Question

The pK, of hypochlorous acid is 7.530. A 55.0 mL solution of 0.116 M sodium hypochlorite (NaOCI) is titrated with 0.252 M HCI. Calculate the pH of the solution after the addition of 9.37 mL of 0.252 M HCL. pH = Calculate the pH of the solution after the addition of 26.6 ml of 0.252 M HCI. pH = Calculate the pH of the solution at the equivalence point with 0.262 M HCI. pH =

Answer 1

Given that concentration of HCl = 0.252 M concentration of NaCl = 0.116 M Volume of Naoci = 55.0 mL pka for HClO = 7.530 Ka for HClO = 10-7.530 = 2.95 x 10-8 pk of clo-=14 – 7.530 = 6.470

a) Addition of 9.37 mL of HCI Number of moles of NaClO = M*V = 0.116 M* 55.0 mL = 6.38 mmol Number of moles of HCI = M*V = 0.252 M* 9.37 mL = 2.36124 mmol Net moles of NaClO = 6.38 - 2.36124 = 4.01876 mmol Kb mmol clo + HC HCIO + c Initial 6. 380 Change - 2.36124 - 2.36124 +2.36124 Equillibrium 4.01876 2.36124 According to Henderson's equation 2.36124 pOH = pK, +log1 =6.470+log 470+log --30127=6.470 -0.23095 [base] 4.01876 pOH = 6.239 pH = 14 - POH =14-6.239 = 7.761

c) Addition of 26.6 mL of HCI Number of moles of NaClO = M*V = 0.116 M* 55.0 mL = 6.38 mmol Number of moles of HCI = M*V = 0.252 M* 26.6 mL = 6.7032 mmol This is after the equivalence point. At this point, the pH of the solution is determined by the amount of unreacted HCl in the solution. Net moles of HCl = 6.7032-6.38 = 0.3232 mmol Total volume of the solution = 55.0 + 26.6 = 81.6 ml Conc. of HCl = [H+] = moles/ volume = 0.3232 mmol / 81.6 mL = 0.003961 M pH =- log [H+] = -log[0.003961 ] = 2.40

At equivalence point Number of moles of NaClO = M*V = 0.116 M* 55.0 mL = 6.38 mmol At the equivalence point, the number of moles of HCl is equal to the number of moles of base. So Number of moles of HCl = 6.38 mmol Volume of HCl needed to reach equivalence point = moles/molarity = 6.38 mmol/0.252 M = 25.32 mL At this point, there is no clo-is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate acid. moles 6.38 mmol [HCIO]= -=0.07943M " Total volume (55.0+25.32) mL K, of HCIO=2.95x10-8 Dissociation of HClO is given by mmol HCIO SCO- 0.07943 Initial Change -X +x Equillibrium 0.0794 - X

K =—x.x = 2.95 x10-8 " 0.07943-X x is small, so 0.07943 – x=0.07943 = 0.07943 = 2.95 x10 3x2 = 2.34x10- 3x = 4.84x10-5 :: [H*] =4.84x10-M pH =-log(4.84*10-) = 4.31