84 0.16 0 -3.84 0 0 +3.84 3.84 net moles 0.16 mmol 0.16 mmol 0.003265 M 49 mL Total volume 25+24 net moles Total volume 3.84 mmol 25124 3.84 mmol 49 mL -0078367M [C,H,NH'Br According to Henderson's equation pOH = pK, + log pOH 10.15 pH 14-pOH-14-10.15-3.85 [acid] Ae] 0.078367 log(1.7x10)+log 8.77 + 1.380211 0.00326.5

d) Addition of 25.0 mL of HBr: Number of moles of C,H,N = M*V = 0.160 M* 25.0 mL = 4.00 mmol Number of moles of HBr= M*V = 0.160 M* 25.0 mL = 4.00 mmol This is the equivalence point, the number of moles of HBr is equal to the number of moles of base. At this point, there is no C5H5N is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate acid. moles 4.00 mmol 4.00 mmol 50.0 mL [C511,NII'Br] 0.08M Total vol ume 25+25 K 1x10 K 1.7x109 14 5.88x10-6 Dissociation of C5H5NH+ is given by mmol C5H5NH C5H5NH + 0.08 Initial Change Equillibrium 0 0 + X + X 0.08 - X

x5.88x10 -6 a 0.08- x is small, so 0.08-x= 0.08 x-= 5.88×10 0.08 x=6.86 x 104M H1-6.86x10 M pH=-log(6.86x10*)= 3.1636795-3161 e) Addition of 29.0 mL of HBr: Number of moles of C5H5N = M*V = 0.160 M* 25.0 mL = 4.00 mmol Number of moles of HBr-M"V-0.160 M* 29.0 mL-4.64 mmol This is after the equivalence point. At this point, the pH of the solution is determined by the amount of unreacted HBr in the solution. Net moles of H Br = 4.64-4.00 = 0.64 mmol Total volume of the solution = 25 + 29 = 54 mL Conc. of H Br = [H+] = moles/ volume = 0.64 mmol / 54 mL = 0.011852 M pHlog [H+log[O.011852] 1.926214 1.93