Question

Calculate the pH for each of the cases in the titration of 25.0 mL of 0.170 M pyridine, C H,N(aq) with 0.170 M HBr(aq). A. before addition of any HBr B. after addition of 12.5 mL of HBr C. after addition of 23.0 mL of HBr D. after addition of 25.0 mL of HBr E. after addition of 35.0 mL of HBr

Answer 1

Given that concentration of HBr = 0.170 M concentration of CzHEN = 0.170 M Volume of CGH N = 25.0 mL Ko of C5H-N = 1.7 x 10-9 a) Before Addition of any acid: CEHEN + H2O = CH3NH+ + OH Initial 0.170 Change -X Equillibrium 0.170 - X Ky = -=1.78x10-9 30.170- x x is small, so 0.170- x = 0.230 0170 =1.78x100 x? = 3.02 x10-10 x=1.74x10-5 :: [OH"]=1.74x10-M pOH =- log (1.74x10- )= 4.76 pH = 14 - 4.76 = 9.24

b) Addition of 12.5 mL of HBr: Number of moles of CEH N = M*V = 0.170 M* 25.0 mL = 4.25 mmol Number of moles of HBr = M*V = 0.170 M* 12.5 mL = 2.125 mmol Net moles of C5H N = 4.25 – 2.125 = 2.125 mmol This is the mid point. Ko mmol CH3 + HBr = CH NH*Br“ Initial 4. 25 0 Change -2.125 -2.125 +2.125 Equillibrium 2.125 2.125 According to Henderson's equation 2.125 pOH = pK, +log = -log(1.7x10-9)+ log2 =8.77+0 [base] 2.125 pOH = 8.77 pH = 14- pOH = 14-8.77 = 5.23

c) Addition of 23.0 mL of HBr: Number of moles of CH N = M*V = 0.170 M* 25.0 mL = 4.25 mmol Number of moles of HBr = M*V = 0.170 M* 23.0 mL = 3.91 mmol Net moles of CH N = 4.25 - 3.91 = 0.84 mmol к, mmol CHEN + HBr = CH3NH+Br Initial 4.25 Change -3.91 -3.91 +3.91 Equillibrium 0.34 3.91 According to Henderson's equation 3.91 pOH = pK, +loga ?= -log(1.7x10-9]+ log -2 =8.77 +1.060698 [base] pOH = 9.83 pH =14 - pOH =14–9.83 = 4.17 0.34

d) Addition of 25.0 mL of HBr: Number of moles of CSH N = M*V = 0.170 M* 25.0 mL = 4.250 mmol Number of moles of HBr = M*V = 0.170 M* 25.0 mL = 4.250 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 CEHEN is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate acid. moles [C,HNH "Br"]=> 4.25 mmol "Total volume (25+25) mL -=0.085M K 1x10-14 K, of [C,H, NH K 1710-9 = 5.88x10-6 Dissociation of CGH NH + is given by mmol CH NH* C6H5NH + Initial 0.085 Change -X Equillibrium 0.085 - x

Ka = -= 5.88x10-6 0.085-3 x is small, so 0.085 - x = 0.085 = 0.035 = 5.88x10-4 = x² = 5.00x10-7 >x=0.0007071 ::[H]=0.0007071M pH =- log(0.0007071) = 3.15 e) Addition of 35.0 mL of HBr: Number of moles of CEH N = M*V = 0.170 M* 25.0 mL = 4.250 mmol Number of moles of HBr = M*V = 0.170 M* 35.0 mL = 5.95 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 HBr = 5.95 – 4.250 = 1.7 mmol Total volume of the solution = 25 + 35 = 60 mL Conc. of HBr = [H] = moles/ volume = 1.7 mmol / 60 mL = 0.02833 M pH =- log (H+) = -log[0.02833] = 1.55