Question

the pka of hypochlorous acid is 7.530. A 60 ml solution of .102M NaCl is titrated with .286 M Hcl. calculate the ph of the solution after

1) after the addition of 6.87ml of .286M Hcl

2)After the addition of 22.7ml of .286 M Hcl

3) at the equivalence point with .286 M Hcl

Answer 1

a) pka of HOCl = 7.53

   no of mol of NaOCl = 60*0.102 = 6.12 mmol

   no of mol of Hcl = 6.87*0.286 = 1.965 mmol

pH = pka + log(NaOCl/HOCl)

     = 7.53 + log((6.12-1.965)/1.965)

    = 7.85

  
b) pka of HOCl = 7.53

   no of mol of NaOCl = 60*0.102 = 6.12 mmol

   no of mol of Hcl = 22.7*0.286 = 6.49 mmol

NaOCl + HCl ----> HOCl + NaCl

concentration of HCl = n/v

                       = (6.49-6.12)/(60+22.7)

                       = 0.0045 M

pH = -log(H3O+)

     = -log(0.0045)

     = 2.35

     = 2.2

c) at equivalence point

     no of mol of NaOCl = 60*0.102 = 6.12 mmol

   no of mol of Hcl = 6.12 Mmol

   volume of HCl required to reach equivalence point = 6.12/0.286 = 21.4 ml

total volume = 60+21.4 = 81.4 ml

concentration of HOCl formed = 6.12/(81.4) = 0.0752 M

   pH = 1/2(pka-logC)

       = 1/2(7.53-log0.0752)

       = 4.33