Question

Calculate the pH after the addition of each of the following volumes when 20.00 mL of 0.125 M NaOCl is titrated with 0.250M HCl. (Ka = 3.0 * 10^-8 for HOCl) (Hint: This is a weak base-strong acid titration)

A) 5.0 mL of HCl added

B)10.0 mL of HCl added

C) 17.0 mL of HCl added

Answer 1

Given that concentration of HCl = 0.250 M concentration of NaClO = 0.125 M Volume of NaClo= 20.00 mL Ka for HClO = 3.0 × 10-8 Ko of clo-=Kw/Ka = 1 10-14/(3.5 x 10-8 ) = 3.33 x 10-7.

a) Addition of 5.00 mL of HCI. Number of moles of NaClO = M*V = 0.125 M* 20.00 mL = 2.5 mmol Number of moles of HCl = M*V = 0.250 M* 5.00 mL = 1.25 mmol Net moles of Nacio = 2.5 – 1.25 = 1.25 mmol Kb mmolclo + HC HCIO + c Initial 2.5 Change -1.25 -1.25 +1.25 Equillibrium 1.25 0 1.25 According to Henderson's equation 1.25 pOH = PK, +log laha! =-lo log (3. acid] - [base] + = 6.48 +0 1.25 pOH = 6.48 pH = 14- pOH = 14 - 6.48 = 7.52

(b) Addition of 10.00 mL of HCI Number of moles of NaClO = M*V = 0.125 M* 20.00 mL = 2.5 mmol Number of moles of HCI = M*V = 0.250 M* 10.00 mL = 2.5 mmol This is equivalence point. At this point, there is no CIO is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate acid. moles 2.5 mmol [ HTOT Total volume (20.00+10.00) mL -=0.083333M K, of HC10=3.0x10-8 Dissociation of Hclo is given by mmol mmol HCIO HCIO 2 clo CIO + H* Initial 0.083333 Change -X Equillibrium 0.08333 - x

x.x Kar -= 3.00 x10-8 0.083333-X x is small, so 0.083333-X=0.083333 X.X 0.083333 =3.00x10-8 3x2 = 2.50 x10-9 x=5.00x10--M ::[H+]=5.00x10-5 pH = -log(5.00x10-) = 4.30 c) Addition of 17.0 mL of HCl : Number of moles of NaClO = M*V = 0.125 M* 20.00 mL = 2.5 mmol Number of moles of HCl = M*V = 0.250 M* 17.0 mL = 4.25 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 = 4.25 -2.5 = 1.75 mmol Total volume of the solution = 20.00 + 17.00 = 37.00 mL Conc. of HCl = [H+] = moles/ volume = 1.75 mmol / 37 mL = 0.047297 M pH =- log [H+] = -log[0.047297 ] = 1.33